ch 07 - 09 (1)

7/17/2019 ch 07 - 09 (1)

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Solutions Chapter 7

In the problems in Chapter 6 it says to assume pipes are Sch. 40. In most problems inthis chapter pipe inside dimensions are given, normally in integral inches. The solutionshere are based on those integral inch diameters, rather than Sch. 40 diameters. I hope

this causes no confusion.In all solutions, except as noted, it is assumed that the flow is in the positive x-direction,and the direction subscripts are dropped unless it would cause confusion. Where a sketchis made, the positive x-direction is from left to right. ______________________________________________________________________

7.1* (a) System: earth, open system d mV ( ) = −V outdmout , ΔV = −V out

dmout

mearth

ΔV = −20ft

s⋅

1 lbm

1025 lbm= −2 ⋅10−24 ft

s

(b) System: earth-ball, closed system

d mV ( ) = 0; V f = V i

It is instructive to plot the earth's velocity and trajectory (perpendicular to the maindirection in its orbit), as shown below

time s t a n c e p e r p

e n

c u a r

t o o r b i t a l p a t h

ball throwntop of flight ball hits earth

time

p e r p e n d

i c u l a r

d i s p l a c e m e n t f r o m

o r b i t a l p a t h

______________________________________________________________________

7.2 (a) d m ;mV V ( )sys = −V outdmout f − mV i( )= −V outdmout

V f = −V out

dmout

mgun

= − 1500ft

s

⎝

⎜

⎠

⎟0.05 lbm

5 lbm

⎝

⎜⎜ ⎠

⎟⎟ = −15ft

s

= −4.57m

s

(b) ;Δ mV ( )= 0 = mV ( )gun + mV ( ) bullet V gun = −V bullet

m bullet

mgun

⎝⎜⎜

⎠⎟⎟

which is the same as in part (a). ______________________________________________________________________

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 1

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7.3 System: duck-bullet, closed system Δ Σm u +V 2

2

⎝⎜⎜

⎠⎟⎟

⎣⎢⎢ ⎦

⎥⎥

= 0; Δmu = −ΔmV 2

2

mV 2

2

∑⎛

⎝

⎜⎜ ⎞

⎠

⎟⎟i

=3 lbm ⋅ 15

ft

s

⎝⎜

⎠⎟

2

2

+0.05 lbm 1000

ft

s

⎝⎜

⎠⎟

2

2

= 337.5 + 25000 = 25337.5ft2

s

2

mV 2

2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

f

=3.05lbm 1.6

ft

s

⎝⎜

⎠⎟

2

2= 3.90

ft2

s2

99,985% of the original KE is converted to internal energy. If we had chosen the initialmasses and velocities so that the final velocity of the duck-bullet system were zero, thenall of the initial KE would have been converted to internal energy. ______________________________________________________________________

7.4 If the bullet stays in the shootee, then the momentum transfer to the shootee is thesame as that to the shooter. We all know that guns "kick", meaning that they transfer asmuch momentum to the shooter as they do to the bullet. For artillery pieces there arelarge and complicated shock absorbers to accept this force without either damaging thegun or its mount, or moving it from its location, thus requiring re-sighting.

The following eyewitness account describes the physical response of Mata Hari, whomthe French executed as a spy, (which she denied), to having 11 rifle bullets hit her chestsimultaneously, .(Rifle bullets almost certainly pass through a human body, and exit withmuch of their momentum; so the total momentum transferred to her body was certainlyless than the total momentum in the entering bullets.)

".. Mata Hari fell. She did not die as actors and moving-picture stars would have us

believe that people die when they are shot. She did not throw up her hands nor did she plunge straight forward or straight back.

Instead she seemed to collapse. Slowly, inertly, she settled to her knees, her headup always, and without the slightest change of expression on her face. For a fraction of asecond it seemed she tottered there, on her knees, gazing directly as those who had takenher life. Then she fell backward, bending at the waist, with her legs doubled up beneathher. ...." Wales, Henry G. International News Service, 19 October, 1917.

Movies and TV are a poor place to learn one's physics. ______________________________________________________________________

7.5* System: as sketched at the right, open

0 = m V in − 0 − F wall : F wall = −F fluid = − m V in

F fluid = + Ým V in = 50kg

s⋅ 80

m

s⋅ N s2

kg m= 4000 N = 899.2 lbf


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Ask the students how one would do this by Bernoulli's equation? Answer; determine theforce by an integral of PdA, determine P by Bernoulli's equation. That can be done bythe methods shown in Chap. 10. It is much more difficult than the method shown here.______________________________________________________________________

7.6 F fluid = + Ým V in = 200 kgs ⋅ 400 ms ⋅ N s

2

kg m = 80 kN = 1.8⋅104 lbf

______________________________________________________________________

7.7 (a) By material balance, m3 = m1 − m2 Substituting this in Eq. 7.BK and

simplifying, we findm2

Ým1

=1 + cosθ

2

(b) By momentum balance in the r direction − F wall, r direction = m1V r , 1 − m2V r , 2 − m3V r , 3

The two rightmost velocities are zero, and V r , 1 = −V 1 sinθ , minus because it is flow in the

minus r direction, so that F wall, r direction = V 1 sinθ

One may make plausibility checks by setting θ = 0 and 90°, finding that the answers arewhat we expect.______________________________________________________________________

7.8 We make an x-directed momentum balance for steady flow

0 = Ým V in − V out( )+ F = ρ in AV in V in −V out( )+ A Pin − Pout( ); V out = V in ρ in

ρ out

⎝⎜⎜

⎠⎟⎟

Pin − Pout = ρ inV in2 ρ in

ρ out

−1

⎝⎜⎜

⎠⎟⎟ ;

ρ in

ρ out

≅3660 o R

528o R = 6.93

Pin

− Pout

= 0.075⋅28

29

lbm

ft3

⎛

⎝⎜

⎠⎟ 2

ft

s

⎛

⎝⎜

⎠⎟

2

6.93 −1( ) ⋅lbf s2

32.2 lbm ft⋅

ft2

144in2

= 3.7 ⋅10−4 psi = 2.56 Pa = 0.019 torr = 0.01 in H2O Contrary to our intuition, in an unconfined steady-state flame the pressure decreases asthe gas flows through the flame. The magnitude of this pressure change is quite small, but it plays a role in shaping the flame.______________________________________________________________________

7.9 We make a z-directed momentum balance for a steady flow system consisting of theelevator and jet inside turning vane − F elevator acting on jet = min (V in − V out )

By BE, V in = V out and V in = V 0 − 2gz so that −F elevator acting on jet = Ýmin ⋅2 V 02 − 2gz

For ground level, z = 0, we have−F elevator acting on jet = 500

lbm

s⋅ 2 200

ft

s

⎛⎝⎜

⎞ ⎠⎟

2

⋅lbf s2

32.2 lbm ft= 6,211 lbf = 27.6 kN

For zero weight of the elevator and its cargo, we have V 02 − 2gz = 0


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z =V 0

2

g=

200ft

s

⎝⎜

⎠⎟

2

2 ⋅32.2ft

s2

= 621 ft = 189 m

The load-height curve is a parabola,

621 ft

6211 lbf

______________________________________________________________________

7.10* See sketch at the right.We make a y-directed steady flowmomentum balance. The motion ofthe boat is neglected (or we takethe coordinate axes based on the boat, and consider the velocity anddirection stated to be those basedon that frame of reference.)

From BE, assuming zero friction,

we see that the wind's absolutevelocity does not change, while itscomponents do.

Wind in at 45°as seen from boat

Wind out at 0°as seen from boat

Boat moves in y direction

Sail turnswind

-F y = Ým V in −V out ( ) y

= Ým V 0 sin45 − −V 0( )( )

= 200kg

s⋅10 m

s−0.707 − −1( )( )⋅

N s2

kg m= +586 N F y

= −586 N = 131.7 lbf

−F x = Ým −V 0 cos45 − 0( )= 200kg

s⋅7 m

s−0.7071( ) = −1414N

F x = +1414 N = 318 lbf


Here we see that the forces we calculate are those exerted by the boat on the wind. The boat moves the wind in the negative y and positive x directions. The wind moves the boat in the positive y and negative x directions. The y component of the force exerted by

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the wind moves the boat along its axis. The x component of the force exerted by thewind is resisted by the hull and keel of the boat, which keep the boat from being moved(significantly) in the minus x direction. ______________________________________________________________________

7.11*(a) V 2 =2(−ΔP)

ρ ⋅(1 − A22 / A1

2 )=

2(40 lbf in2 )

62.3lbm

ft3⋅(1 − 0.252 )

⋅32.2 lbmft

lbf s2 ⋅ft2

144 in2 = 79.7ft

s

V 1 = V 2 A2

A1

= 79.7ft

s

3 in2

12 in2 = 19.9ft

s

Ým = ρ V 1 A1 = 62.3lbm

ft3 19.9ft

s⋅12 ft

144= 103.4

lbm

s

(b) −F support = Ýmin V in − V out( )+ P1 A− P2 A2

= 103.4lbm

s

⋅ 19.9 − 79.9( ) ft

s

⋅lbf s2

32.2 lbm ft

+ 40lbf

in2 ⋅12 in2 − 0

lbf

in2 ⋅3 in2

= −192 + 480 − 0 = 288 lbf A pressure force of 480 lbf is exerted on the fluid and nozzle. Of this 192 lbf is used toaccelerate the fluid. The remaining 288 are resisted by the bolts.______________________________________________________________________

7.12 (a) By straightforward application of the methods in Ch. 6, R = 8.1·105, e/D =0.0017, f = 0.0059, - dP/dx = 18.1 psi/ft. For 5 ft of pipe -ΔP = 90.6 lbf. (b) -F bolts = +F onflange = Ým V in − V out( )+ Pflange A

For this constant-diameter pipe there is no change in velocity from in to out, so the first

term on the right is zero, and F flange = 90.6

lbf

in2 ⋅

π

4 1.049 in( )

2

= 78.3 lbf = 348 N (b) The force is transmitted as a shear force on the walls of the pipe

P ⋅π

4 D

2 = τ ⋅π DL

______________________________________________________________________

7.13 This is a simple remake of that example with different numbers. As in thatexample the two components of the support force have the same magnitude, and point inthe negative x and y directions. Then, from Appendix A.2, A = 0.05130 ft2 and

Ým = 62.3lbm

ft3 ⋅ 8ft

s⋅ 0.05130 ft2 = 25.57

lbm

s= 11.6

kg

s

By x directed, steady-state momentum balance0 = Ým(V in − V out ) x + F x∑ = Ým(V in − V out ) x + P1 A1 + F x , support

−F x, support = m(V in − V out ) x + P1 A

= 25.57lbm

s⋅(8 − 0)

ft

s⋅

lbf s2

32.2 lbm ft+ 30

lbf

in2 ⋅ 0.05130 ft2 ⋅144 in2

ft2

= 6.35 + 221.62 = 227.8 lbf = 1.014 MN


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We see, as in the example that the pressure term is much larger than the change-of-momentum term. Most of the examples I know of industrial accidents due to pipeinternal failure have been due to pressures greater than the pipe could withstand ratherthan to the change-of-momentum term, e.g. Flixboro. ______________________________________________________________________

7.14 By z directed, steady-state momentum balance

−F z, support = m(V in − V out) z + P1 A1 + P2 A2 ; V =600

gal

minπ

4ft4

⎛⎝⎜

⎞ ⎠⎟

2 = 27.2ft

s

Ým =600gal

min⋅

ft3

7.48 gal⋅62.3 lbm

ft3 = 83.3lbm

s

F support = 83.3lbm

s⋅ 27.2 − −27.2( )( )ft

s⋅

lbf s2

32.2 lbm ft⋅

min

60 s+ π

4

ft

4

⎝⎜

⎠⎟

2

5 + 3( ) lbf

in2 ⋅144in2

ft2

= 140.9+ 56.5 = 197.4 lbf = 878N Students have a hard time with the idea that both pressure forces act in the minus z

direction. If you look ahead to Prob. 7.17, you can show an example where the pressureforces act in opposite directions. Make clear to them that the pressure forces act inward,wherever the pressure on the system boundary is not atmospheric. ______________________________________________________________________

7.15* This is practically the same problem as Prob. 7.14

-F supp = 50ft

s

⎛⎝⎜

⎠⎟π

4

ft

4

⎛⎝⎜

⎠⎟

2

62.3lbm

ft

3 50 − −50( )( )ft

s

⋅lbf s2

32.2 lbm ft+π

4ft4

⎛⎝⎜

⎞ ⎠⎟

2

30 + 20( ) lbf in2

⋅144in2

ft2

-F supp = 475 + 353 = 828 lbf = 3.68 kN

The support force acts in the minus x direction

______________________________________________________________________

7.16 This is practically the same as the preceding two problems.−F x, support = m(V in − V out ) x + P1 A1 + P2 A2 ; (V in − V out) x = 2V in, x

m = ρ AV in, x ; −V in, x2 =F

x, support

+ P1

A1

+ P2

A2

2 ρ A

−V in, x2 =

−45 +18 + 20( ) lbf

2 ⋅62.3lbmft3

⋅ft2

144

⋅32.2 lbmft

lbf s2 = − 16.1ft

s

⎝⎜

⎠⎟

2

;

V in, x =16.1ft

s= 4.92

m

s


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People have made all kinds of bizarre flow meters, but I have never heard of this one. Itwould have the drawback that its answer depends on the algebraic sum of the readings ofthree different instruments, and might depend on the small differences of large numbers,leading to doubtful accuracy. The advantage of this fanciful meter is that there is nodisturbance of the flow, and little chance of leakage. The commercially available meter

which has that set of advantages is the coriolis meter, in which the fluid to be testedflows through one or more coils of pipe, and the coriolis force on the pipe is measuredwith strain gauges. These are not cheap, but they have the advantages mentioned above. ______________________________________________________________________

7.17*−F support = Ýmin V in − V out( )+ P1 A − P2 A2

A1 = π

4⋅

ft

3

⎛⎝⎜

⎞ ⎠⎟

2

= 0.0873ft2; A2 = π

4⋅

ft

4

⎛⎝⎜

⎞ ⎠⎟

2

= 0.04909ft2

V 1 =230

gal

min⋅

ft3

7.48gal⋅min

60 s0.0873ft 2 = 5.87

ft

s; V 2 = 5.87

ft

s⋅

0.0873

0.04909=10.44

ft

s

Ým = 62.3 lbmft3 ⋅ 5.87 f t

s⋅0.0873 ft2 = 31.9 lbm

s

−F support = 31.9lbm

s5.87 −10.44( ) ft

s⋅

lbf s2

32.2lbmft+

+ 10lbf

in2 ⋅ (144 ⋅0.0873 in2 ) − 50lbf

in2 ⋅(144 ⋅ 0.04909 in2 )

= −4.5 +125.7 = 353.4 = −232.2 lbf = -1.04 kN

(b) The fore must be directed along the x axis, in the direction from the inlet to the outletof the pipes.

