che212 fluid mechanics 2017-18 homework 3...

CHE212 FLUID MECHANICS

2017-18 HOMEWORK 3 SOLUTION KEY

Q1. A fluid flowing in laminar flow in the x-direction between two parallel plates. 2yo is the distance between the plates,

y is the distance from the center line and vx is the velocity in the x-direction at y.

Derive the momentum balance equation and velocity profile equation. Simplify the equation using necessary assumptions.

Write the boundary conditions.

Solution 1.

rateof momentum in bymolecular transport rateof momentumoutbymolecular transport

rateof momentum inbyconvectivetransport rateof momentumoutbyconvectivetransport

sumof the forcesactingonthesystem momentum accumula

tion

{(𝜏𝑦𝑥∆𝑧 ∆𝑥|𝑦 − 𝜏𝑦𝑥∆𝑧 ∆𝑥|𝑦+∆𝑦)} + {(𝜌∆𝑧 ∆𝑦(𝜗𝑥)̅̅ ̅̅ ̅̅ 𝜗𝑥|𝑥

− 𝜌 ∆𝑧 ∆𝑦 (𝜗𝑥)̅̅ ̅̅ ̅̅ 𝜗𝑥|𝑥+∆𝑥

)}

+ {(𝑃𝑥 ∆𝑧 ∆𝑦 |𝑥) − (𝑃𝑥 ∆𝑧 ∆𝑦|𝑥+∆𝑥)} + 𝜌∆𝑥 ∆𝑦∆𝑧 𝑔𝑥 = 𝜌∆𝑧 ∆𝑦 ∆𝑥 𝑑𝜗𝑥

𝑑𝑡= 0 = 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒

𝜌 ∆𝑧 ∆𝑦(𝜗𝑥)̅̅ ̅̅ ̅̅ 𝜗𝑥|𝑥

− 𝜌 ∆𝑧 ∆𝑦 (𝜗𝑥)̅̅ ̅̅ ̅̅ 𝜗𝑥|𝑥+∆𝑥

= 0 ( 𝜗𝑥|𝑥 = 𝜗𝑥|𝑥+∆𝑥)

𝑔𝑥 = 0 → 𝜌∆𝑧 ∆𝑦 𝑔𝑥 = 0

{(𝜏𝑦𝑥∆𝑧 ∆𝑥 |𝑦 − 𝜏𝑦𝑥∆𝑧 ∆𝑥|𝑦+∆𝑦)} + {(𝑃𝑥 ∆𝑧 ∆𝑦|𝑥) − (𝑃𝑥 ∆𝑧 ∆𝑦|𝑥+∆𝑥)} = 0

Divide Δx ΔyΔz and take the lim Δx→ 0 and lim Δy→ 0

lim(𝜏𝑦𝑥|

𝑥− 𝜏𝑦𝑥|

𝑥+∆𝑥)

∆𝑦+

( 𝑃𝑥|𝑥) − ( 𝑃𝑥|𝑥+∆𝑥)

∆𝑥= 0

𝑑𝜏𝑦𝑥

𝑑𝑦= −

𝑑𝑃

𝑑𝑥

𝑑𝑃

𝑑𝑥=

∆𝑃

∆𝐿=

𝑃𝐿 − 𝑃𝑜

𝐿

𝑑𝜏𝑦𝑥

𝑑𝑦= −


𝐿

𝜏𝑦𝑥 = − 𝑃𝐿 − 𝑃𝑜

𝐿 𝑦 + 𝐶1

𝜏𝑦𝑥 = −𝜇𝑑𝜗𝑥

𝑑𝑦= −


𝐿 𝑦 + 𝐶1

𝑑𝜗𝑥

𝑑𝑦=


𝜇𝐿 𝑦 + 𝐶1

𝜗𝑥 = 𝑃𝐿 − 𝑃𝑜

2𝜇𝐿 𝑦2 + 𝐶1 𝑦 + 𝐶2

B.C

y= 0 ,vx =0 → C2 = 0

y= 2y0 vx =0 → 𝐶1 = −𝑃𝐿−𝑃𝑜

𝜇𝐿 𝑦0


2𝜇𝐿 (𝑦2 − 2𝑦0 𝑦)


2𝜇𝐿 𝑦0

2 (𝑦2

𝑦02

− 2𝑦

𝑦0)

