thermal science ch 04

31

FOUR

The First Law of Thermodynamics4.1FE B

4.2FE A

4.3FE C Because the container is insulated we assume the heat transfer is zero. For a rigid container:Q electric vW mc T

2 2( 400 0 60) 10 10.085( 20). 29 CT T

Table B.2 was used to find vc .

4.4FE A For a rigid container the boundary work is zero:

q w 2 1 2961 2553.6 407 kJ/kgu u u

where we used 2 1v v and interpolated for 2u in Table C.3 as follows:

2 12

2

0.46252961 kJ/kg

400 Cv v

uT

4.5FE C For this adiabatic process we use

W = m(u2 uand( 1) /

0.285722 1

1293 6 488.9 K

k kp

T Tp

2 1( ) 2 0.717 (489 293) 281 kJvW mc T T

4.6FE C This is an adiabatic process so that

W = m(u2 u 2 1( )vW mc T T Then

( 10 100 10 50) (5 60) 2 717 . 313.8 CT T

4.7FE D lnV

Q W mRT 2

V 11 0.287 373 ln 0.5 74.2 kJ .

The minus sign indicates that heat is rejected.

4.8FE D 5 0.717 100 358.5 kJvW mc T

4.9FE A p2 = p1 = 400 kPa. Use Table C.3 for this superheated steam:

V 31 1 0.7726 0.7726 m . V 2 V 3 3

1 22 0.05 0.6726 m and 0.6726 m / kg.v

Interpolate at 400 kPa and 0.6726 m3/kg:

20.7726 0.6726

400 (400 300) 315 C0.7726 0.6548

T

4.10FE D Use Table C.3 at 315C and interpolate to find h2:

1 (3098 3273) 175 kJQ m h

32

4.11FE A Use Tables C.3 and C.2 with W = 0:

3 2( ) 1 (1245 2829)Q m u u = 1584 kJ

We interpolated at 315C in Table C.3 to find u2 = 2829 kJ/kg and used v2 = v3 =0.672 m3/kg which is in the quality region. Then using Table C.2,x3 = 0.3963 and u3 = 1245 kJ/kg.

4.12FE D If T = const,

1

2

100ln 10 0.287 333 ln 1987 kJ

800p

Q W mRTp

4.13FE C For p = const,

2 1 3273 604.7 2668 kJ/kgq h h

4.14FE C For p = const,

2 1 2 2 2( ). 170 1 ( 3108). 3278 and 402 CQ m h h h h T

We interpolated at 320C in Table C.3 to find h1 and then to find T2.

4.15FE D For p = const,

2 1( ) 1 400 (0.7749 0.6784) 38.6 kJW mp v v

We interpolated in Table C.3 to find v1 and v2 and the two temperatures.

4.16FE D In Table C.3 we observe that u =2949 kJ/kg is between p = 1.6 MPa and 1.8 MPa butcloser to 1.6 MPa so the closest answer is 1.7 MPa.

4.17FE C Interpolate in Table C.3 and find h = 3459 kJ/kg.

4.18FE C Use a central difference for accuracy:3213.6 2960.7

2.53 kJ/kg 蚓400 300p

hc

T

4.19FE C We use q because the mass is not known:

2 1( ) 2.254 300 676.2 kJ/kgpq c T T

4.20FE B The copper gains heat and the water losses an equal amount of heat:

, 2 , 2 2 2( 0) (30 ). 20 0.38 10 4.18 (30 ).c p c w p wm c T m c T T T 2 25.4 CT

4.21FE B First the ice melts and then the ice water heats up:

, ,( ) .i i p w i w p w wm h c T m c T

2 210 [333 4.18( 0)] 60 4.18 (20 )T T . 2 = 5.85蚓T

4.22FE C For process 1: Q = W =100 = a.