______________________________________________________________________

7.18 −F = Ým V in − V out( ) = 200kg

s0 − 4000

m

s

⎝⎜

⎠⎟ ⋅

Ns

kg m= −800 kN

This is the force which the rocket exerts on the fluid. The fluid exerts an equal andopposite force on the rocket, which is the thrust of the rocket. ______________________________________________________________________

7.19* I sp =F

Ým=

325 MN

1000kg

s

= 3.25kN

k g / s= 331.4

lbf

lbm / s

In Ex. 7.9 we can compute the I sp by a simple conversion of units from the exhaustvelocity, because the exhaust pressure is atmospheric. Here we cannot do that, becausethe exhaust pressure is not atmospheric. So we must compute it from the thrust and thefuel flow rate. ______________________________________________________________________


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7.20 −F = − ÝmV out = − ρ AV out2 ; By BE

V out =2 5

lbf

in2

⎛⎝⎜

⎠⎟

62.3lbm

ft3

⋅32.2 lbm ft

lbf s2 ⋅144 in2

ft2 + 2 32.2ft

s2

⎛⎝⎜

⎞ ⎠⎟ 8 ft( ) = 35.5

ft

s=10.8

m

s

F = 62.3lbm

ft3

⎝⎜

⎠⎟ 1 in2( ) 35.5

ft

s

⎝⎜

⎠⎟

2

⋅lbf s2

32.2 lbm ft⋅

ft2

144 in2 =16.9 lbf = 75.3 N

This is the force which the rocket exerts on the fluid. The fluid exerts an equal andopposite force on the rocket, which is the thrust of the rocket. This wouldn't be a veryinteresting rocket. However toy rockets of the same design, much smaller, with fittingsfor using a bicycle pump to get a much higher air pressure have been sold and playedwith. ______________________________________________________________________

7.21 We choose as our system the balloon, initially at rest (so that we need not considerair resistance) and make an x directed momentum balance, dropping the x subscriptsd (mV )

dt = Ým V ( )in − ÝmV ( )out + F ∑

There are no forces, nor flow in. We expand the differential on the left and write

mdV

dt + V

dm

dt = − ÝmV ( )out ;

dV

dt =

− Ýmout V out − V ( )msystem, instantaneous

This is correct for any rocket if summation F = 0. Here at the start, V = 0 so thatdV

dt =

− Ýmout V out( )msystem, instantaneous

=− ρ AV out

2

m

From BE we know that V = 2ΔP ρ which we substitute to find dV

dt = −2 AΔPm . The

mass of the balloon is 0.0075 lbm, and that of the air inside it 0.0049 lbm, so the totalmass is 0.0124 lbm. Thus

dV

dt =

−2 ⋅π

4⋅

0.5 ft

12

⎝⎜

⎠⎟

2

⋅1lbf

in2

0.0124 lbm⋅144 in2

ft2 ⋅32.2 lbm ft

lbfs2 = −1019ft

s2 = −311m

s2

This seems like a large number, but the observation is that such a balloon acceleratesquite rapidly at first. Air resistance soon becomes significant. The minus signs indicateacceleration in the minus x direction.______________________________________________________________________

7.22*(a) We choose as our system the cylinder, initially at rest (so that we need notconsider air resistance) and make an x directed momentum balance, dropping the x subscripts

−F = Ým 0 − V out( )− Pexit Aexit = − ρ AV 2( )

exit− Pexit Aexit


= −7 lbm

ft3 ⋅1in2 ⋅ 975ft

s

⎝⎜

⎠⎟

2

⋅lbf s2

32.2 lbm ft⋅

ft2

144in2 − 1060 −14.7( ) lbf

in2 ⋅1 in2

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−F = −1435−1045 = −2480 lbf = -11.0 kN

(b) It is very worthwhile. Lives have been lost by failing to do it.(c) The collars or caps have the same purpose as chaining cylinders to the wall; theyminimize (but do not entirely prevent) accidents in which the cylinder valve is broken

off, with disastrous consequences.______________________________________________________________________

7.23 The error is the statement that the pressure everywhere inside the container exceptfor the exit is constant at 2000 psi. In the nozzle region, as the velocity is increasing the pressure must be falling (see BE) so that there is a region near the outlet with pressuresignificantly less than 2000 psi.

In principle, one could solve the problem this way, calculating the x component of the pressure force around the whole interior of the container, taking the decreasing pressurein the area of fluid acceleration into account. The result would have to be the same as inProb. 7.22. But that would be difficult; (a problem for graduate students?). ______________________________________________________________________

7.24 (a) Yes. (b) No. If it were perfectly straight the force would be directly down itsaxis, and it would not move. However this is the "pushing a rope" problem. In principleone should be able to push a perfectly straight rope, but in practice the smallest deviationfrom straightness in the rope leads to buckling instability, and it is very hard to push arope.(c) The maximum force would occur if the hose were bent completely back on itself, asin Fig. 7.32 In that case we could calculate the force as

−F = Ýmin V in − V out( )= Ýmin 2V in = 2 ρ AV in2

−F = 2 62.3 lbmft3

⎛⎝⎜ ⎞

⎠⎟π

4⎛⎝⎜ ⎞

⎠⎟ ⋅

3

412

ft

⎝

⎜⎜⎜⎜

⎠

⎟⎟⎟⎟

2

10 fts

⎛⎝⎜ ⎞

⎠⎟

2

⋅ lbf s2

32.2 lbm ft= 1.18 lbf = 5.28 N

This seems like a small force, but it is acting on a small mass (the nozzle and end of thehose). Most of us have observed that the resulting motion can be substantial. ______________________________________________________________________

7.25 This is solved as a rocket. By BE the exit velocity is

V = 2 −ΔP

ρ

− gΔ z⎛

⎝

⎜⎜ ⎞

⎠

⎟⎟ = 21000

lbf

in2

50lbm

ft3

⎝

⎜⎜

⎜⎜ ⎠

⎟⎟

⎟⎟

⋅144 in2

ft

2 ⋅32.2 lbm ft

lbf s

2 + 2 ⋅ 20 ft ⋅32ft

s

2

= 432ft

s= 132

m

s

and

−F = Ým V in − V out( )= ρ AV out2 = 50

lbm

ft3

⎝⎜

⎠⎟π

4⋅9ft 2

⎝⎜

⎠⎟ 432

ft

s

⎝⎜

⎠⎟

2 lbf s2

32.2 lbm ft


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= 2.05 ⋅10 6 lbf = 9.1 MN The rupture of a main cooling line is the "maximum credible accident" for typicalAmerican PWR nuclear reactors. The force generated this way is one of the problems to be dealt with in designing to contain that kind of accident.

______________________________________________________________________

7.26* (a) We choose as our system the whole rocket, initially at rest (so that we need notconsider air resistance) and make a z directed momentum balance, dropping the z

subscripts−F = Ým −V out( )+ (Pout − Patm ) Aout

= 300lbm

s

⎝⎜

⎠⎟ ⋅ 4600

ft

s

⎝⎜

⎠⎟ ⋅

lbf s2

32.2 lbm ft+ 40 −14.7( ) lbf

in2 ⋅ 1.5⋅144 in2( )

= 42,857 + 5464( ) lbf = 48321 lbf = 215 kN (b) We choose as our system the nozzle, initially at rest (so that we need not considerair resistance) and make an z directed momentum balance, dropping the z subscripts

−F = Ým −V out( )+ (Pout − Patm ) Aout

−F = 300lbm

s⋅ 300 − 4600( ) ft

s⋅

lbf s2

32.2 lbm ft

+ 300 −14.7( ) lbf

in2 5 ⋅144( )in2 − 40 −14.7( ) lbf

in2 1.5 ⋅144( )in2

= -4.006⋅10 4 + 2.054 ⋅10 5 - 5464 = 1.653⋅105 lbf = 736 kN

This is the force tending to tear the nozzle off the rocket. The pressure term is muchlarger than the momentum term.

______________________________________________________________________

7.27 We write BE from the place where V ≈ 0 (relative to the engine) and P = Patm tothe engine inlet

Pinlet − Patm = Pinlet, gage = − ρ V 2

2

= 0.075lbm

ft3 ⋅500

ft

s

⎝⎜

⎠⎟

2

2⋅

lbf s

32.2 lbm ft⋅

ft2

144 in2 = −2.03lbf

in2

Taking the new system compared to that in Ex. 7.10 we see that two new terms appear,

andPinlet, gagedAwhole inlet nozzle

∫ Ým V ( )inlet The first is negative, the second positive. Their

algebraic sum is zero. I put this in because when I asked the Pratt and Whitney engineerwho got me the copy of Fig. 7.15 what the inlet and outlet velocities were, he respondedwith the values in Ex. 7.10. After some discussion it became clear that the inlet velocitywas based on Pinlet, gage = 0.

______________________________________________________________________


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7.28 For the stated assumptions, the only term in the momentum balance which increasesis m V ( )out which increases to 1.02% of its initial value. The calculated thrust is proportional to this value (for the system and assumptions in Ex. 7.10) so the thrust becomes 1.02% of that in Ex. 7.10. This is normally ignored in simple textbook

examples, but not in real-life jet -engine calculations. ______________________________________________________________________

7.29 Instead of having P2 acting in the downstream direction over area A1a we wouldhave the pressure increase steadily in the downstream direction, so there would be moreforce acting in the downstream direction, hence a higher outlet pressure, and a lower pressure loss.If we plotted P vs distance, for the sudden expansion we would have no change in pressure from upstream to down (check that by applying BE from 1 to 2), while for thegradually tapering nozzle we would have a gradual transition from the lower pressureupstream to the higher one downstream.

______________________________________________________________________

7.30 It is. For example if we take D1 / D2 = 0.5 then from Eq. 7.28

K = 1 − D1

2

D22

⎣⎢⎢ ⎦

⎥⎥

2

= 1 − 0.52[ ]2= 0.5625

As best I can read Fig. 6.16 K ≈ 0.56. The check is comparably good for other values. ______________________________________________________________________

7.31 The spreadsheet is shown below. It shows both the solution to this problem and tothe next. The first column lists the variables. Comparing it to Tab. 7.2 we see that itincludes several additional rows, not included there, but useful in computing the average

molecular weight of the mixture.. The second and third columns are the solution to Ex.7.12, shown in Table 7.2 The fourth column is the solution to part (b) of this problem,which shows that introducing an orifice coefficient of 0.6 changes the flow rates and P2 but has negligible effect on A/F . That is one of the great merits of this type of device;once the dimensions and the upstream gas pressure and properties are set, the A/F is practically constant. The final two columns correspond to the next problem.

Ex 7.12 Ex. 7.12 Pb. 7.31(b) Pb. 7.32(a) Pb. 7.32(b) First guess Solved

D tube, inches 0.425 0.425 0.425 0.425 0.425 D gas jet, inches 0.035 0.035 0.035 0.035 0.035 A tube, in sq 0.14186 0.14186 0.14186 0.14186 0.14186 A jet, in sq 0.00096 0.00096 0.00096 0.00096 0.00043P2, guessed, psi -0.01 -4.432E-05 -1.592E-05 -4.470E-05 -9.427E-06 ρ air, lbm/ft3, at 2 0.0749 0.0750 0.0750 0.0750 0.0750

ρ gas, lbm/ft3, at 2 0.0414 0.0414 0.0414 0.1138 0.1138P gas, in H2O 4 4 4 11 11 same, psi 0.1446 0.1446 0.1446 0.3977 0.3977


7/17/2019 ch 07 - 09 (1)


same, Pa 0.9974 0.9974 0.9974 2.7428 2.7428Orifice Coef, C 1 1 0.6 1 1V gas, ft/s 186.21 180.05 108.02 180.04 180.03 same, m/s 56.75 54.88 32.92 54.87 54.87V air, ft/s 35.18 2.34 1.40 2.35 1.08

m air, lbm/s 2.597E-03 1.730E-04 1.037E-04 1.737E-04 7.977E-05m gas, lbm/s 5.145E-05 4.978E-05 2.986E-05 1.369E-04 6.159E-05m air, lbmol/s 8.956E-05 5.965E-06 3.575E-06 5.990E-06 2.751E-06m gas, lbmol/s 3.215E-06 3.111E-06 1.867E-06 3.111E-06 1.400E-06total molar flow 9.278E-05 9.076E-06 5.442E-06 9.101E-06 4.151E-06total mass flow 2.649E-03 2.228E-04 1.335E-04 3.106E-04 1.414E-04avg M, lbm/lbmol 28.5494 24.5435 24.5408 34.1276 34.0590 ρ mix, lbm/ft3 0.0738 0.0635 0.0635 0.0883 0.0881V 3 35.9523 3.5195 2.1102 3.5344 1.6213P2 calculated, psi -4.386E-

04-4.432E-05 -1.594E-05 -4.470E-05 -9.431E-06

P2 calc/P2guessed 0.0439 1.0000 1.0009 1.0000 1.0004 A/F lbm/lbm 50.4844 3.4748 3.4716 1.2690 1.2951

______________________________________________________________________

7.32 See the spreadsheet in the preceding problem.(a) The two columns on the right correspond to this problem. P2 and V 3 are practicallyunchanged, but the air-fuel ratio is substantially reduced.(b) P2 and V 3 are reduced, but the air-fuel ratio is practically unchanged. The massflow rate of gas is reduced by about a factor of 2, becoming practically the same as fornatural gas (column 3). The result is that the burner has practically the same heat outputand flame characteristics as for natural gas. ______________________________________________________________________


7.33 (a) Using the spreadsheet in the preceding two examples,one finds that the R of the jet, incoming air, and mixture are:3256, 514 and 772.