Q2. For a layer of liquid flowing in laminar flow in the z-direction down a vertical plate or surface, the velocity profile is

𝜗𝑧 =𝜌𝑔𝛿2

2𝜇[1 − (

𝑥

𝛿)

2

]

where δ is the thickness of the layer, x is the distance from the free surface of the liquid toward the plate, and 𝜗𝑧 is the

velocity at a distance x from the free surface.

a) What is the maximum velocity 𝜗𝑧,𝑚𝑎𝑥 ? (Ans : 𝜗𝑧,𝑚𝑎𝑥 = 𝜌𝑔𝛿2/2𝜇)

b) Derive the expression for the average velocity 𝜗𝑧,𝑎𝑣𝑔 and also relate it to 𝜗𝑧,𝑚𝑎𝑥. (Ans: 𝜗𝑧,𝑎𝑣𝑔=2/3 𝜗𝑧,𝑚𝑎𝑥)

Solution 2.

Assumptions:

1) Steady state flow

2) Incompressible fluid (ρ = constant)

3) Newtonian fluid




tion

{(𝜏𝑥𝑧∆𝑧 ∆𝑦|𝑥 − 𝜏𝑥𝑧∆𝑧 ∆𝑦|𝑥+∆𝑥)} + {(𝜌∆𝑥 ∆𝑦(𝜗𝑧)̅̅ ̅̅ ̅̅ 𝜗𝑧|𝑧=0

− 𝜌 ∆𝑥 ∆𝑦 (𝜗𝑥)̅̅ ̅̅ ̅̅ 𝜗𝑥|𝑧=𝐿

)} + 𝜌∆𝑥 ∆𝑦∆𝑧𝑔𝑧 = 𝜌∆𝑧 ∆𝑦 ∆𝑥 𝑑𝜗𝑧

𝑑𝑡

= 0

𝜗𝑧|𝑧=0 = 𝜗𝑧|𝑧=𝐿

Divide Δx ΔyΔz and taking the lim Δx→ 0

lim∆𝑥→0

(𝜏𝑥𝑧|𝑥 − 𝜏𝑥𝑧|𝑥+∆𝑥)

∆𝑥+ 𝜌𝑔𝑧 = 0

−𝑑𝜏𝑥𝑧

𝑑𝑥= −𝜌𝑔𝑧

𝜏𝑥𝑧 = −𝜇𝑑𝜗𝑧

𝑑𝑥

⟹ −𝑑 (−𝜇

𝑑𝜗𝑧𝑑𝑥

)


⟹𝑑 (𝜇

𝑑𝜗𝑧𝑑𝑥

)


𝑑𝜗𝑧

𝑑𝑥= −

𝜌𝑔𝑧

𝜇𝑥 + 𝐶1

𝜗𝑧 = −𝜌𝑔𝑧

2𝜇𝑥2 + 𝐶1𝑥 + 𝐶2

BC 1: x = -δ z = 0

BC 2: 𝑥 = 0 𝑑𝜗𝑧

𝑑𝑥= 0 (𝑧 = 𝑚𝑎𝑥)

From BC 2:

0 = −𝜌𝑔𝑧

𝜇0 + 𝐶1

⟹ 𝐶1 = 0


2𝜇𝑥2 + 𝐶1𝑥 + 𝐶2 = −

𝜌𝑔𝑧

2𝜇𝑥2 + 𝐶2

From BC 1:

0 = −𝜌𝑔𝑧

2𝜇(−𝛿)2 + 𝐶2

⟹ 𝐶2 =𝜌𝑔𝑧

2𝜇𝛿2


2𝜇𝑥2 +

𝜌𝑔𝑧

2𝜇𝛿2 =

𝜌𝑔𝑧

2𝜇𝛿2 (1 −

𝑥2

𝛿2)

at 𝑥 = 0 𝑧 = 𝑚𝑎𝑥

𝒎𝒂𝒙 =𝜌𝑔𝑧

2𝜇𝛿2 (1 −

02

𝛿2) =𝝆𝒈𝒛

𝟐𝝁𝜹𝟐

𝑄 = ∫ ∫ 𝜗𝑧𝑑𝑥𝑑𝑦𝛿

0

=𝑊

0

∫ ∫𝜌𝑔𝑧

2𝜇𝛿2 (1 −

𝑥2

𝛿2) 𝑑𝑥𝑑𝑦𝛿

0

𝑊

0

𝑄 =𝑊𝜌𝑔𝑧𝛿3

3𝜇

𝝑𝒛,𝒂𝒗𝒈 =𝑄

∫ ∫ 𝑑𝑥𝑑𝑦𝛿

0

𝑊

0

=𝝆𝒈𝒛𝜹𝟐

𝟑𝝁

𝝑𝒛,𝒂𝒗𝒈 =𝟐

𝟑𝒎𝒂𝒙

Q3. Derive the equation for steady-state laminar flow inside the annulus between two concentric horizontal pipes. Calculate

the rmax.