For process 2: b = 40 + 60 = 100.

or 100 100 40 100 60 . 80Q W c c

33

4.23FE B For the T = const process:2

1-2 11

0.8ln 0.287 278.7 ln 166.4 kJ/kg0.1

vw RT

v

We used pv = RT to find T1 = (800)(0.1)/0.287 = 278.7C.

4.24FE C For the adiabatic process:

3-1 1 3( ) 0.717(278.7 121.3) 113 kJ/kgvw c T T 1 0.4

13 1

3

0.1where 278.7 121.3 K0.8

kvT Tv

.

4.25FE D The net heat transfer is equal to the net work. The work for the 2-3 process is zero.Therefore, using the results from the above two problems, the net q is

net 1-2 2-3q w w 3-1w = 166.4 113 = 53 kJ/kg

4.26FE A

4.27FE D1/

0.285722 1

1293 8 530.7 K or 258 C

k kp

T Tp

.

4.28FE B For the Q = 0 process:

2 1( ) 2 0.717 (622 373) 358 kJvW mc T T 1/

0.285722 1

1where 373 6 622.3 K

k kp

T Tp

.

4.29FE A100 400

or 0.3 15 60 (10 20 3 1.2) 1.003600pQ mc T T

14.3 CT

We assumed a p = const process since there are openings where air is allowed to escapeor enter.

4.30FE D

4.31FE D This term includes the flow work rate term due to the pressure force moving due to thevelocity.

4.32FE B

4.33FE C2 2 2 2

2 1 20 2000 . 1.00 . 19.8 C

2 2 1000pV V

c T T T

.

The factor 1000 is needed to convert J to kJ since cp includes the kJ unit.

4.34FE D 2 1 23500.9 kJ/kg. 0.8 MPa.h h p Interpolate in Table C.3 at 0.8 MPa and findT2 = 508C

34

4.35FE D1/1

0.285722 1

1293 8 530.7 K or 258 C

kp

T Tp

.

4.36FE B 211 1 1 1 1 1 1

1

100. 2 0.1 53.5 m/s

0.287 293p

m A V A V V VRT

4.37FE D The heat that leaves the steam enters the cooling water:

s s w p wm h m c T

Use hfg from Table C.2:

10 2373 400 4.18 . 14.2 CT T .

4.38FE C Assume p = const in the tube:

100 1.00 (25 20) 500 kJ/minpQ mc T

4.39FE C For liquid:

6000 1004 23.6 kW

1000

Pp

W m

4.40FE B The enthalpy of the fluid when it enters the tank is assumed equal to the enthalpyof the fluid in the pipe on the upstream side of a valve connecting the tank and pipe.

4.41FE C 1 1 1 1 1 2 2 2 2 2and c cQ h A T t Q h A T t . But, 2 1 2 1 2 1 2 1, 2 , , and . Q Q h h T T A A

So,

2 2 1 1 1 12

1 1 2 2 2

1. 20 s

2 2

t Q h A T tt

t Q h A T

4.42FE C 2 1 2 1 2 1 2 1or or4

A A B B A A B BA B A B

A B A B

T T T T T T T Tq q k k

L L L L

If 1 1 2 2/ 2 and 2 and 2 B A B A B AL L T T T T then

2 1 2 12( )2

A A A A

A A

T T T TL L

and 1 = 1 so C is the answer.