(b) The jet is probably turbulent, the other two flows laminar.

(c) The probable shape is sketched at the right. It is perturbedfrom the simple parabola we expect for laminar flow by thehigh velocity jet in its center. It is further complicated by the

fact that for this short a tube, L/D ≈ 9, this is an entrance regionflow.

(d) Eq. 7.BL assumes fully developed laminar flow, which isnot completely correct for this entrance region flow.According to Fig 15.8, for this flow the observed f might be ≈ 3times that used here. Ignoring that for the moment, we makethe substitution, finding that the rightmost term becomes .

V e l o c i t y

7/17/2019 ch 07 - 09 (1)


8πμ V 3Δ x = 8π ⋅1.21⋅10−5 lbm

ft s⋅3.52

ft

s⋅3.75ft

12⋅

lbfs2

32.2lbm ft= 6.2 ⋅10−6 lbf

The gas jet force is

Ýmgas V 3 − V gas, 2

( )= 5.0 ⋅10

−5 lbm

s 3.5 −180( )

ft

s ⋅

lbfs2

32.2 lbm ft = −9.8⋅10

−5

lbf

and their ratio isÝmgas V 3 − V gas, 2( )

8πμ V 3Δ x=

−9.8 ⋅10−5 lbf

6.2 ⋅10−6 lbf = −15.8

Which makes dropping the wall shear term plausible, if not perfect. ______________________________________________________________________

7.34 In the Bunsen burner example, there were no surfaces (except the minuscule surfaceof the gas jet) which were not parallel to the flow. Thus the integral was

zero, and did not appear in Eq. 7.29. For real ejectors, jet pumps, etc, as sketched in Fig.7.36 this integral is not zero; the back surface, the contracting surface and the expandingsurface all have components to this integral. This is the reason that the simple analysisgiven for Bunsen burners, while illustrating what is going on in jet ejectors, etc, cannot be used directly for them.

PdAcomponent inthe flow direction

∫

The current Perry's (7e, p 10-57) shows that as if its publication date (1997) no one had published a complete solution to such flows, comparable to the example in this book forthe Bunsen burner, and presents a design chart, based on theory and experimental data. ______________________________________________________________________

7.35

d mV ( )sys

dt = 0 = V

dm

dt + m

dV

dt ;

dV

dt = −

V

m

dm

dt

The fluid is being accelerated in one direction, the rocket in the other, but theirinstantaneous algebraic sum is not being accelerated.______________________________________________________________________

7.36d mV ( )

sys

dt = m

dV

dt + V

dm

dt = − ÝmoutV out − mg

mdV

dt = − Ýmout V out + V sys( )− mg = +

dm

dt V rel − mg ; dV = V rel

dm

m− gdt

V f − V i = V rel lnm f

mi− g t f − t i( )

Moral; do it as quickly as you can! However you don't want the really high velocities to be reached until you are above most of the atmosphere, and hence above most of thefriction drag and friction heating.

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


7.37 See the solution to Prob. 7.36

V f = V rel lnm f

m f

− gΔt = -13850ft

s

⎝⎜

⎠⎟ ⋅ ln 0.1− 32.2

ft

s2 ⋅ 60 s

= 31900 -19 00 = 30 000ft

s= 9130

m

s

The "escape velocity" from earth's gravity is ≈ 36 880 ft/s, so this is close to the speed ofan interplanetary rocket. As far as I know, no one tries that with a one-stage rocket, norone which burns its fuel this quickly. But this problem shows what the effect of gravityis on the velocity of vertically-firing rockets. ______________________________________________________________________

7.38* See the solution to Prob. 7.37

V f = − −4000m

s

⎝⎜

⎠⎟ ⋅ ln

100000 kg

20000kg− 9.81

m

s⋅ 50 s

= 6438- 490 = 5947m

s= 19,500

ft

s

______________________________________________________________________

7.39 For either case d mV ( )

sys

dt = −mV out = m

dV

dt +V

dm

dt ; mout = −

dmsys

dt

mdV

dt = −

dm

dt V sys + V out( )= −

dm

dt V rel ; V rel = 2gh

msys = π

4 D

2 ρ h + M T ; dmsys =

π

4 D

2 ρ dh

dV sys =2gh dh

h + M T

π

4 D2

ρ

=2gh dh

h + β

(a) Here M T = 0, so dV sys =2g

hdh

V f = −2 2ghi = −2 2( ) 32.2ft

s2

⎛⎝⎜

⎠⎟ 10 ft( ) = −50.8

ft

s= −15.4

m

s

(b) Here to get this into the form of an integral I can find in my table I must substituteh = a2. Then using two integrals in the tables, and substituting back for h

V f = 2 2g h − β tan −1 h

β

⎝

⎜⎜

⎠

⎟⎟h=10ft

h= 0

= 2 2g h1 − β tan −1 h1

β

⎝

⎜⎜ ⎟⎟

β = M T

π

4 D2 ρ

=3000 lbm

π

410ft( )2 ⋅ 62.3

lbm

ft3

= 0.613 ft

V f = −2 2 ⋅ 32.2ft

s2 10 ft − 0.613 ft ⋅ tan−1 10

0.613

⎝⎜⎜

⎠⎟⎟ = −34.1

ft

s= −10.4

m

s

The minus sign indicates that the tank is moving in the minus x direction.


7/17/2019 ch 07 - 09 (1)


______________________________________________________________________

7.40 We make an x-directed momentum balance, taking the cart and the part of theturning vane enclosed by it as the system. For this system there is flow in and out, butthe mass of fluid contained in the system does not change with time. The complication

here is that the mass flow rate into the cart is not a constant, but decreases as the cartspeed approaches the jet speed.

mcart

dV cart

dt = Ýmin V in − V out( ) = 2 ÝminV rel = 2 Ýmin V jet − V cart( );

mcart

dV cart

dt = 2 ρ A jet V jet − V cart( )2

; dV cart

V jet −V cart( )2 =

2 ρ A jet

mcart

dt

−−1

V jet − V cart( )⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

start

time t

=2 ρ A jet

mcart

t

−−1

V jet − V cart

⎣

⎢

⎢ ⎦

⎥

⎥start

time t

=1

V jet − V cart, time t

−1

V jet

⎣

⎢

⎢ ⎦

⎥

⎥=

2 ρ A jet

mcart

t

The term in brackets simplifies to1

V jet − V cart, time t

−1

V jet

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=V cart, time t

V jet V jet − V cart, time t ( )

⎣

⎢⎢

⎦

⎥ then

a little algebra leads to V

⎥

cart, time t =

V jet2 ⋅

2 ρ A jet

mcart

⎝⎜⎜

⎠⎟⎟t

1 +2 ρ A jet

mcart

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅ V jet ⋅ t

; Here

2 ρ A jet

mcart =

2 Ým jet / V jet

mcart =

2 ⋅100kg

s/ 50

m

s2000 kg = 0.002 / m so that

V cart, time t =50

m

s

⎛⎝⎜

⎠⎟

2

⋅0.002

m

⎛⎝⎜

⎠⎟ t

1 +0.002

m

⎛⎝⎜

⎞ ⎠⎟ ⋅ 50

m

s

⎛⎝⎜

⎞ ⎠⎟ t

The most common error is to assume that min is a constant. That leads to

V cart = V jet ⋅ 1− exp−2 Ýmint

mcart

⎝⎜⎜

⎠⎟⎟

⎝⎜⎜

⎟⎟

Both of these functions are shown in the following plot


7/17/2019 ch 07 - 09 (1)


0

10

20

30

40

50

0 20 40 60 80 100

V e l o c i t y , m / s

Time, seconds

Correct solution

Constant mass flow rate solution

______________________________________________________________________

7.41 This is the same as the preceding problem but the 2 in the solution disappears and

we have V cart, time t =

V jet2 ⋅

ρ A jet

mcart

⎝⎜⎜

⎠⎟⎟ t

1 + ρ A jet

mcart

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅ V jet t

which can be rearranged to

t =

V cart

V jet

⎝⎜⎜

⎠⎟⎟

1-V cart

V jet

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅

Ým jet

mcart

=0.8

(1 − 0.8) ⋅0.05

s

⎛⎝⎜

⎞ ⎠⎟

= 80 s

______________________________________________________________________

7.42

d

dV blade

dW

dm

⎝⎜

⎠⎟ =

d

dV blade

2 V jetV blade − V blade2( )= 2 V jet − 2V blade( )

We set this equal to 0, and find that the right hand side an only be zero if V jet = 2V blade

which is the same result we found in the text by considering the ke of the fluid. ______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


7.43(a) Taking the kinetic energy in the exiting fluid into account changes the steady

state velocity to V ∞ =

2 P2 − P3( ) ρ

1 + 4 f L

D

One may compute here that the 4fL/D term = 327, so adding 1 to it makes a change of0.3%. If the pipe entrance from the tank is not smooth, then we must add another 0.5, but these changes are certainly small compared to our uncertainty in f .

(b) Here, in Eq. 7.344 we delete the friction term, but we must add another term, to takeaccount of the fact that the pressure just inside the pipe is not the same as the pressure inthe tank. Assuming a rounded pipe entrance, we can say that the pressure just inside the

pipe is P2a = P2 − ρ V 2

2

Eq. 7.44 becomes ρ

π

4 D2

LdV = P2 − P3( )π

4 D2

− ρ V 2

2

π

4 D2

⎣⎢⎢ ⎦⎥⎥dt

dV =1

L

P2 − P3

ρ −

V 2

2

⎝⎜⎜

⎠⎟⎟dt = V ∞

2 − V 2( ) 1

2 Ldt

This is the same as Eq. 7.46 except that the 2 f/D is replace by a 1/2 L. Thus the solutionis the same as in Eq. 7.48, with the first term on the right being 2 L / V ∞ /

Here, the steady state velocity is V ∞ = 2gh = 2 ⋅ 9.81m

s2 ⋅100 m = 44.3m

s= 145

ft

s

In Ex. 7.13 (Table 7.3) we get to 2.4/2.45 = 0.9796 of the steady state velocity in 17 s.Here we get to that fraction of the steady state velocity in

t = 2 ⋅1 ft

44.3m

s⋅ 3.281

ft

m

lnV ∞ + 0.9796V ∞

V ∞ − 0.9796V ∞= 0.063s

We see that starting behavior is normally of interest for long pipes, but not for short ones.

______________________________________________________________________

7.44 With less and less compressible fluids the pressure rise on instantaneous stoppingincreases. It becomes infinite if the fluid is absolutely incompressible. There are noabsolutely incompressible fluids, but fluids like water or mercury are about half ascompressible as fluids like gasoline. ______________________________________________________________________

7.45* The SI values of the speed of sound and density are 668.1 m/s and 498.3 kg/m 3 Then ΔP = c ρ ΔV

= 668.1m

s

⎝⎜

⎠⎟ ⋅ 498.3

kg

m3

⎝⎜

⎠⎟ ⋅ 2.45

m

s

⎝⎜

⎠⎟ ⋅

Pa m2

N⋅ N s2

kg m= 816 kPa = 118 psi

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


7.46 (a) Δt =2 L

V sound wave

=2 ⋅3000 m

1520m

s

= 3.95s

(b) When it has all been brought to rest, it is at a much higher pressure than that in thereservoir, and will expand backwards into the reservoir. The "reflected wave" travels

back to the reservoir, and is then followed by another forward wave, etc. You may thinkof this as the equivalent of dropping a ball on a trampoline. When the ball has reachedzero velocity, the trampoline is compressed, and will send the ball back into the air. Ifthere were no friction in the system the ball would bounce up and down forever. Thefriction in the system slowly brings the ball to rest. The analogous thing happens here.There are a series of such waves, in and out of the pipe, each weaker than the precedingone, until they become negligible. None are as powerful as the one before, so wenormally only worry about the first one. ______________________________________________________________________

7.47 In the starting problem, the acceleration is proportional to (1/ the mass of liquid being accelerated). In the stopping problem the length of time to stop the fluid is also proportional to the pipe length, but the pressure (which we are more interested in) is not, because it depends only how much pressure is needed to stop the first increment of fluid.

If the students don't like the text comparison of starting and stopping an egg, theycan think of an auto. The fastest drag racer cannot accelerate fast enough to hurt itsoccupants on starting. Any car, run into a solid obstacle at even moderate speed candecelerate fast enough to kill its occupants. We can easily produce decelerations muchstronger than the accelerations we normally produce. The exceptions are guns, whichregularly produce accelerations in tens of thousands of g's. ______________________________________________________________________

7.48** z2 = −2 ft2

±2 ft2

⎛⎝⎜ ⎞

⎠⎟ + 2

50ft

s

⎛⎝⎜

⎞ ⎠⎟

2

⋅ 2 ft

32.2ft

s2

⎝

⎜⎜⎜⎜⎜

⎠

⎟⎟⎟⎟⎟

= −1 ±17.65 = 16.65 ft = 5.07 m

V 2 = V 1 z1

z2

= 50ft

s⋅

2 ft

16.65 ft= 6.01ft = 1.83 m

______________________________________________________________________

7.49 The minimum velocity is that for which z1= z2, i.e. a jump of negligible strength.Substituting in Eq 7.52

z2 = z1 = − z12 ± z1

2

⎛⎝⎜

⎞ ⎠⎟

2

+ 2 V 12

z1g

⎛

⎝⎜⎜ ⎞

⎠⎟⎟ ; 32 z1 = ± z1

2

⎛⎝⎜

⎞ ⎠⎟

2

+ 2 V 12

z1g

⎛

⎝⎜⎜ ⎞

⎠⎟⎟

9

4 z1

2 =1

4 z1

2 + 2V 1

2 z1

g; 2 z1

2 = 2 z1V 1

2

g; V 1 = gz1

Any lower value of V 1 leads to a negative value of Δ z , i.e. a jump from a higher to alower elevation, which is thermodynamically impossible. ______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


7.50* We must have V so that V 12 / gz ≥ 1.0 1 = gz = 20 ft ⋅32.2

ft

s2 = 25.4ft

s= 7.73

m

s

If the velocity is less than this, the surface of the pool formed behind the obstruction willremain flat, and its elevation will rise gradually until it overflows the obstruction.