Solution 3.




tion

( *2 *2 ) ( * * )

( )[ *2 * ( *2 *2 ) 2 *

rx rx x xr r r x x x

xx x x x

r x r x m m

r r x g P r r P r r r r xt

(1)

* *2xm r r

( *2 *2 ) ( * *2 * * *2 * )

( )[ *2 * ( *2 *2 ) 2 *

rx rx x x x xr r r x x x

xx x x x

r x r x r r r r

r r x g P r r P r r r r xt

(2)

Assumptions:

steady state condition, there is no accumulation term in the momentum balance( )

2 * xr r xt

=0

there is no gravity force in x direction *2 * xr r x g =0

νx=f(r) νx≠f(x) one dimensional flow

( *2 *2 ) ( * *2 * * *2 * )

( *2 *2 ) 0

rx rx x x x xr r r x x x

x x x

r x r x r r r r

P r r P r r

(3)

( *2 *2 ) ( *2 *2 ) 0rx rxr r r x x xr x r x P r r P r r

(4)

Both side of the Eq.4 is divided by 2 r x and0 0

lim limr x

and

0( )rx Lr P PP P Pr

r x x x L

(5)

If we integrate the Eq.(5)

2

01

0 1

( )

2

2

rx

Lrx

Lrx

r Pr

r x

P P rr c

L

P P cr

L r

(6)

In order to calculate integral constant “c1” in Eq.6 we need to boundary conditions:

Schematic representation in the above 𝒓 = 𝒓𝒎𝒂𝒙 ν=νmax so 0 0xrx

d

dr

If we rearrange the Eq.(6);

−𝜏𝑟𝑥 = (𝑃𝐿 − 𝑃𝑂

𝐿)

𝑟𝑚𝑎𝑥

2+ 𝐶1 = 0

𝐶1 = − (𝑃𝐿 − 𝑃𝑂

𝐿)

𝑟𝑚𝑎𝑥

2

−𝜏𝑟𝑥 = (𝑃𝐿 − 𝑃𝑂

𝐿)

𝑟

2−

(𝑃𝐿 − 𝑃𝑂

𝐿 )𝑟𝑚𝑎𝑥

2

𝑟

−𝜏𝑟𝑥 = (𝑃𝐿−𝑃𝑂

2𝐿) 𝑟𝑚𝑎𝑥 [

𝑟

𝑟𝑚𝑎𝑥−

𝑟𝑚𝑎𝑥

𝑟] (7)

𝜏𝑟𝑥 = −𝜇𝑑𝑣𝑥

𝑑𝑟 (8)

In order to determine velocity profile equation 7 and 8 are solving with together:

𝜏𝑟𝑥 = −𝜇𝑑𝑣𝑥

𝑑𝑟= − (

𝑃𝐿 − 𝑃𝑂

2𝐿) 𝑟𝑚𝑎𝑥 [

𝑟

𝑟𝑚𝑎𝑥−

𝑟𝑚𝑎𝑥

𝑟]

𝑑𝑣𝑥

𝑑𝑟= − (


2𝜇𝐿) 𝑟𝑚𝑎𝑥 [

𝑟

𝑟𝑚𝑎𝑥−

𝑟𝑚𝑎𝑥

𝑟]

𝑣𝑥 = (𝑃𝐿 − 𝑃𝑂

4𝜇𝐿) 𝑟2 − (


2𝜇𝐿) 𝑟𝑚𝑎𝑥

2 ln 𝑟 + 𝐶2

𝑣𝑥 = (𝑃𝐿−𝑃𝑂


2 [(𝑟

(𝑟𝑚𝑎𝑥))

2

− 2 ln 𝑟] + 𝐶2 (9)