35

4.1 net netQ W100 9.81 3. 3.398 kgm m

4.2 2 2net net net

1 1. 1500 30 675000 J

2 2Q W Q mV

4.3 atm0. 0F pA mg p A

2 20.05 10 9.81 100000 0.05 . 112 490 Pap p

2. 300 (112 490 0.05 ) 0.2 123.3 JQ W U U

4.4 2 778 [ 600 (2 1/ 24)] 381 ft-lbf or 0.49 BtuU Q W

4.5 a) 2 220 5 15 kJ 7. 22 kJE Q W E E

b) 2 26 3 3 kJ. 8 6. 14 kJQ E W E E

c) 40 (30 15) 25 kJ. 30 15 15 kJW Q E E

d) 1 110 20 30 kJ. 20 10 . 10 kJW Q E E E

e) 8 6 10 4 kJ. 8 6 14 kJQ E W E

4.6 1-2 1-2 200 0 200 kJa W Q U

cycle( ) 0. 0 400 1200 0. 800 kJU c c

2-3 2-3 800 800 0 kJb W Q U

3-4 3-4 400 600 1000 kJd Q U W

4-1 4-1 0 ( 1200) 1200 kJe W Q U or we could use net netQ W

4.7 1-2 1-2 100 100 200 kJa Q E W

cycle( ) 0. 100 200 0. 100 kJE c c

2-3 2-3 100 50 50 kJb Q E W

3-1 3-1 100 ( 200) 300 kJd W Q E or we could use net netQ W

36

4.8 First, find the initial pressure in the cylinder:

atm 2

60 9.81100000 119000

0.1W

p pA

Pa

Now, apply the 1st law adding the work of the pressure force and the work tocompress the spring:

21. ( )

2Q W U Q pA h Kx U

2 21200 (119000 0.1 0.05 50000 0.05 ) 49 J

2U

4.9 1-2 6 5 (20 60) 36000 J.W VI t

1-2Q 1-2 2-1 2-1W Q W cycle 2-1 1-2( ) 0. 36 kJU Q W

4.10 400000 12 3 (6 60 60) 377 600 J or 377.6 kJQ U W

4.11 300000 12 10 (30 60) 84000 J or 84 kJ Q U W

4.12 8000 110 15 (2 60 60) /1055 3260 BtuQ U W where the factor 1055 converts J to Btu’s.

4.13 The only transfer of energy across the boundary of the system is via theelectrical wires leading to the refrigerator. The 1st law is

Q . ( ) 2 0.746 (30 60) 2686 kJW U U W VI t

4.14 Q WVm u 2 1

0.3( ) (3655.3 2621.9) 1505 kJ0.206

u uv

4.15 a) Q WVm u 2 1( )u uv

20.2

1000 { [669.9 0.8(2567.4 669.9)]}0.0011 0.8(.3157 .0011)

u

2 3452 kJ/kgu and 32 0.2528 m /kgv . We must find where in Table C.3

this state exists. After checking the table we interpolate for the following:

2 2 2 22

2 2 2 2

1.5 MPa, 0.2528, 3216 : 556 C686 C

2.0 MPa, 0.2528, 3706 : 826 Cp v u T

Tp v u T

b) TK solution:Rule Sheet

;This is a closed, constant-volume system so no work is done and the first law isQ = m * (u2 -u1) ; First lawV1 = m * v1 ; Definition of specific volumev2 = v1 ; For a constant-volume system; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George S.Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW

37

Variable Sheet

Input Name Output Unit Comment` Thermal Sciences, Potter & Scott

*STEAM8SI.TKW Steam, 1-8 States, SI UnitsP4-15.tkw Problem 4.15

T1 159 C Temperature600 p1 kPa Pressure

h1 2340 kJ/kg Enthalpys1 5.79 kJ/(kg*K) Entropyv1 0.253 m^3/kg Specific Volume

0.8 x1 Qualityphase1 'SAT Phaseu1 2190 kJ/kg Internal energy

T2 688 C Temperature (starting guess needed)p2 1750 kPa Pressureh2 3890 kJ/kg Enthalpys2 7.99 kJ/(kg*K) Entropy

0.253 v2 m^3/kg Specific Volume Transfer value to input)x2 'mngless Qualityphase2 'SH Phaseu2 3450 kJ/kg Internal energy

*THUNITS.TKW Units for thermo1000 Q kJ Heat transfer

m 0.791 kg Mass of steam0.2 V1 m^3 Volume

4.16 a) Let’s first find the mass. We use Table C.1E:

31 0.016 0.5(203 0.016) 101.5 ft /lbm. Vv m

2 0.0197 lbm101.5v

The 1st law is Q W 2 1( )m u u or 2 18 0.0197( ) :u u

1 87.99 0.5(961.9) 568.9 Btu/lbmu . 2 975 Btu/lbmu

This is less than gu so that state 2 is in the wet region with 32 101.5 ft /lbm.v

This requires trial-and error:

At 2 140 F :T 2 2

2 2

101.5 0.016 (122.9 0.016). 0.826

975 108 948.2 . 0.914

x x

x x

38

At 2 150 F :T

2 2

96.99. slightly superheat975 118 941.3 . 0.912

gvx x

State 2 lies between 140 F and 150 F . Since the quality is insensitive to theinternal energy, we find 2T such that 3

2 101.5 ft /lbm:v

2101.5 96.99

150 10 148 F122.88 96.99

T

A temperature slightly less than this provides the answer: 2 147 FT .b) TK solution:

Rule Sheet;This is a closed, constant-volume system so no work is done and the first law isQ = m * (u2 -u1) ; First lawV1 = m * v1 ; Definition of specific volumev2 = v1 ; For a constant-volume system

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW

Variable SheetInput Name Output Unit Comment

` Thermal Sciences, Potter & Scott*STEAM8E.TKW Steam, 1-8 States, English unitsP4-16.tkw Problem 4.16

120 T1 F Temperaturep1 1.69 psi Pressureh1 601 B/lbm Enthalpys1 1.05 B/(lbm*R) Entropyv1 101 ft^3/lbm Specific Volume

0.5 x1 Qualityphase1 'SAT Phase

T2 144 F Temperature (starting guess needed)p2 3.21 psi Pressureh2 1030 B/lbm Enthalpys2 1.73 B/(lbm*R) Entropy

101 v2 ft^3/lbm Specific Volumex2 0.913 Qualityphase2 'SAT Phase

*thunits.tkw Units for thermodynamics8 Q B Heat transfer

m 0.0197 lbm Mass of steamu2 975 B/lbm Internal energyu1 569 B/lbm Internal energy

2 V1 ft^3 Initial volume

39

4.17 1 0.0179 0.95(3.73 0.0179) 3.544. Vv m 100 / 1728 0.01633 lbm

3.544v

2 1120 144 100

( ) 0.01633{1371.5 [312.3 0.95(796)]} (7.208 )778 1728

Q m u u W

6.28 Btu

4.18 First, find the mass:Vm 1

1

0.004 0.0302 kg0.1325v

. The 1st law is 2 1( )Q W m u u

which takes the form

2 2 2 240 1500 0.03019( 0.1325) 0.0302( 2598) or 124.4 45.3 0.0302v u v u

The above equation has two variables but the steam tables represent the 2nd equation.This requires trial-and-error (let '

2u be the 2u from the above equation):'

2 2 2 2 2'

2 2 2 2 2 2'

2 2 2 2 2

At 600, 1.5 : .2668, 3294. 3720At 780, 1.5 : .3230, 3622. 3636 785 CAt 800, 1.5 : .3292, 3627. 3720

T p v u uT p v u u TT p v u u

4.19 a) 2 1100 /1728

( ) {1531.5 [312.7 0.95(878.5)]} 6.27 Btu3.544

Q m h h


;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressurebecomes; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) orQ = m * (h2 - h1)v1 = V1/m ; Definition of v1

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8E.TKW

Variable Sheet


*STEAM8E.TKW Steam, 1-8 States, English unitsP4-19.tkw Problem 4.19

T1 341 F Temperature120 p1 psi Pressure

h1 1150 B/lbm Enthalpys1 1.53 B/(lbm*R) Entropyv1 3.54 ft^3/lbm Specific Volume


1000 T2 F Temperature120 p2 psi Pressure

40

h2 1530 B/lbm Enthalpys2 1.9 B/(lbm*R) Entropyv2 7.2 ft^3/lbm Specific Volumex2 'mngless Qualityphase2 'SH Phase

*THUNITS.TKW Units for thermoQ 6.28 B Heat transfer

100 V1 in^3 Initial volumem 0.0163 lbm Mass of steam

4.20 a) 2 20.004

. 40 ( 2797). 4120 kJ/kg0.1325

Q m h h h .