______________________________________________________________________

7.51 If, instead of using gauge pressure, we used absolute pressure, then there would be three pressure terms, one at 1, one at 2, and one being the x component of theatmospheric pressure on the region between 1 and 2. Assuming that the change inatmospheric pressure over these short vertical distance is negligible, then we can say thatthis would add Patm ⋅ z1 ⋅ l to the pressure force at 1, Patm ⋅ z2 ⋅ l to the pressure force at 2,and add a term, equal to Patm ⋅ ( z2 − z1) ⋅ l for the x component of the atmospheric pressureon the surface of the region between 1 and 2. The algebraic sum of these three terms iszero.______________________________________________________________________

7.52 In Eq. 7.AY the appropriate value of P is P =PdA∫ A

=gl ρ

z2

2lz

= g ρ z

2

substituting this twice in Eq.. 7.AY and solving for F .

−F = g z2 − z1( )+V 2

2 − V 12

2

Then solving Eq. 7.52 for V 1, and substituting

−F = g z2 − z1( )+g

4 z1

z22 + 2 z2 z1( ) z1

2

z22 −1

⎝⎜⎜

⎠⎟⎟

Placing this over a common denominator, multiplying out and grouping like terms onefinds

F =g z2 − z1( )3

4 z1 z2

This shows that must always be positive, i.e. flow from a lower elevation to a

higher, because if it were negative we would have a negative value of F . which would bea second law violation.

z2 − z1( )

______________________________________________________________________

7.53 If we take the viewpoint of someone riding on the jump, with velocity V j we findthat the mass balance becomes

V 1 + V J ( ) z1 = V J z2

and Eq. 7.52 becomes

z2 = − z1

2+

z1

2

⎛⎝⎜

⎞ ⎠⎟

2

+ 2 z1

gV 1 + V J ( )2

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 7, page 19 solving these both for ( z2/ z1) and equating the results we find

7/17/2019 ch 07 - 09 (1)


V 1 + V J

V J

= −1

2+

1

4+

2 V 1 + V J ( )2

2 z1

This is a cubic equation as is Eq. B.2-2. In the case of Eq. B.2-2 there is some neatalgebra to reduce it to a quadratic. If any of my readers know how to reduce this to aquadratic, I would be grateful if they would tell me. Although there exist algorithms foralgebraic solution of cubics, they go much more easily by trial and error. In this case wecan easily solve by trial and error on a spreadsheet, finding

V 1 + V J

V J

= 8.45, V J =1.34ft

s, z2 = 0.845 ft

______________________________________________________________________

7.54* Using the result from Prob. 7.26 F =g

4 z1 z2

z2 − z1( )3

z2 = −

10

2 ft +

10

2 ft

⎛

⎝⎜

⎞

⎠⎟

2

+

2 ⋅ 50ft

s

⎛⎝⎜

⎞ ⎠⎟

2

⋅10 ft

32.2 fts

= 34.72 ft

F =32.2

ft

s4( )⋅ 10 ft( ) ⋅ 34.72 ft( )

⋅ 34.72 ft −10 ft( )3= 350.3

ft 2

s2 = 32.5m2

s2

ΔT =Δu

C V

=F

C V

=350.3

ft2

s

1BTU

lbm oR

⋅lbf s2

32.2 lbm ft⋅

BTU

778 ft lbf = 0.014 o F = 0.0077°C

This shows what we all know; the friction heating in most flows of liquids is enough to be very significant in the flows, but not enough to raise the fluid temperature much. ______________________________________________________________________

7.55 Closing the sink drain causes the value of z2 (the fluid depth downstream of the jump) to increase as the liquid accumulates in the sink. From Eq. 7.52 it is clear that as z2 becomes larger, V 1 must become larger. Thus, the jump must move to a place wherethe incoming velocity is higher, which means moving radially inward. (By material balance on can show that in radial, constant-depth flow, the velocity decreases withincreasing radius). So the jump moves inward, eventually engulfing the region ofshallow, fast flow, and "drowning" the jump. After this we have a vertical jet of water

flowing into a pool of water with a practically flat surface. ______________________________________________________________________

7.56* It is much easier on a cold day, because the air is denser. The highest commercialairport I know of is at Cuzco, Peru, approximately 11,000 ft above sea level. Theirtakeoffs and landings are all scheduled in the morning, to take advantage of the cooler air(and the low atmospheric turbulence in the morning). ______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


7.57 For steady state we use Euler's turbine equation Γ = Ým rV θ ( )out

Ým = 5gal

min

⎝⎜

⎠⎟ 8.33

lbm

gal

⎝⎜⎜

⎠⎟⎟ = 41.6

lbm

min= 0.694

lbm

s= 0.315

kg

s

V θ =5 galmin

⎛⎝⎜ ⎠⎟ 231 in

3

gal

⎝⎜⎜ ⎠⎟⎟

2 ⋅π

4⋅

14

in⎛⎝⎜

⎞ ⎠⎟

2 ⋅min

60 s⋅

ft

12 in= 16.3

ft

s= 5.0

m

s

Γ = 0.694lbm

s⋅ft

2⋅16.3

ft

s⋅

lbf s2

32.2 lbm ft= 0.176 ft lbf = 0.24 N m

______________________________________________________________________

7.58 For steady state we use Euler's turbine equation Γ = 0 = Ým rV θ ( )in

− rV θ ( )out[ ]

Here V θ in = 0 so that V θ out must also be zero. This appears odd, but if you observe a

sprinkler of this type, you will see that the liquid jets leaving it move practically radiallyoutward. They have a small backward (opposite to the direction of rotation) component,due to the internal friction and air resistance of the rotating part, but that is generallysmall. The streams in the air appear to curve backwards mostly because as the sprinklerrotates the direction in which the stream emerges changes with time.

V θ out = V θ rel + V θ system = 0 = 16.3ft

sfrom prob. 7.58( )+ r ω

ω =-16.3

ft

s0.5 ft

= −32.6rad

s= 5.19

rev

s= 311 rpm,

The minus sign indicates that the rotor is moving in the direction opposite to the jet. ______________________________________________________________________

7.59 Yes, it rotates, the same way as it does in air. One can see that by applying Euler'sturbine equation, and noting that the properties of the surrounding fluid do not enter it.The higher density and viscosity of water compared to air would make the assumption ofno resistance from the surrounding fluid poorer, but other than that the result is the sameas in Prob. 7.56. If you don't believe this result, put this kind of sprinkler in a pool ortank of water, turn it on, and watch the result. ______________________________________________________________________

7.60 The rotor encounters air resistance, and thus exerts a torque, which tend to spin the

helicopter in the direction opposite the direction of rotation of the rotor.In propeller-driven airplanes the corresponding torque, which is much smaller, tendsto turn the plane over about its long axis. Normally there are "trim tabs" on the plane'scontrol surfaces, which exert and equal and opposite torque, and thus prevent the planefrom rotating about its long axis.

Students who know about flying tell me that you couldn't take off the most powerful propeller fighters in the second world war (e.g. the Spitfire or P-51) at fullthrottle, because the torque would flip them over. Once they got enough air speed, the


7/17/2019 ch 07 - 09 (1)



trim tabs balanced the torque, and they were OK. I have also read that the Japanese zerofighter had a very big propeller and hence a high torque. The US recovered one inalmost mint condition in June 1942 and tested it. They discovered it could turn left muchfaster than it could turn right. They also found that the carburetor had a momentary stallwhen it went into a steep dive. After that US fighter pilots were told to escape from azero "on their tail" by diving and then turning to the right.

The same students tell me that one of the most exciting stunt maneuvers is a whipstall, in which the propeller-driven plane dives to gain speed, and then turns straight up.It loses speed, and eventually stops, at which point the propeller torque makes the plane begin to rotate in the direction opposite the direction of propeller rotation. Don't try thisat home!______________________________________________________________________

7/17/2019 ch 07 - 09 (1)


Solutions Chapter 8______________________________________________________________________

8.1*− ΔP

Δ x=

4 f

D ρ

V 2

2

a( ) − ΔPΔ x

= 4 0.005( )2

12f t

⎛⎝⎜

⎞ ⎠⎟

⋅ 62.3 lbmft3⎛

⎝⎜ ⎞

⎠⎟ ⋅

1000ft

s

⎝⎜

⎠⎟

2

2⋅ lbfs

2

32.2 lbm ft⋅ ft

2

144 in2 = 807 psift

= 18.2 MPam

b( )−ΔP

Δ x= above answer ⋅

0.075

62.3= 0.97

psi

ft= 22.0

kPa

m

The point of this problem is to show that this high a velocity requires an extremely large pressure gradient for a liquid, but only a modest one for a gas. ______________________________________________________________________

8.2 T 2T 1

= P2

P1

⎛

⎝⎜⎜

⎞

⎠⎟⎟

k −1

k

= 14.70114.7⎛⎝⎜ ⎞ ⎠⎟

0.4

1.4 = 1.00002 ; T 2 = 528°R ⋅ T 2T 1

= 528.010 o R

ΔT = T 2 − T 1 = 0.01o R This shows why we do not perceive the change in temperature; it is quite small. ______________________________________________________________________

8.3 (a) P = P0 ⋅10 dB/ 20( ) = 2.9⋅10−9 psi ⋅10 1/20( ) = 3.2 ⋅10−9 psi = 2.2 ⋅10−5 Pa

(b) P = P0 ⋅10 dB/ 20( ) = 2.9⋅10−9 psi ⋅10 60/20( ) = 2.9⋅10−6 psi = 0.02 Pa

(c) P = P0 ⋅10 dB/ 20( ) = 2.9⋅10−9 psi ⋅10 140/20( ) = 0.029 psi = 200 Pa

______________________________________________________________________

8.4* c =K

ρ =

1.5⋅106 lbf

in2

60lbm

ft3

⋅32.2 lbm ft

lbf s2 ⋅144 in2

ft2 =10,800ft

s= 3.28

km

s= 2.04

mi

s

It is almost certainly different with the grain than across the grain. ______________________________________________________________________

8.5c =K

ρ =0.110 ⋅1010Pa

1.05 gcm 3

⋅

1000kg

g

106 cm 3

m3

⋅ N

Pa m2 ⋅kg m

N s2 = 1024m

s = 3360ft

s = 0.63mi

s

This is about 70% of the speed of sound in water. Most organic liquids are not as stiff aswater, and thus have lower speeds of sound. ______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


8.6* In the K and K tables at 212° F we look up, at 11.526 psia, v = 0.016634 ft3 / lbm and at 200 psia, (v − v11.56psia ) = −1.1⋅10−5 ft3 / lbm The corresponding densities are

60.118 and 60.158 lbm/ft3. Then

K = ρ

∂P

∂ ρ

⎛

⎝⎜⎜

⎞

⎠⎟⎟ T ≈ ρ

ΔP

Δ ρ

⎛

⎝⎜⎜

⎞

⎠⎟⎟ T ≈ 60.1

lbm

ft3 ⋅

(200 −11.526)lbf

in2

60.158 − 60.118( ) lbmft3

= 2.85 ⋅10

5

psi = 1.99 ⋅10

4

bar

In the NBS tables at 100° C we look up, at P = 1.1 bar, ρ = 958.40 kg / m3 and at P =100 bar, ρ = 958.40 kg / m3 ρ = 962.98 kg / m3

K = ρ ∂P

∂ ρ

⎝⎜⎜

⎠⎟⎟

T

≈ ρ ΔP

Δ ρ

⎝⎜⎜

⎠⎟⎟

T

≈ 960kg

m3 ⋅(100 −1.1)bar

962.98 − 958.40( ) kg

m3

= 2.07⋅104 bar = 3.00 ⋅105 psi

The difference between the two values is certainly round-off error.

______________________________________________________________________

8.7 c =kRT

m= 223

ft

s

lbm

lbmol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2 1.667 ⋅ 960o R

4lbm

lb mole

= 4460ft

s= 1.36

km

s= 0.644

mi

s

Mostly the low molecular weight leads to this value, which is much higher than thecorresponding value for air. ______________________________________________________________________

8.8* c =

kRT

M = 223

ft

s

lbm

lbmol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2 1.2 ⋅660o R

352 lbmlb mole

= 334

ft

s = 102

m

s

______________________________________________________________________

8.9ΔPE

ΔKE=

gΔ z

ΔV

2

2

=32.2

ft

s2⋅1 ft

2000fts

⎛⎝⎜

⎞ ⎠⎟

2

− 1fts

⎛⎝⎜

⎞ ⎠⎟

2⎛

⎝⎜⎜

⎞

⎠⎟⎟

2

= 1.6⋅10−5

This shows why we normally leave the gravity terms out of high velocity gas flow

calculations. ______________________________________________________________________

8.10*T R

T 1= 2.0( )2

⋅1.667 −1

2

⎝⎜

⎠⎟ + 1 = 2.333

T 1 =T R

2.333=

528o R

2.333= 226.3o R = −233.7o F = -97.9°C

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


8.11 Example 8.3 is for any gas with k = 1.4, so its results are usable for hydrogen.

Then c = 223ft

s

lbm

lb mol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2 1.4 ⋅ 293o R

2lbm

lb mole

= 3194ft

s

V = cM = 3194ft

s⋅ 2.0 = 6388

ft

s= 2007

m

s

______________________________________________________________________

8.12*T R

T 1= 22 1.6667 −1

2

⎝⎜

⎠⎟ +1 = 2.333

T 1 =530°R

2.333= 226°R = −234°F = −147°C

P R

P1

=T R

T 1

⎝⎜⎜

⎠⎟⎟

1.667

1.667 −1= 2.333( )1.667

0.667= 8.317; P1 =

30

8.317= 3.60 psia = 24.9 kPa

ρ R

ρ 1

= 2.333( ) 11.667 −1

= 3.564

ρ R = MP

RT

4lb

lb mole

⎝⎜⎜

⎠⎟⎟ ⋅ 30 psia( )

10.73 psi ft3

lb mol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅ 530 oR ( )