To calculate the second integral constant “c2” boundary conditions are used

Boundary conditions

r=r1 ν=0 (stationary pipe) & r=r2ν=0 (stationary pipe)

If we solve equation 9 according to the boundary conditions

(𝑃𝐿 − 𝑃𝑂


2 [(𝑟1

(𝑟𝑚𝑎𝑥))

2

− 2 ln 𝑟1] + 𝐶2 = (𝑃𝐿 − 𝑃𝑂


2 [(𝑟2

(𝑟𝑚𝑎𝑥))

2

− 2 ln 𝑟2] + 𝐶2

𝑟𝑚𝑎𝑥 = √1

ln(𝑟2

𝑟1)

𝑟22 − 𝑟1

2

2

Q4. Water having a density of 998 kg/m3 is flowing at the rate of 1.676 m/s in a 7.8 cm diameter horizontal pipe at a pressure

P1 of 68.6 kPa absolute. It then passes to a pipe having an inside diameter of 5.25 cm.

a) Calculate the new pressure P2 in the 5.25 cm pipe. Assume no friction losses. (Ans: P2=63.2kPa)

b) If the piping is vertical and the flow is upward, calculate the new pressure P2. The pressure tap for P2 is 0.457m above

the tap for P1. (Ans: P2=58.6 kPa)

Solution 4.

D1= 3.068 m and D2=2.067 m

From Table A.5-4

𝜌 = 998 𝑘𝑔 𝐻2𝑂/𝑚3, 𝜗1 = 1.676𝑚

𝑠, 𝑃1=68.9 kPa

A1= 4.769*10-3 m2 A2= 2.165*10-3 m2

a)

𝜗1𝐴1 = 𝜗2𝐴2

𝜗2 = 1.6764.769

2.165

𝜗2 = 3.7 𝑚/𝑠

𝑃1

𝜌+

𝜗12

2+ 𝑧1𝑔 =

𝑃2

𝜌+

𝜗22

2+ 𝑧2𝑔

𝑧1 = 𝑧2

𝑃1 − 𝑃2

998+

1.6762

2+ 0 =

3.72

2

𝑃1 − 𝑃2 = 5430 𝑃𝑎 = 5.43 𝑘𝑃𝑎

𝑷𝟐 = 𝟔𝟖. 𝟔 − 𝟓. 𝟒𝟑 = 𝟔𝟑. 𝟐 𝒌𝑷𝒂

b) Vertical pipe

𝑃1 − 𝑃2

998+

1.6762

2+ 0 =

3.72

2+ 0.457(9.980)

P2 = 58.6 kPa

Q5. Calculate the pressure drop (because of friction) in pascal for olive oil at 293 K flowing through the commercial pipe

having an inside diameter of 0.0525 m and a lenght of 76.2 m. The velocity of the fluid is 1.22 m/s.

(Ans: 90.65 kN/ m2)

Solution 5.

T = 293 K; d = 0.0525 m; ΔL = 76.2 m.

For olive oil:

𝜌 = 919 𝑘𝑔

𝑚3, 𝜇 = 84𝑥10−3

𝑘𝑔

𝑚. 𝑠

𝑁𝑅𝑒 =𝐷𝜗𝜌

𝜇=

0.0525(𝑚)1.22 (𝑚𝑠 ) 919 (

𝑘𝑔𝑚3)

84𝑥10−3= 700.7

Hence the flow is laminar.

𝑓 =16

𝑁𝑅𝑒=

16

700.7= 0.02283

∆𝑃𝑓 = 4𝑓𝜌∆𝐿

𝐷

𝜗2̅̅ ̅

2

∆𝑃𝑓 = 4(0.02283)911 (𝑘𝑔

𝑚3)

76.2 𝑚

0.0525 𝑚

(1.22 𝑚𝑠 )

2

2

∆𝑃𝑓 = 90650 𝑁

𝑚2= 90.65

𝑘𝑁

𝑚2

Q6. A pipeline laid cross country carries oil at the rate of 795 m3/day. The pressure of the oil is 1793 kPa gage leaving

pumping station 1. The pressure is 862 kPa gage at the inlet to the next pumping station, 2. The second station is 17.4 m

higher than the first station. Calculate the lost work (ƩF, friction loss) in J/kg mass oil. The oil density is 769 kg/m3. (Ans:

1040 J/ kg)

Solution 6.