= 0.0212lbm

ft3

ρ 1 =0.0212

lbm

ft3

3.564= 0.0059

lbm

ft3

______________________________________________________________________

8.13 From Tab. A.4 for M = 0.6,P

P R

= 0.7840,T

T R= 0.9328

T = 560 o R 0.9328( ) = 522 o R = 62o F = 2 9 0 K

;P = 60 psia 0.7840( ) = 47 psia = 324 kPa c = 2231.4 ⋅ 522

29=1119

ft

s

V = cM =1119ft

s⋅ 0.6 = 672

ft

s= 204

m

s

______________________________________________________________________

8.14*T R

T 1= 0.6( )2 1.667 −1

2

⎝⎜

⎠⎟ +1 =1.120 ; T 1 = 560o R

1.120= 500o R = 4 0°F = 4.6°C

P R

P1

= 1.120( )1.667

0.667 =1.3275 ; P =60 psi

1.3275= 45.2 psia = 311 kPa

c1 = 223ft

s

1.667⋅ 500

4= 3219

ft

s= 1932

m

s


7/17/2019 ch 07 - 09 (1)


V 1 = c1M1 = 3219 ⋅ 0.6 =1931ft

s= 589

m

s

______________________________________________________________________

8.15 T * = 0.8333T R = 0.8333 ⋅500°R = 417 o R ; c* = 2231.4( ) 417( )

29= 1000

ft

s

V

c *=

1300

1000=1.30 In Tab. A.4 we see that for M = 1.40, V/c* =1.2999 ≈ 1.30

so we use the values from that row of the table,T

T R= 0.7184,

P

P R

= 0.3142

;

T = 500 o R ( ) 0.7184( ) = 359 o R = -101°F = 199.6 K

P = 60 psia( ) 0.3142( ) =18.9 psia = 130 kPa______________________________________________________________________

8.16*T R

T 1

= 12 ⋅1.667 −1

2

⎝

⎜

⎠

⎟ + 1 = 1.3335; T 1 =560

1.3335

= 420o R=-40°F = -40°C

P R

P1

= 1.3335( )1.667

0.667= 2.053 ; P1 =

14.7

2.053= 7.16 psia = 49.4 kPa

c1 = V 1 =223ft

s

1.667 ⋅ 420

4= 2950

ft

s= 899

m

s

ρ 1 =

4lb

lb mole

⎝⎜⎜

⎠⎟⎟ ⋅ 7.16 psia( )

10.73 psi ft3

lb mol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅ 420o R ( )

= 0.00636lbm

ft3 = 0.102kg

m3

______________________________________________________________________

8.17P

P R

=14.7

125= 0.1176 From Eq. 8.18

T

T R=

P

P R

⎛

⎝⎜⎜

⎠⎟⎟

(k −1/ k )

= 0.11760.4/1.4 = 0.543;

T = 530( ) 0.543( ) = 288o R = −172o F

This is a startlingly low temperature. If the compressed air retained thistemperature as it came to rest, it would certainly form a condensation cloud. Normallywe do not see such a cloud. The reason is that the flow is small enough that the kineticenergy is converted to internal energy in a small enough space that the flow comes totemperature equilibrium with its surroundings before it can form a cloud.

This is true of almost all discharges of high pressure gases to the atmosphere.Only under very unusual circumstances (very large discharges, very high relativehumidity) is such a cloud visible. See, NdeN, "Cloud formation on vapour or liquidmixing with air", J. Loss Prev. Process Ind., 5, 205-210 (1992) ______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


8.18*T *

T R= 0.8333; ; T * = 560o R ⋅ 0.8333 = 467o R

c* = 223ft

s

1.4( ) 467( )29

= 1058.4ft

s;

V

c *=

1200

1058.4=1.1337

By interpolation in Table A.4, M 1 ≈ 1.17. Using the spreadsheet's root-finder engine on

Eq. B.6-12, we find M 1 =1.1679. Using that value we findT R

T 1= 1.2728 and

P R

P1

= 2.326

Then T 1 =560o R

1.2728= 440o R = −20o F and P1 =

14.7 psia

2.326= 6.3 psia = 43.6 kPa

______________________________________________________________________

8.19T *

T R= 0.8333 ; T R = 520o R ( ) 0.8333( ) = 433o R ;

c* = 223

1.4( ) 433( )2 = 3885

ft

s ;

V

c * =

6000

3885 =1.545

Interpolating in Tab. A.4, we find M ≈1.82. ______________________________________________________________________

8.20* From Table 8.6, for steam, k =1.30

T R

T 1= 2.0( )2 1.30 − 1

2

⎝⎜

⎠⎟ + 1 = 1.600; T 1 =

1060o R ( )1.600

= 662.5o R = 202.5°F

P R

P1

= 1.600( )1.30

0.30 = 7.665 ; P1 =50 psia

7.7665= 6.52 psia = 45.0 kPa

c = 2331.30( ) 662.5( )

18=1543

ft

s; V 1 = 1543

ft

s

⎝⎜

⎠⎟ 2.0( )= 3085

ft

s= 940

m

s

______________________________________________________________________

8.21T R

T = 12 1.667 −1

2

⎝⎜

⎠⎟ + 1 = 1.3335; T R = 500o R ( )1.3335( )= 667°R

P R

P=

T R

T

⎛⎝⎜

⎞ ⎠⎟

k −1=1.3335

1.667

0.667 = 2.05; P R = 20 psia( ) 2.05( ) = 41 psia = 283 kPa

______________________________________________________________________

8.22* The maximum flow rate corresponds to sonic flow at the nozzle. Using Eq. 8.27

Ým

A *=

P R

T R1

2

Mk

R

⎛⎝⎜

⎞ ⎠⎟

1

2 1

k −1( )2

+ 1⎡

⎣⎢

⎤

⎦⎥

k +1( )2 k −1( )


7/17/2019 ch 07 - 09 (1)


=30

lbf

in2

⎛⎝⎜

⎞ ⎠⎟

530o R ( )1/ 2 ⋅

29lbm

lb mol⋅1.4

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2

223ft

s

lbm

lb mol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2

⋅1

1.2( )2.4

0.8

⋅32.2 lbm ft

lbf s2 = 0.69lbm

s in2

For a 1 in2 nozzle, the mass flow rate is 0.69 lbm/s = 0.31 kg/s..______________________________________________________________________

8.23 (a) Qa = 2 ⋅ 0.00154 ⋅ 480 ⋅96 = 142 scfm

(b) Using Eq. 5.H with an orifice coefficient of 0.6

V 2 = 0.6⋅2 ⋅

10.73 psi⋅ ft3

° R ⋅ lbmol

⎛

⎝⎜⎜

⎠⎟⎟ ⋅ 528°R

(49.7psi)(29 lbm/ lbmol)⋅ 480

lbf

in2 ⋅144 in2

ft2 ⋅32.2 lbm ft

lbf s2

⎣

⎢⎢⎢⎢⎢

⎦

⎥⎥⎥⎥⎥

1/ 2

= 809ft

s

Aorifice =Q

V =

142ft3

min

809ft

s

⋅min

60 s⋅144 in2

ft2 = 0.0131 in2

Dorifice =4

π ⋅0.0131in2⎛

⎝⎜

⎠⎟

1/2

= 0.129 in

(c) The required mass flow rate is

Ýmrequired = Q ρ =142std ft3

min⋅ 0.075

lbm

ft3 ⋅528°R

520°R =10.81

lbm

min

From Eq. 8.27

Ým

A *=

494.7lbf

in2

(528°R)1/2

29lbm

lbmol⋅1.4

(223)2 lbm

lbmol °R⎝

⎜⎜⎜⎜

⎠

⎟⎟⎟⎟

1/2 32.2 lbmft

lbfs2

1.728= 11.12

lbm

s in2

Arequired = A* =Ýmrequired

Ým / A *=

10.81lbm

min

11.12lbm

s in2

⋅min

60 s= 0.0162 in2

Dorifice =4

π ⋅0.0162 in2

⎝⎜

⎠⎟

1/2

= 0.143in

(d) The required area is (1/0.84) times that in part (c) or 0.0193 in2 and the required

diameter is Dorifice =4

π ⋅0.0193 in2⎛

⎝⎜

⎠⎟

1/ 2

= 0.157 in

(e) The diameter actually used is ≈ the same as the answer to part (d). The simple BEsolution gives a much smaller diameter, the frictionless compressible flow solution iscloser to the value used. For this high an initial pressure and corresponding high gas


7/17/2019 ch 07 - 09 (1)


velocity BE is very misleading, and the equations developed in this chapter shouldalways be used.(f) The standard valve on small propane cylinders (not including the 1 lbm throwawaycontainers) screws into a boss in the top of the container. On one side it has a connectionused to fill it with liquid and then discharge vapor to the appliance being served through

some kind of flexible hose (the "service opening"). On the other side the relief valve is placed, closer to the container than the service opening. Normally it will have "375 psi"on the valve body near it. At its back is a retaining cap or insert, which is screwed in farenough to set the relief valve spring, and then "staked" in place by punching the side ofthe valve body to lock it in the proper position.______________________________________________________________________

8.24 Here the flow will be choked at the valve as soon as the vacuum pump empties theline between it and the valve. So

dm

dt = V

d ρ

dt = −mout = −

A * P R

T R1

2

etc.[ ]=VM

RT RdP R where V is the volume of the

system, not some velocity.dP R

P R

= − A * T R

1

2

V

kR

M

⎛⎝⎜

⎞ ⎠⎟

1

2 1

1.23 dt ;

Δt =V

A

M

kRT R

⋅1.23 lnP0

P

=100ft3

10−4 ft 2 ⋅1

223ft

s

29lbm

lb mol1.4 ⋅ 530o R

⋅o R lbm

lbmol

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

1

2

1.23 ln1

0.1= 3.52 ⋅103s ≅1 h r

Here if we did not have the choked flow in the valve, we could treat this the same as Ex.3.7, finding Δt =

100 ft3

10ft3

min

⋅ ln1atm

0.1atm= 23min. If one ignored the problem of choked

flow in the value one would make a very poor estimate of the pump down time. Vacuumsystems are full of choked flow. System designers try to avoid it, but whenever youreduce the pressure by a factor of 2 you will likely have sonic flow. At the pressures ofvacuum systems the continuum assumption which underlies all of this book becomesinaccurate; one must use models which treat individual molecules. They are described in books on vacuum engineering.______________________________________________________________________

8.25* Here the pressure at the throat of the nozzle is the same as the pressure in thedownstream reservoir. That would not be true for a converging-diverging nozzle, but it

is true for the converging nozzle shown here. Thus,P1

P R

= 0.800 We could enter Table

A.4 and interpolate, finding M 1 ≈ 0.57 and A1

A *≈1.22 . However with our spreadsheet


7/17/2019 ch 07 - 09 (1)


we can solve Eq. 8.20 finding M 1 ≈ 0.5737. Then we can use this value in Eq 8.24,

finding A1

A *≈1.2213 from which A* =

2 in2

1.2213=1.6376 in2 . Then from Eq. 8.R we find

m

A *= 2.25

lbm

s in2 and Ým = 2.25lbm

s in2 ⋅1.6376 in2 = 3.685lbm

s

______________________________________________________________________

8.26 For M = 2.0 we read from Tab. A.4 thatP2

P R

= 0.1278 and A

A *=1.6875

thus P R =1atm

0.1278= 7.82 atm and A* =

1 in2

1.6875= 0.592 in2

Ým =

0.592 in2( ) 7.82 ⋅14.7lbf

in2

⎛⎝⎜

⎞ ⎠⎟

29 lbm

lb mol⋅1.4

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2

⋅32.2 lbm ft

lbf s2

530

o

R ( )

1

2

223

ft

s

lbm

lbmol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2

1.2

3

= 1.574lbm

s= 195

lbmol

hr

Po = Ým RT lnP2

P1

= 195lbmol

hr ⋅1.987

Btu

lbmol oR ⋅ 530o R ⋅ ln7.82 ⋅

hp hr

2545 Btu=166 hp

This large power requirement shows why there are not many wind tunnels delivering asupersonic jet to the surroundings in common use. One may also calculate the outlettemperature, finding - 165°F. That shows why if one has such a tunnel one normallyeither dries the air very thoroughly or preheats the air, to prevent fog or ice fog formationin the supersonic section. ______________________________________________________________________

8.27 As the following table shows, the answers are identical. They must be; Table A.4was made up from the same equations as those examples.

Example # From Example From Tab. A.4

8.3 T R / T 1 = 1.80 =1/0.5556 = 1.808.5 P R /P1 =7.82 =1/0.1278 = 7.828.5 ρ R / ρ 1 = 4.35 =1/0.2300 = 4.35

______________________________________________________________________

8.28T

T R=

1

0.8( )2 1.667 −12

⎛⎝⎜ ⎞ ⎠⎟ +1= 0.8241;

P

P R

= 0.8241( )1.667

0.667 = 0.6166

ρ

ρ R

= 0.8241( )1

0.667 = 0.7482 ; A

A *Eq. 8.24( )=

1

0.8

1

0.82410.667

2+1

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

2.667

2 0.667( )

=1.035


7/17/2019 ch 07 - 09 (1)


For V/c* we use Eq. B.6-12,V

c *

⎝⎜

⎠⎟

2

=0.8( )2 2.667( )

0.8( ) 0.667( )+ 2= 0.7033 ;

V

c *= etc. = 0.8386

______________________________________________________________________

8.29 Using Eq. 8.15

V =2 Rk

M k −1( )T R − T ( )=

2 RkT R

M k − 1( )1 −

T

T R=

2 RkT R

M k −1( )1 −

P

P R

⎛

⎝⎜⎜

⎞

⎠⎟⎟

k −1k

which is the equation used to make up the table. As a check, for the last row in Table5.1,

V = 223ft

s

2lbm

lbmol °R

⎝⎜⎜

⎠⎟⎟ ⋅1.4 ⋅ 528°R

29lbm

lbmol1.4 −1( )

1 −14.7

19.7

⎛⎝⎜

⎞ ⎠⎟

0.4

1.4= 713

ft

s

______________________________________________________________________

8.30 (a) From the K and K tables, at the reservoir conditions

h1 = 1329.3Btu

lbm; the KE at 2000 ft/s = 79.8

Btu

lbm, so that at 2000 ft/s (and

adiabatic flow) h2 = 1249.5Btu

lbm. For isentropic flow s2 = s1 . Using the large Mollier

diagram in that steam table, we find that T 2 ≈ 427o F and P2 ≈ 47 psia. Then

c = 2231.30( ) 427 + 460( )

18=1786

ft

s and M =

2000

1786=1.1199

(b)T R

T *

= 12 1.30 −1

2

⎝

⎜

⎠

⎟ + 1 = 1.150; T * =1060o R

1.150

= 921.7o R

c* = 2231.30( ) 921.7( )

18=1819

ft

s ;

V

c *=

2000

1819=1.099

Using Eq. B.6-12V

c *

⎝⎜

⎠⎟

2

=M1

2 (k +1)

M12 (k −1) + 2

after some algebra

M12 =

2V

c *

⎝⎜

⎠⎟

2

k 1−V

c *

⎛⎝⎜

⎞ ⎠⎟

2⎛

⎝⎜⎜

⎞

⎠⎟⎟ +1 +

V

c *

⎛⎝⎜

⎞ ⎠⎟

2 =2 ⋅1.0992

1.30 ⋅ 1 −1.0992( )+1 +1.0992 = 1.2473

M = 1.1168

T R

T 1 = 1.1168( )2 1.30 −1

2

⎝⎜ ⎠⎟ + 1 =1.187 ;

T =1060

1.187= 893o R = 433°F

P R

P= 1.187( )

1.30

0.30 = 2.103 ; P =100

2.103= 47.6 psia


7/17/2019 ch 07 - 09 (1)


The values computed both ways are very close but not identical. The assumption of aconstant value of k for this region is the probable cause of the differences. The K and Kand NBS/NRC tables both give plots of k as a function of T and P. In the low T and P corner of the plot k has values between 1.29 and 1.31. The 1.30 shown in Table 8.6 is agood average value for this region, but any treatment with any constant value gives only

approximate values.

______________________________________________________________________

8.31* From Table A.4 and the problem statement we make up the following table

Mach no P/P R ρ / ρ R P, psia ρ , lbm/ft30.5 0.843 0.8852 20 0.102

2 0.1278 0.23

Then, P2 = P1 ⋅P2 / P R

P1 / P R

= 20 psia ⋅0.1278

0.8430= 3.03 psia and

ρ 2 = ρ 1 ⋅ ρ 2 / ρ R ρ 1 / ρ R

= 0.102 lbmft3 ⋅ 0.2300

0.8852= 0.0265 lbm

ft3

______________________________________________________________________

8.32 Write Eq. 8.17 twice, divide out T R

T R

T 1T R

T 2

=T 2

T 1=M1

2 k −1

2

⎝⎜

⎠⎟ + 1

M22 k −1

2

⎛⎝⎜

⎞ ⎠⎟ + 1

=0.5( )2 1.4 − 1

2

⎝⎜

⎠⎟ +1

22 1.4 −1

2

⎛⎝⎜

⎞ ⎠⎟ + 1

= 0.5833

T 2 = 298.15 K ( ) 0.5833( )= 171K Which is the same as in Example 8.9. ______________________________________________________________________

8.33* From Chapter 7 we know that

HereF = Ým V rel out( )+ Pex − Patm( ) Aexit

Aexit

A *= 1.90 We can go to Table A.4 and

interpolate for the values we need, but with a spreadsheet we can calculate them, finding

M T/T R P/P R ρ / ρ R A/A* V/c*

2.1389 0.5222 0.1029 0.1971 1.9000 1.6932

Then Pexit = 0.1029⋅ 200 psia = 20.58 psia; T * = 0.8333( ) 2000o

R ( )= 1667o

R

c*=2231.4( ) 1667( )

20= 2409

ft

s; V exit = 1.6932( ) 2409( ) = 4078

ft

s

Ým

A *=

2000

223

20 ⋅1.4

2000⋅

1

1.728⋅ 32.2 =1.977

lbm

sec in2

Ým = 1.977lbm

s in2 ⋅144 in2 = 284.8lbm

s


7/17/2019 ch 07 - 09 (1)


Thrust = -F =4078

ft

s

⎝⎜

⎠⎟ 284.7

lbm

s

⎝⎜

⎠⎟

32.2lbm ft

lbf s2

+ 20.58 −14.7( ) lbf

in2 ⋅1.90 ⋅144 in2

= 36,177 +11609 = 37787 lbf which we should probably report as ≈ 38,000 lbf

= 0.26 MN.

______________________________________________________________________

8.34 See the following spreadsheet.Prob. 8.33 Prob. 8.34(a) Prob. 8.34(b)

A throat, ft2 1 1 1A exit, ft2 1.9 1.9 1.9Pr, psia 200 200 400Tr, deg R 2000 4000 2000k 1.4 1.4 1.4

M, lbm/lbmol 20 20 20A/A* 1.9 1.9 1.9P exit, psia 20.58 20.58 41.16T*, deg R 1666.6 3333.2 1666.6c*, ft/s 2408.63 3406.31 2408.63V/c* from table 1.6932 1.6932 1.6932V exit 4078.29 5767.57 4078.29mdot/A, lbm/(s in2) 1.98 1.40 3.95m dot, lbm/s 284.75 201.35 569.50momentum term 36177.34 36177.34 72354.67 pressure term 1608.77 1608.77 7239.46Thrust, lbf 37786.10 37786.10 79594.13 , N 260610.75 260610.75 548960.69

(a) To our surprise, the thrust does not change. The outlet velocity increases, but themass flow rate decreases, and their product remains constant. While there is noimprovement in thrust, the rate of fuel and oxidizer consumption drops to 1/sq. rt 2 of theoriginal value. For this reason one chooses the highest chamber T one can get (and canwithstand).(b) The thrust slightly more than doubles. If the exit pressure were atmospheric so thatthere were no pressure term, then it would exactly double. The mass flow rate doubles,so that going up in pressure does not increase the specific impulse very much. It does getthe fuel and oxidizer burned up faster, thus getting more instantaneous thrust out of a

given size rocket. Generally we choose as high a pressure as we can withstandmechanically and thermally.______________________________________________________________________

8.35 (a) Here at the exitP R

P=

200 psia

14.7 psia=13.605 = Mexit

2 k −1

2+1

⎛⎝⎜

⎞ ⎠⎟

k

k −1


7/17/2019 ch 07 - 09 (1)


Mexit2 =

P R

P

⎛⎝⎜

⎞ ⎠⎟

−1

k

−1

k −1( ) / 2=

13.605( )0.4

1.4 −1

1.4 −1( ) / 2= 5.54; Mexit = 2.353

T R

T exit

= 2.3532 1.4 −1

2

+1 = 2.1082 ; T exit =2000°R

2.1082

= 948.7°R

cexit = 223⋅1.4 ⋅948.7

20= 1817.2

ft

s; V exit = 2.354 ⋅1817.2

ft

s= 4277.7

ft

s

−F momentum = 284.75lbm

s⋅1817.2

ft

s⋅

lbfs2

32.2 lbmft= 37828lbf

In this case there is no pressure term, because the exhaust plane is at atmospheric pressure. The increase in thrust −ΔF momentum = 37 828 lbf - 37 786 lbf = 42 lbf issurprisingly small, indicating that once the exhaust pressure is lowered close to thesurroundings pressure, further lowering it is of little value. ______________________________________________________________________

8.36 A bigger pump would be useless, because the valve, with a pressure ratio of 0.5 ischoked. We need a bigger control valve. ______________________________________________________________________

8.37 A choked orifice (sonic orifice) will be much cheaper than your coworker's proposal. As long as the pressure ratio of downstream to upstream is less than 0.5283,fluctuations in the downstream pressure will not influence the flow rate. ______________________________________________________________________

8.38 Three ways of solving the problem are shown here, because all three regularlyappear in class if you assign this problem.

1-Eq. 8.30 is transcendental (contains both M 2 and its ln) so it cannot have an analytical

solution. But it can be easily solved on a spreadsheet. Here N = 2.196, k = 1.4 and M 2 =

1.00. The numerical solution leads to M 1 = 0.4064.

2-On Fig 8.12 we see that for N = 2 (≈ 2.196) the curve becomes vertical, correspondingto choked flow at the outlet -- further pressure reduction at the outlet does not increase

the mass flow rate, atm / A( )Ým / A( )*

≈ 0.65. If we read vertically upward from that value to

the N = 0 curve we see that the intersection occurs at P3 / P0 ≈ 0.88 . On Fig 8.12, N = 0

corresponds to the outlet of the frictionless nozzle, (location 1 on Fig 8.11) so we knowthat at the outlet of the nozzle P3 / P R ≈ 0.88. From Tab. A.4 we see that this corresponds

to M 1 = 0.43 ≈ 0.4064 Within chart-reading accuracy this is the same as the answer bythe first method shown above.


7/17/2019 ch 07 - 09 (1)


3-Starting the same as method 2, we find thatm / A( )Ým / A( )*

≈ 0.65 But here we can cancel

the mass flow rates, and find A

A *≈

1

0.65=1.538 This step looks funny, and puzzle

students. But it is correct. The flow rate is 65% of the flow rate one would have if one

had only the frictionless nozzle and had choked flow there. So the area at the end of thenozzle must be 65% of the area the nozzle would have for the same mass flow rate andchoked flow. Using this value of the area ratio we can enter Tab. A.4, and againinterpolate to find M 1 = 0.43 ≈ 0.4064,

______________________________________________________________________

8.39 See method 2 of the preceding problem. The halfway point in the pipe

corresponds to N =2.196

2≈ 1 on Fig 8.12 so if we repeat that procedure, reading the

intersection of a vertical line at

m / A( )

Ým / A( )* ≈ 0.60 with the N = 1 line, we find

P

P0 ≈ 0.77 Thus the pressure at the midpoint is P = 0.77⋅ 30 psia = 23.1psia

As an internal check we can find that the pressure at the outlet of the nozzle is26.6 psia, so the pressure drop in the first half of the pipe is (26.6-23.1 = 3.5 psi), whilethat in the second half of the pipe is (23.1 -18.0 = 5.1 psi). Because of the increase invelocity down the pipe, the pressure drop in the second half of the pipe must be morethan that in the first, as this example shows.

______________________________________________________________________

8.40 Lines of constant M outlet are straight lines through the origin. The three lines

shown on the figure for different values of k are the lines for M outlet = 1. Corresponding

lines for lower values of the M outlet also pass through the origin, but lie above the linesfor sonic flow at the outlet. One may also show that those are lines of constant T outlet /T R.

______________________________________________________________________

8.41 It is not a coincidence; it is required by the fluid mechanics and thermodynamics.For a given M outlet, T outlet /T R. is fixed, by Eq. 8.17, which applies to any adiabatic flow,

frictionless or frictional. Thus for a fixed M outlet, T outlet, coutlet and V outlet are all fixed.many point = moutlet = ρ AV ( )

outlet Two of the terms in the bracket at right are fixed,

independent of N and because of the fixed outlet temperature, the ρ outlet ∝ Poutlet Thusmany point ∝ Poutlet

which is the description of a straight line through the origin on Fig. 8. 12. ______________________________________________________________________

8.42 Here P3/P0 sets the mass flow rate for the whole pipe. If we draw a vertical lineupward from the intersection of the N = 2 ≈ 2.196 curve and the P3/P0=0.8 line, and read


7/17/2019 ch 07 - 09 (1)


the values of P3/P0 at various values of N , those correspond to the pressures which wouldexist for pipes of those values of N , for that mass flow rates. So, for example, by this procedure, we read that for N = 0, P3/P0 = 0.94 and for N =1, P3/P0= 0.87. Thus, for thismass flow rate, the pressure at the end of the smooth nozzle (station 1) would be 30·0.94= 28.2 psia, and at the mid-length of the pipe it would be 30·0.87 =26.1 psi. We see that

the values found that way for parts (b) and (c) are practically identical, because in part(b) the flow is almost at its choked value at the outlet, after which lowering thedownstream pressure does not change any of the values inside the pipe.

This problem is easier using Lapple's version of this plot [2], because the graph iseasier to read in the region of interest.

Following this procedure, we produce the plot shown below

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

x/L

P / P 0

(a)

(b and c)

On this plot the points correspond to the readings of Fig 8.12 described above. Thecurves are simple smooth curve fits, produced by my plotting program. ______________________________________________________________________

8.43*P3

P0

=15psia

150 psia= 0.1 From Fig. 8.12, as best I can read it, that corresponds to

N ≈ 50 =4 f Δ x

D

, so Δ x ≈50 D

4 f

=50 ⋅ (1.049ft / 12)

4 ⋅0.0055

= 199ft = 61 m

______________________________________________________________________

8.44 For N = 0, the choked condition corresponds to a P3/P0 of 0.53 (chart readingaccuracy). From App. A.5 it should correspond to a value of 0.5283.


7/17/2019 ch 07 - 09 (1)


For M = 0.5, from App .A.4, P3/P0 = 0.8430. Reading the N = 0 line for thatvalue of the pressure ratio, we find a mass flow rate ratio of 0.74 (chart readingaccuracy). From App. A.4, for this M , A/A* = 1.3398, so the value we read should be itsreciprocal = 0.75.

Within chart reading accuracy, the N = 0 curve on Fig 8.12 corresponds to the

values in App .A.4. ______________________________________________________________________

8.45 Using the values in Ex. 8.6Ým

A*= 0.62

lbm

in2 s

660°R

560°R = 0.673

lbm

in2 s

Ým* = m

A *⋅ Aexit = 0.673

lbm

in2 s⋅ 0.864 in2 = 0.582

lbm

s

Here the outlet pressure is P3/P0 =14.7/30= 0.49. So we read the values of the mass flow

ratio corresponding to that value and the various values of N . For example for N = 200 Iread 0.08. The other values, reading to the right are 0.13, 0.17, 0.27, 0.36, 0.47, 0.64,

0.74 and 1.00. Then, for N = 200, Ým = Ým ⋅ m

Ým= 0.582

lbm

s⋅ 0.08 = 0.0465

lbm

s and

N = 4 f Δ x / D ; 2.196 = 4 f ⋅ 8 f t / D Δ x = N

2.196⋅8 ft =

200

2.196⋅8 ft = 728.6 ft

These values are the lower right round point on the plot below. The other points weremade the same way. (b) In this case N = 4 f Δ x / D + K

Δ x = N − K

2.196⋅ 8 ft =

200

2.196⋅8 ft -

0.5

2.196⋅8 ft = 726.6 ft

So that the L for each of these flow rates is ≈ 2 ft less than the L in part (a) This is thelower right square point on the plot below.