𝑃1

𝜌+

�̅�12

2+ 𝑔𝑧1 − 𝑊𝑠 =

𝑃2

𝜌+

�̅�22

2+ 𝑔𝑧2 + ∑ 𝐹

�̅�1 = �̅�2

Ws = 0

z1 = 0 and z2 = 17.4 m

𝑃1

𝜌=

𝑃2

𝜌+ 𝑔𝑧2 + ∑ 𝐹

∑ 𝐹 =1793𝑥1000 𝑃𝑎

769 𝑘𝑔𝑚3

− 9.81 (𝑚

𝑠2) 17.4(𝑚) −

862𝑥1000 𝑃𝑎

769 𝑘𝑔𝑚3

⟹ ∑ 𝑭 = 𝟏𝟎𝟒𝟎 𝑱

𝒌𝒈

Q7. Air flows steadily from a tank, through a hose of diameter D = 0.03 m, and exits to the atmosphere from a nozzle of

diameter d = 0.01 m as shown in the figure. The pressure in the tank remains constant at 3 kPa (gage) and the atmospheric

conditions are standard temperature and pressure. (Assume air is incompressible. ρair = 1.26 kg/m3)

a) Determine the flow rate (Ans : 5.42*10-3 m3/s)

b) Determine the pressure in the hose. (Ans :2.96 kPa)

Solution 7.

a) If the flow is assumed steady, inviscid and incompressible. We can apply the Bernoulli equation aong the

streamline from (1) to (2) to (3) as

𝑃1

𝜌+

�̅�12

2+ 𝑔𝑧1 =

𝑃2

𝜌+

�̅�22

2+ 𝑔𝑧2 =

𝑃3

𝜌+

�̅�32

2+ 𝑔𝑧3 … (1)

For horizontal hose;

z1 = z2 = z3

�̅�1 = 0 (𝑙𝑎𝑟𝑔𝑒 𝑡𝑎𝑛𝑘) P3 = atmospheric pressure (P3 = 0 (gage))

From Eq. (1)

𝑃1

𝜌=

�̅�32

2⟹ �̅�𝟑 = √

𝟐𝑷𝟏

𝝆

and

𝑃1

𝜌=

𝑃2

𝜌+

�̅�22

2 ⟹ 𝑃2 = 𝑃1 −

𝜌�̅�22

2

The density of the air in the tank is obtained from the perfect gas low, using standard absolute pressure and

temperature, as

Thus, we find that

�̅�3 = √2 (3𝑥103 𝑁

𝑚2)

1.26 𝑘𝑔𝑚3

= 69 𝑚

𝑠

Or

𝑄 = 𝐴3�̅�3 =𝜋

4𝑑2�̅�3 =

𝜋

4(0.01 𝑚)2 (69

𝑚

𝑠)

𝑄 = 0.00542 𝑚3

𝑠

b) 𝑄 =𝜋

4𝑑2�̅�2 → 0.00542 =

𝜋

40.032�̅�2

�̅�2 = 7.67 𝑚/𝑠

𝑃2 = 𝑃1 −𝜌�̅�2

2

2

𝑷𝟐 = 𝟐. 𝟗𝟔 𝒌𝑷𝒂

Q8. Water stored in a large, well-insulated storage tank at 21⁰C and atmospheric pressure is being pumped at steady state

from this tank by a pump at the rate of 40 m3/h. The motor driving the pump supplies energy at the rate of 8.5 kW. The

water is used as a cooling medium and passes through a heat exchanger, where 255 kW of heat is added to the water. The

heated water then flows to a second large, vented tank, which is 25m above the first tank. Determine the final temperature

of the water delivered to the second tank. (Ans: T2 =26.63⁰C)

Solution 8.

𝐻2 − 𝐻1 +1

2𝛼(𝜗2

2 − 𝜗12) + 𝑔∆𝑧 = 𝑄 + 𝑊

𝜗1 = 𝜗2 = 0 and ∆𝑧 = 25 𝑚

From table 4, 𝜌𝑤𝑎𝑡𝑒𝑟 = 998𝑘𝑔

𝑚3 𝑎𝑡 21𝑜𝐶

𝑄 = 40𝑚3

ℎ=

1

90(

𝑚3

𝑠)