7/17/2019 ch 07 - 09 (1)


1 10 100 10000

0.1

0.2

0.3

0.4

0.5

Pipe length, L . ft

m a s s f l o w r a t e , m ,

l b m

/ s

.

(a)

(b)

This plot assumes a constant f . For the 800 ft pipe the mass flow rate is ≈ 0.1 times thatin the example, so the R should also be 0.1 times the value there. For importantcalculations one would have to take the changing value of f into account. ______________________________________________________________________

8.46 The spreadsheet is shown below

1 st Guess 2nd Guess Ex 8.10 Prob 8.46Pzero, psia 30 30 30 30To, deg R 660 660 660 660 N 2.196 2.196 2.196 2.196P2.psia 18 18 18 12M1, guessed 0.5 0.3 0.371 0.4045mdot/A*, Eq. 8.24 0.622 0.622 0.622 0.622A1/A* Eq 8.21 1.340 2.035 1.692 1.576mdot/AI lbm/s in2 0.464 0.306 0.367 0.395M2,guessed 0.800 0.363 0.553 0.858 N, eq. 8.27 0.997 2.196 2.196 2.196

T2, deg R 585.106 643.052 621.983 575.358c2, ft/s 1185.188 1242.490 1221.966 1175.273V2, ft/s 948.151 451.035 675.527 1007.973rho 2 based on P2 0.083 0.076 0.078 0.056mdot/A 2 lbm/s ft2 78.834 34.122 52.837 56.819same, per in2 0.547 0.237 0.367 0.395


7/17/2019 ch 07 - 09 (1)


Ratio mdot2/mdot1

1.180 0.776 0.999 1.000

It took me 5 guesses of M 1 to converge on the solution shown. One may see on Fig. 8.12that going from P3/P0 = 0.6 to 0.4 moves us to the right on the N = 2 curve, leading to a

higher M outlet and a higher M 1 .

______________________________________________________________________

8.47 V 2 ≈ V 1 ρ 1

ρ 2

⎝⎜⎜

⎠⎟⎟ ≈ V 1

P1

P2

⎝⎜⎜

⎠⎟⎟ ≅ 20

ft

s

750

500

⎛⎝⎜

⎠⎟ = 30

ft

s

ρ avg =

18lbm

lb mole

⎝⎜⎜

⎠⎟⎟ 625 psia( )

10.73 psia ft3

lb mol oR ⋅530o R

= 1.98lbm

ft3

First term

Second term= ρ avgΔ

V 2( )

2ΔP

=

1.98lbm

ft3

⎛⎝⎜

⎠⎟ 30

ft

s

⎛⎝⎜

⎠⎟

2

− 20ft

s

⎛⎝⎜

⎠⎟

2

⎝⎜⎜

⎠⎟⎟

2

250lbf in2

⋅144 in2

ft2

⋅lbf s2

32.2 lbm ft

= 4.2 ⋅10−4 = 0.04% ______________________________________________________________________

8.48 If we integrate Eq 8.33 we find P22 − P1

2 =−4 f

2

⋅ RT

DM

⋅Ým

A

⎛⎝⎜

⎠⎟

2

⎝

⎜⎜

⎠

⎟⎟Δ x For a given

mass flow rate, pipe diameter, and an assumed constant friction factor, this is the plot ofa parabola, as sketched below. For a constant-density fluid, the corresponding plot is astraight line, as also sketched below

x

P

Eq. 8.33

P

x

Constant-density flow

Intuitively, what is going on is that as the gas pressure decreases, the density mustdecrease, with the result that the velocity must increase. It is counter-intuitive thatfriction causes an increase in velocity, but that is exactly what happens in compressibleflows. _____________________________________________________________________


7/17/2019 ch 07 - 09 (1)


8.49 First f =0.0080

36( )1

3

= 0.00242

Then, by Eq 8.33

Ým =750

2

− 5002

( ) psi( )2

3 ft( )5

18lbm

lb mol

π

4

⎛⎝⎜

⎞ ⎠⎟

2

4( ) 0.00243( ) 60⋅ 5280 ft( ) 10.73 psi ft3

lb mol oR

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ⋅ 530o R

⋅32.2 lbm ft

lbf s2 ⋅144in2

ft2

ÝmWeymouth = 473lbm

s= 215

kg

s

Then we compute the upstream and the average densities

ρ upstr = MP

RT =

18lbm

lb mole

⎝⎜⎜

⎠⎟⎟ 750 psi( )

10.73 psi ft3

lbmolo

R

⋅ 530o R = 2.37

lbm

ft3 ;

ρ avg = 2.37625

750

⎝⎜

⎠⎟ =1.978

lbm

ft3

Then, by Bernoulli's equation, using the upstream density

Ým = ρ AV = π

4 D

2 2 ρ ΔP

1 + 4 fL / D

= π

43 ft( )2

2( ) 2.37lbm

ft3

⎛⎝⎜

⎠⎟ 250

lbf

in2

⎛⎝⎜

⎠⎟

1+ 4 0.00243( ) 60 ⋅52803

⎛⎝⎜ ⎞ ⎠⎟⋅144 in2

ft2 ⋅

32.2lbm ft

lbf in2 = 518

lbm

s= 235

kg

s

and then using the average density

Ým = 518lbm

s

1.978

2.37= 473

lbm

s= 215

kg

s

This surprised me, so I checked. If 4 fL/D >> 1, then the Weymouth equation is identicalwith Bernoulli's equation, based on the average density. (Factor the ΔP2 term, andrearrange). For this flow, at the average velocity and density, R ≈ 9105 and by Eq. 6.21

we would calculate f ≈

0.0032, suggesting that the friction factor used in the Weymouthequation is somewhat smaller than one would predict from Eq. 6.21. ______________________________________________________________________

8.50 M y

2 =1.52 +

2

0.42 1.4( )0.4

⎡

⎣⎢

⎤

⎦⎥1.52 −1

= 0.7011; P y

P x

=2 1.4( ) 1.5( )2 − 1.4 − 1( )

1.4 +1= 2.458


7/17/2019 ch 07 - 09 (1)


T y

T x=

1.5( )2 0.4

2

⎝⎜

⎠⎟ + 1

0.42

1+1.52 0.4

2

⎛⎝⎜

⎞ ⎠⎟

1.4⋅1.5

2

−

0.4

2

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣

⎢⎢⎢

⎢

⎤

⎦

⎥⎥⎥

⎥

+ 1

=1.320 ; ρ y

ρ x

=

P y

P x

T y

T x

=2.458

1.320=1.862

These all agree with the values in App. A.5, which were made up using these equations. ______________________________________________________________________

8.51 Shapiro, [4], page 120 shows that this equation can be put in a somewhat moretractable form,

M

Rs y − s x( )=

k

k −1ln

2

(k +1)M x

2 +k − 1

k + 1

⎣⎢⎢ ⎦

⎥⎥

+1

k −1ln

2k M x

2

(k +1)−

k −1

k +1

⎣⎢⎢ ⎦

⎥⎥

The values of this function for k = 1.40 and 1.67 are sketched below. The curves are practically identical, and show a decrease in entropy for upstream Mach numbers less

than 1 (rarefaction shocks) and an increase in entropy for upstream Mach numbers morethan one, (compression shocks).

-0.08

-0.06

-0.04

-0.02

0

0.02

0.4 0.6 0.8 1 1.2 1.4 1.6

M

Rs y − s x( )

M x2

k =1.40

k =1.67

______________________________________________________________________

8.52 Because A y

*

A x

* =P Rx

P Ry

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


8.53* First we switch to the satellite as the frame of reference. In this frame of referencethe satellite is standing still and

T 1 = 225 K = 405 °R ; c1 =1126405

528= 986

ft

s

M1 =V 1

c1

=

30,000 kmhr

⎛⎝⎜

⎠⎟ ⋅ 3281 ft

km⎛⎝⎜

⎠⎟ ⋅ hr

3600 s

⎝⎜⎜

⎠⎟⎟

986ft

s

= 27.72

T R

T 1= 27.72( )2

⋅1.4 −1

2

⎝⎜

⎠⎟ + 1 = 154.7

T R = 405 ° R( )154.7( ) = 62,700o R = 62,200°F = 34,500°C The startlingly high value here is unreasonable, because at these temperatures the airdissociates and its "heat capacity" becomes much higher than the value used. But the

temperatures do become high enough for micrometeorites, which are small bits of rock,to become "shooting stars" and for heat protection to be a major problem in spaceshuttles and other space vehicles. ______________________________________________________________________

8.54* As in the previous problem, we switch frame of reference to have the fighter stand

still. In that frame of reference. For Mach 3 = 2.0T 1

T R= 0.5556,

P1

P R

= 0.1278

T R =460o R

0.5556= 828o R ; P R =

4 psia

0.1278= 31.3 psia Then from Eq. 8.27

Ým

A *=

31.3lbf

in2

223ft

s

lbm

lbmol o R

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1

2

29lbm

lbmol

⎛

⎝⎜ ⎠⎟ 1.4( )828o R

11.728

⋅32.2 lbm ft

lbf s2 = 0.579lbmin2 s

Ým = 0.579lbm

m2s⋅2.0ft2 ⋅

144 in2

ft2 = 167lbm

s= 75.7

kg

s

______________________________________________________________________

8.55 The thermometer will read the reservoir temperature (approximately) which is

T T R

= 3.0( )2 1.4 −12

⎝⎜ ⎠⎟ + 0.1 = 2.8 ; T R = 300o R ⋅ 2.8 = 840o R = 380°F = 467 K

______________________________________________________________________

8-56* The trial-and-error solution, is similar to that in Ex. 8.15, except that the goal is to pick the upstream Mach no for which the velocity change across the shock is 500 ft/s.The following solution, based on that in Zucker [9] is easier. Starting with Eq. B.6-19inverted


7/17/2019 ch 07 - 09 (1)


V y

V x=M x

2 k −1( )+ 2

M x

2k +1( )

subtract 1 from both sides and rearrange to

k +1

2

V x − V y

c x

⎛

⎝⎜⎜

⎞

⎠⎟⎟ =

M x

2 −1

M x

2 =2.4

2

500ft

s

1128ft

s

⎝

⎜⎜⎜

⎜ ⎠

⎟⎟⎟

⎟

= 0.5319

;M x

2 − 0.5319M x −1 = 0 M x = 1.301 From Table A.5 M y = 0.7860

V x = 1.301( )⋅ 1128ft

s

⎝⎜

⎠⎟ = 1467

ft

s;

T y

T x=1.191; T y = 530o R ⋅1.191 = 631o R =171o F

P y

P x

=1.805 ; P y =10 psia ⋅1.805 = 18.05 psia

c y = c x

T y

T x

= 1128ft

s

⎛

⎝

⎜

⎠

⎟ 1.1909 = 1231ft

s

V y = c yM y = 1231ft

s⋅ 0.7860 = 967.5

ft

s

V x −V y = 1476 − 976.5 = 499.5ft

s≈ 500

ft

s

From the frame of reference of the moving shock, the gas approaches at 1467 ft/s andleaves at 967.5 ft/s. From the viewpoint of a stationary observe, the gas downstream ofthe shock is standing still, so the shock must be moving upstream at 967.5 ft/s.______________________________________________________________________

8.57 (a) If the flow is sonic at the throat, then A/AT = A/A* = 1.5, and M = 1.86 or 0.43.These correspond to the values on curves AC or AG in Fig. 8.16.(b) If the flow at the throat is not sonic, then there is only one possible (isentropic) Machnumber at this point. Using App. A.4 we make the following table

M throat Athroat/ A* A1/ A* M table interp. M Eq.8.24. 0.1 5.8218 8.7327 0.065 0.066444080.5 1.3398 2.0097 0.31 0.304246530.9 1.0089 1.51335 0.43 0.42544824

For M throat =0.1 we read the value of Athroat/ A*=5.2812. Then A1/ A* = 1.5 times this

value = 8.7327. We can estimate M 1 by interpolating in Tab. A.4, finding M table interp≈

0.065, or we can use the spreadsheet's numerical engine to find the value from Eq. 8.24,M Eq.8.24.= 0.0664. The next two rows show the corresponding values for the other two

values of M throat. These correspond to curves like AB on Fig 8.16. The point of interestcould be upstream or downstream of the throat; in subsonic, adiabatic frictionless flowall the parameters of the flow depend only on M 1, not on whether the flow is upstreamor downstream of a throat and whether it is accelerating or decelerating.


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______________________________________________________________________

8.58 For subsonic flow throughout, M exit goes from 0 to 0.33, while P/P R goes from 1to 0.93, corresponding to curves AA through AC on Fig. 8.16.

For sonic flow at the throat, M exit goes from 0.33 to 0.555 while P/P R goes from

0.33 to 0.53, corresponding to curve AD through AE on Fig. 8.16.For isentropic flow throughout, M exit = 2.14 and P/P R = 0.1027, corresponding

to curve AG on Fig. 8.16. The behavior is sketched below, not to scale.

0 0.1027 0.53 0.93 1.0

2.14

0.33

0.53

0.00

P /Pexit R

M a c h N o a t e x

i t

Not to scale

Sonic atthroat

Shock atoutlet

Shock innozzle

Supersonicexit

P r e s s u r e a d j u s t s

o u t s i d e n o z z l e

______________________________________________________________________

8.59 (a) When the flow is subsonic M can take any value between 0 and 0.43, for whichrange the T/T R can take any value from 1 to 0.9643, and hence the temperature can beany value from 530 to 510 deg R. There is only one possible supersonic condition, withM = 1.86, T/T R =0.5910, and T = 313 deg R.(b) All of the answers to part (a) are also answers to part (b). In addition, we have the possibility of normal shocks anywhere between the throat and the point in question. Ifthe shock is at the throat, then it is of vanishing strength, and the temperature at point x will be 510 deg R. As the shock moves downstream its strength increases, so the

temperature at point x falls. For the shock at point x theM =

1.86, T 2/T 1 = 1.577, so T =313 ·1.577 = 495 deg R. So the range of possible values is from 530 to 495 deg R, and313 deg R. ______________________________________________________________________

8.60 (a) Aexit

A *= 2.035 ;

AT

A *=

2.035

1.5=1.36

The flow at throat is subsonic, so we can look up M throat = 0.48.