�̇� = 𝜌 𝑄 = 998 ∗1

90= 11.09 𝑘𝑔/ 𝑠

𝐻2 − 𝐻1 + 𝑔∆𝑧 = 𝑄 + 𝑊

∆𝐻 = 𝑄 + 𝑊 − 𝑔∆𝑧

∆𝐻 = 255𝑘𝐽

𝑠∗

1

11.09

𝑠

𝑘𝑔+ 8.5

𝑘𝐽

𝑠∗

1

11.09

𝑠

𝑘𝑔− 9.81

𝑚

𝑠2∗ 25

∆𝐻 = 23.55 𝑘𝐽/𝑘𝑔

From Table 8, H1= 88.14 kJ/ kg

∆𝐻 = 𝐻2 − 𝐻1 = 23.55𝑘𝑗

𝑘𝑔

𝐻2 = ∆𝐻 + 𝐻1 = 88.14 + 23.55

𝐻2 = 111.69 𝑘𝐽/𝑘𝑔

Temperature at point 2 is found by making interpolation using the data from Table 8.

Temperature (⁰C) H (kJ / kg)

25 104.89

27 113.25

T2 = 26.63 ⁰C

Q9. At 20 ⁰C water is pumped from one tank (1st) to another tank (2nd) which is 15 m higher than 1st tank. The volumetric

flow rate of the water is 5 m3/s. All pipes are made of st40 steel. There is a gate valve in the half open position at the

pump inlet. Calculate pump power. (Ans: Wp = 170.47 J/kg)

Solution 9.

𝑃1

𝜌+

𝜗12

2+ 𝑧1𝑔 + 𝑛𝑊𝑝 =

𝑃2

𝜌+

𝜗22

2+ 𝑧2𝑔 + ∑ 𝐹

𝜗1 = 𝜗2 = 0

𝑃1 = 𝑃2

∑ 𝐹 = (𝑧2 − 𝑧1)𝑔 − 𝑊𝑝

ρ = 998.2 kg/ m3 , μ= 1.005 *10-3 Pa.s

4 inc D=0.1023 m → S= 8.219*10-3 m2

2 inc D=0.0525 m → S= 2.165*10-3 m2

Velocity in pipe

�̅� 4𝑖𝑛𝑐 =𝑄

𝑆=

5. 10−3

8.219 10−3= 0.608 𝑚/𝑠

𝑅𝑒4𝑖𝑛𝑐 =𝐷𝜗𝜌

𝜇=

(0.1023) ∗ (0.608) ∗ (998.2)

(1.005 ∗ 10−3)= 6.2 104 (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡)

�̅� 2𝑖𝑛𝑐 =𝑄

𝑆=

5. 10−3

2.165 10−3= 2.309 𝑚/𝑠


𝜇=

(0.0525) ∗ (2.309) ∗ (998.2)

(1.005 ∗ 10−3)= 1.2 106 (𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡)

∑ 𝐹 = 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑒𝑥𝑖𝑡 + 𝑔𝑎𝑡𝑒 𝑣𝑎𝑙𝑣𝑒 + 2 𝑒𝑙𝑏𝑜𝑤 + 𝑐𝑜𝑛𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑓𝑟𝑜𝑚 4𝑖𝑛𝑐 𝑡𝑜 2 𝑖𝑛𝑐 𝑝𝑖𝑝𝑒

+ 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑘 + 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 4 𝑖𝑛𝑐 𝑝𝑖𝑝𝑒 + 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 2 𝑖𝑛𝑐 𝑝𝑖𝑝𝑒

1) Contraction at the pipe exit

𝐹𝑑 = 𝐾𝑑 (1 −𝑆𝑏

𝑆𝑎)

𝜗2

2𝛼

𝑆𝑏

𝑆𝑎= 0 (𝑛𝑒𝑔𝑙𝑒𝑐𝑡) and Kd=0.4 (turbulant flow)

𝐹𝑑 = 0.4 (1)(0.608)2

2= 0.0739 J/kg

2) Gate valve (Half open)

𝐹𝑏 = 𝐾𝑏 𝜗2

2𝛼

𝜗 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦𝑢 𝑡ℎ𝑎𝑡 𝑒𝑛𝑡𝑒𝑟𝑠 𝑡ℎ𝑒 𝑣𝑎𝑙𝑣𝑒 = �̅� 4𝑖𝑛𝑐

Kb=4.5

𝐹𝑏 = 4.5 (0.608)2

2= 𝟎. 𝟖𝟑𝟏 𝐽/𝑘𝑔

3) 2 Elbows (90⁰)