7/17/2019 ch 07 - 09 (1)


(b) AT

A *=

1.3389

1.5= 0.893

This is based on the assumption of no shock waves. It is clearly wrong, so there must bea shock wave, and the flow at the throat must be sonic, M throat = 1.00.. ______________________________________________________________________

8.61 m1 = m2 and both are sonic, so thatÝm

A*

⎛⎝⎜

⎠⎟

1

= 1.01Ým

A2*

⎝⎜⎜

⎠⎟⎟

A2*

A1* =1.01 =

P Rx

P Ry

. From App. A.5,P Rx

P Ry

=1

1.01= 0.990 corresponds to

______________________________________________________________________

M x ≈ 1.22

8.62* The first three columns of the following table are the same as Table 8.5. Thesame spreadsheet program used to solve Ex. 8.15 was used to solve this problem. The

solution is shown in the rightmost column.

The value of P2 / P R1 is set equal to 0.70, and the spreadsheet search engine finds the

value of the upstream Mach number which makes the ratio at the bottom of the table =1.000. The various values can be read from the table.

Trial number 1 2 Prob 8.56P2/Pr1 0.8022 0.8022 0.7Mach No x, guessed 1.1000 1.4878 1.7109Mach No y, Eq F.16 0.9118 0.7055 0.6377Pry/Prx, Eq 8.33 0.9989 0.9336 0.8512

P2/Pr2 0.8031 0.8592 0.8224Aexit/A*, from Pry/Prx 1.4984 1.4004 1.2767M exit from P2/PR2 and eq 8.17 0.5686 0.4706 0.5360Aexit/A*, from M exit

Eq 8.21 1.2282 1.4004 1.2767Ratio 1.2200 1.0000 1.0000

Ax/Athroat 1.1680 1.3480

The lowest outlet pressure at which one can perform this calculation is the one at whichthe shock wave is just at the outlet, for which we know that A / A* = 1.5 . From Table 8.4

we see that, upstream of the shock this corresponds to M ≈ 1.85. At that value we haveP / P R = 0.1612 and for a shock at that Mach number, we have P y / P x = 3.8263 so that

P y

P R x

=P x

P R x

⋅P y

P x

≈ 0.1612 ⋅3.8263 = 0.617

I show the answer as ≈ because I did it with table reads and M ≈ 1.85 is not exact.Substituting 0.617 in the same spreadsheet used for the above table gives A/A* = 1.4978≈ 1.50.


7/17/2019 ch 07 - 09 (1)


______________________________________________________________________

8.63P Ry

P Rx

=P Ry

P y

⋅P y

P x

⋅P x

P Rx

; For ,M = 3P y

P x

=10.33

For ,M

= 0.4752

P Ry

P y = 0.855

For ,M = 3P Rx

P x

= 1+1.4 −1

232( )⎣⎢ ⎦⎥

= 36.73

P Ry

P Rx

=1

0.855⋅10.33⋅

1

36.73= 0.329

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


Solutions Chapter 9

______________________________________________________________________

9.1 (a) Inertia and surface forces, hence the Weber number(b) Inertia and gravity forces, hence the Froude number. See Bird et. al., p 108.

(c) Here we add viscous forces, so in addition to the Froude number, the Reynoldsnumber is probably important.(d) Gravity, surface, viscous, perhaps inertia. Froude, Reynolds, Weber numbers. Theratio of the gas and liquid densities may also appear.(e) Gravity and surface forces, the Eötvös number.(f) Pressure and viscous forces, the Stokes number

(g) Same as (f) with the gravity force added, g ρ L3 /µV

______________________________________________________________________

9.2 If there is only one π, then the relation must be of the formV 2

V 1

⎛

⎝⎜⎜

⎞

⎠⎟⎟

F 2

F 1

⎛

⎝⎜⎜

⎠⎟⎟

a

= const

If there are two π's, then the relation is probably of the form V 2

V 1= f

F 2

F 1

⎝⎜⎜ ⎟⎟

______________________________________________________________________

9.3* π 2 ; L0t

0F

0 ≡ F a

Ft 2

/ L4( )b Lc

Ft / L2( )d

; 00 = −4b + c − 2d = 2b + d ; and 0 = a + b + d

We select d = 2 to avoid fractional exponents, then d = 2; a = −1 b = −1 c = 0 and

π 2 =2

F ρ =

2

R ⋅ Pressure coefficient( )2

π 3 ; L0t

0F

0 ≡ F a

Ft 2

/ L4( )

b

Lc L

d

; 0 and 00 = −4b + c + d = a + 2b = 2b These lead to a = 0 b = 0 c = −d ;

π 3 = L1 / L2

______________________________________________________________________

9.4 The variables and their dimensions are

Variable Dimension

A mL/t 2

B1 L/t

B2 m/L3

B3 m/Lt

B4 L

B5 L

π 1 = m0 L

0t

0 = L

t

⎛⎝⎜

⎠⎟

am

L3

⎛⎝⎜

⎞ ⎠⎟

b

L( )c mL

t 2

⎛⎝⎜

⎠⎟

d


7/17/2019 ch 07 - 09 (1)


0 = b + d

0 = a − 3b + c + d

0 = −a − 2d

d = 1

a = −2

b = 1

c = −2

π 1 =F

V 2 ρ L

2 =1

π 1 in Ex. 9.3

Proceeding the same way we find that π 2 =V ρ L

=1

R and that π 3 =

L1

L2

.

______________________________________________________________________

9.5 The variables and their dimensions are

Variable Dimension

A F

B1 L/t

B2 m/L3

B3 Ft/L2

B4 L

B5 L

gc mL/Ft 2

π 1 = F 0m

0 L

0t

0 = L

t

⎝⎜

⎠⎟

am

L3

⎝⎜

⎠⎟

b

L( )cF ( )d mL

Ft 2

⎝⎜

⎠⎟

e

0 = a − 3b + c + e

0 = d − e

0 = b + e

0 = −a − 2e

d = 1

a = −2

b = −1

c = −2e = 1

π 1 =Fgc

V 2 ρ L

2 =gc

π 1 in Ex. 9.3

Here gc = dimensionless 1.00. It plays no role in a dimensionless group, and may be

deleted. Proceeding the same way we find that

π 2 =2

F ρ gc

=2

L2V

2 ρ

2g

c

⋅ L2V 2 ρ

F =

1

R2 ⋅ pressure coefficient

and that π 3 = L1

L2

.

______________________________________________________________________

9.6 Repeating; B1, B2, B4, others A, B3, B5,

π 1 =F 0 L

0t

0 = L

t

⎛⎝⎜

⎞ ⎠⎟

aFt 2

L4

⎝⎜⎜

⎠⎟⎟

b

L( )cF ( )d

⇒ π 1

V 2F

ρ L2

=1

2 ⋅ pressure coef.


7/17/2019 ch 07 - 09 (1)


π 2 =F 0 L

0t

0 = L

t

⎛⎝⎜

⎞ ⎠⎟

aFt 2

L4

⎝⎜⎜

⎠⎟⎟

b

L( )c Ft

L2

⎛⎝⎜

⎞ ⎠⎟

d

⇒ π 2 = μ

ρ VL=

1

R

π 3 = F 0 L

0t

0 = L

t

⎛⎝⎜

⎞ ⎠⎟

aFt 2

L4

⎝⎜⎜

⎠⎟⎟

b

L( )c L( )d

⇒ π 3 = L1

L2

______________________________________________________________________

9.7 If a velocity, a density a viscosity and two lengths are believed significant, then wewould expect the viscous, inertia and pressure forces to be involved. From Tab. 9.1 wewould select the Reynolds number, and the pressure coefficient. We would also select alength ratio, when two lengths are involved. ______________________________________________________________________

9.8

Course Dimensionless

group

Ratio of

ElementaryThermodynamics qualitycompressibilityfactor, z efficiency

Air fuel ratio

mass vapor/total massvolume/volume of an ideal gas

work produced/best possiblework producedmass air/mass fuel

Heat and MaterialBalances

humidityrelative humiditymole fractionvapor-liquidequilibrium coefspecific gravity

mass water/mass airhumidity/saturated humiditymols/molesmol fraction/mol fraction

density/density

These are all ratios of one thing to the same thing under different circumstances. Thatexplains why little is said about dimensional analysis in those courses. In FluidMechanics, Heat Transfer and Mass Transfer we encounter the dimensionless groupswhich are not simple ratios of this type. ______________________________________________________________________

9.9* If the viscosity and density are kept constant, then the product of the length andvelocity must be kept constant. This means that a 1/10 scale model must have a speed of600 miles/hr, which is quite impractical, because the Mach number becomes too high. Ifone tests in water, then

V2 = V 1

L1

L2

ρ 1

ρ 2

μ 2

μ 1 = 60

mi

hr ⋅10 ⋅

0.075

62.3 ⋅

1.002

0.018 = 40.2

mi

hr

______________________________________________________________________


7/17/2019 ch 07 - 09 (1)


9.10 Taking the velocity as the tip velocity, V rotor tip = r ω = D

2 and the rotor diameter,

D as the length dimension, we have R pump = D ⋅ D

2ν It follows that if we are to have the

same Reynolds number in the full size pump and the model, then

ω model

ω full size

= Dfull size

Dmodel

⎛⎝⎜⎜ ⎞

⎠⎟⎟

2

= 102 = 100; model =100 ⋅1800rpm =180 000 rpm

This is an impossibly high rotation speed. One would never do this. The properdimensionless groups for rotating pumps, compressors and turbines are discussed in Ch.10. They do not include the Reynolds number.

The Reynolds number is one of the most important, if not the most important ofthe dimensionless groups in fluid mechanics. Experienced engineers always think aboutthe Reynolds number in any fluid flow. But this example and the next show that in somemodel studies we cannot practically match Reynolds numbers, and must work our wayaround that difficulty. ______________________________________________________________________

9.11 It is not possible to scale both Reynolds and Froude numbers simultaneously. If themodel and the real boat are to have the same Reynolds numbers, then the velocity of themodel must be 10 times the velocity of the real boat. If the Froude numbers are to be thesame, then the velocity of the model must be the square root of (1/10) = 0.316 times thevelocity of the real boat.

The common practice in model tests of boat hulls is to keep the Froude numberconstant. The viscous drag is calculated for the model, using methods like those shownin Chapter 17, and subtracted from the total resistance of the boat, to find the "form drag"which is a function of the Froude number. Then the viscous drag for the real boat iscalculated and added to the form drag from the model, scaled to the size of the real boat,

to make the best estimate of the resistance of the real boat.In current racing boat practice, most of the design is done with computers, but a

scale model of the final hull design is still tested in a towing basin to make sure it really behaves as predicted.

The Reynolds number is one of the most important, if not the most important ofthe dimensionless groups in fluid mechanics. Experienced engineers always think aboutthe Reynolds number in any fluid flow. But this example and the previous show that insome model studies we cannot practically match Reynolds numbers, and must work ourway around that difficulty.

______________________________________________________________________

9.12* We have added one new variable, so the number of pi's must increase by one, to 4.Using the same repeating variables as in Ex. 9.3, we have

π 4; L0t

0F

0 ≡ F a

Ft 2

/ L4( )b

Lc

1 / t ( )d

; 00 = −4b + c − 2d = 2b − d ; and 0 = a + b + d

We select a = 1 , then a = 1; b = −1; d = −2 c = −4 and


7/17/2019 ch 07 - 09 (1)


π 4 =F

ρ L4ω

2 =F

ρ L2V

2 ⋅V 2

L2ω

2 = 0.5 ⋅Pressure

coefficient

⎝⎜

⎠⎟ ⋅

Strouhal

number

⎝⎜

⎠⎟−2

The Strouhal number is involved in all aerodynamic vibration studies. See anydiscussion of the failure of the Tacoma Narrows Bridge.

______________________________________________________________________

9.13 Presumably only gravity and inertia, so the Froude number is the only importantone, and if it is the only one then it should be a constant. That leads to

V = const. 2gL . For deep ocean waves the constant appears to be 1.00, so, for

example if the depth is 1000 m (a typical deep ocean depth) then

V ≈ 2 ⋅9.81m

s2 ⋅1000 m =140

m

s= 504

km

hr

This explains why tsunami waves can cross a whole ocean in a few hours.In a small

shallow pond gravity and inertia are still dominant, but surface tension and viscosity begin to play a role. For very small waves surface tension become dominant, so wewould expect the Weber and Eötvös numbers to become important. ______________________________________________________________________

9.14 For a small particle only the gravity and viscous forces should be important so we

would expect their ratioF G

F V

=g ρ L2

μ V to be dominant. If it is the only dimensionless

number then it should be a constant. If we take the L in this number to be the particlediameter, then the constant is 1/18, and the relation is the same as Eq. 6.57______________________________________________________________________

9.15 As the particle becomes larger we would expect the inertial force to come into

play. Among the three forces, we can form three ratios, R, , the Froude

number and

F G, F V and F I

g ρ L2

μ V . But this latter ratio (which does not appear to have a common name)

is equal to R, / Froude number, so that there are only two independent dimensionless

groups. We can writeFroude

number

⎝⎜

⎠⎟ =

V 2

gD= f (R ) where we have dropped the 2 in the

Froude number. Then if we take the square root of both sides and substitute

f (R

) =

1

C D we will have Eq. 6.56 (times some constants). The relation between R,and C D is the drag coefficient plot, Fig 6.24.

______________________________________________________________________

9.16 Ar = ρ f gD

3 ρ p − ρ f ( )μ

2 =4

3C D R p

2


7/17/2019 ch 07 - 09 (1)


This number is popular in calculation of settling velocities in the region too large to beeasily represented by Stokes' law. ______________________________________________________________________

ch 07 - 09 (1)

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