𝐹𝑏(4𝑖𝑛𝑐)= 𝐾𝑏

𝜗2

2𝛼

Kb=0.75

𝐹𝑏(4𝑖𝑛𝑐)= 0.75

(0.608)2

2= 0.138 𝐽/𝑘𝑔

𝐹𝑏(2𝑖𝑛𝑐)= 𝐾𝑏

𝜗2

2𝛼

Kb=0.75

𝐹𝑏(2𝑖𝑛𝑐)= 0.75

(2.309)2

2= 1.999 𝐽/𝑘𝑔

∑ 𝐹𝑏 = 0.138 + 1.999 = 𝟐. 𝟏𝟑𝟕 𝑱/𝒌𝒈

4) Contruction from 4inc pipe to 2inc pipe

𝐹𝑑 = 𝐾𝑑 (1 −𝑆2𝑖𝑛𝑐

𝑆4𝑖𝑛𝑐)

𝜗2𝑖𝑛𝑐2

2𝛼

Kd=0.4

𝑆2𝑖𝑛𝑐

𝑆4𝑖𝑛𝑐=

2.165 10−3

8.219 10−3= 0.263

𝐹𝑑 = 0.4 (1 − 0.263)(2.309)2

2= 𝟎. 𝟕𝟖𝟓 𝑱/𝑘𝑔

5) Expansion at the entrance of the tank

𝐹𝑔 = 𝐾𝑔 𝜗2𝑖𝑛𝑐

2

2𝛼

𝐾𝑔 = (1 −𝑆2𝑖𝑛𝑐

𝑆𝑡𝑎𝑛𝑘)2

𝑆2𝑖𝑛𝑐

𝑆𝑡𝑎𝑛𝑘= 0, 𝐾𝑔 = 1

𝐹𝑔 = 1 (2.309)2

2= 𝟐. 𝟔𝟔𝟓 𝐽/𝑘𝑔

6) 4 inc straight pipe

𝐹4𝑖𝑛𝑐 = 4𝑓𝐿

𝐷

𝜗2

2𝛼

L=50+5+5=60 m

D=0.1023 m

�̅� 4𝑖𝑛𝑐 = 0.608𝑚

𝑠

휀 = 4.6 10−5 휀

𝐷= 0.00045


𝜇= 6.2 104

𝑅𝑒 ↔ 𝑓

f=0.0051 from table

𝐹4𝑖𝑛𝑐 = 4 ∗ 0.0051 ∗60

0.1023∗

0.6082

2

𝐹4𝑖𝑛𝑐 = 𝟐. 𝟐𝟏𝟏𝐽

𝑘𝑔

7) friction loss in 2 inc straight pipe

𝐹2𝑖𝑛𝑐 = 4𝑓𝐿

𝐷

𝜗2

2𝛼

L=20 m

D=0.0525 m,

�̅� 2𝑖𝑛𝑐 = 2.309𝑚

𝑠

휀 = 4.6 10−5 휀

𝐷= 0.00087


𝜇= 1.2 106

𝑅𝑒 ↔ 𝑓

f=0.0036 from table

𝐹4𝑖𝑛𝑐 = 4 ∗ 0.0036 ∗20

0.0525∗

2.3092

2

𝐹4𝑖𝑛𝑐 = 𝟏𝟒. 𝟔𝟐𝟑𝐽

𝑘𝑔

∑ 𝐹 = 0.0739 + 𝟎. 𝟖𝟑𝟏 + 2.137 + 0.785 + 𝟐. 𝟔𝟔𝟓 + 𝟐. 𝟐𝟏𝟏 + 14.623 = 23.32 𝐽/𝑘𝑔

𝑊𝑝 = ∑ 𝐹 + (𝑧2 − 𝑧1)𝑔

𝑊𝑝 = 15 ∗ 9,81 + 23.32

𝑾𝒑 = 𝟏𝟕𝟎. 𝟒𝟕 𝑱/𝒌𝒈

𝑃 = �̇� 𝑊𝑝

�̇� = 𝜌 𝜗 𝑆 = 998 ∗ 0.608 ∗ (8.219 10−3) = 4.991 𝑘𝑔/𝑠

𝑷 = 𝟎. 𝟗𝟓 𝒌𝑾

che212 fluid mechanics 2017-18 homework 3...

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