At 2 1.5 MPap we interpolate 2 787 C.T


;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or Q = m * (h2 - h1)v1 = V1/m ; Definition of v1

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW

Variable Sheet



200 T1 C Temperature1500 p1 kPa Pressure

h1 2800 kJ/kg Enthalpys1 6.45 kJ/(kg*K) Entropyv1 0.132 m^3/kg Specific Volumex1 'mngless Qualityphase1 'SH Phase

4 V1 L Initial volumem 0.0302 kg Mass of steamT2 785 C Temperature (starting guess is needed)

1500 p2 kPa Pressureh2 4120 kJ/kg Enthalpys2 8.28 kJ/(kg*K) Entropyv2 0.324 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH Phase

*THUNITS.TKW Units for thermo40 Q kJ Heat transfer

41

4.21 From Table D.1 we find 1 62 0.8(190.3) 214.2 kJ/kg.h Then

2 1 2 2. 80 214.2. 294.2 kJ/kgq h h h h We interpolate at 2 0.4 MPap in Table D.3:

2294.2 291.8

(10) 50 52.5 C301.5 291.8

T

4.22 a) 2 1( ) 2(3278 209) 6138 kJQ m h h

b) 2 1( ) 2(4044 3278) 1531 kJQ m h h

c) TK solution:Rule Sheet

;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressure becomesQ = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) for each process orQ12 = m * (h2 - h1) ; for the process 1-2Q23 = m * (h3 - h2) ; for the process 2-3

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and George S.Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW

Variable Sheet




h1 209 kJ/kg Enthalpys1 0.704 kJ/(kg*K) Entropyv1 0.00101 m^3/kg Specific Volumex1 'mngless Qualityphase1 'CL Phase





42

*THUNITS.TKW Units for thermo2 m kg Mass of steam

Q12 6140 kJ Heat transfer, process 1-2Q23 1530 kJ Heat transfer, process 2-3

4.23 1 11008 0.8(1796) 2445. 0.0012 0.8(0.06668 0.0012) 0.5359h v Vm 1

1

1.2 22.39 kg0.05359v

a) 2 1( )Q m h h

2 23000 22.39( 2445). 2579h h

22

2

2579234 C

3 MPah

Tp

b) 2 1( )Q m h h . 2 230000 22.39( 2445). 3785h h

22

2

3785645 C

3 MPah

Tp

4.24 31 0.0012 0.9(0.1274 0.0012) 0.1148 m /kgv and

32 3 0.1148 0.3444 m /kgv

1 850.6 0.9(2595.3 850.6) 2421 kJ/kgu

Interpolate in Table C.3:2

22

0.3444 0.3444 0.26080.8 (0.2) 0.617 MPa200 C 0.3520 0.2608

vp

T

Also, 22

2

200 C 0.617 0.62638.9 (2638.9 2630.6) 2638 kJ/kg

0.8 0.60.617T

up

To find the heat transfer we must know the work. It is found by using graph paperand plotting p vs. v and graphically integrating. The work is twice the area sincem = 2 kg. Doing this, we find

2 12 228 456 kJ. ( ) 2(2638 2421) 456 890 kJW Q m u u W

4.25 a)3489 3076.5

2.06 kJ/kg K200p

hc

T

b)3488.1 3074.3

2.07 kJ/kg K200p

hc

T

c)2821.4 2151

13.4 kJ/kg K50p

hc

T

T

v

1 (a)

(b)

43

d) TK solution:Rule Sheet

; cp = the slope of a constant pressure line on an hT diagram or approximately.;cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 398C and T1 = 402C; at 10 kPa. Likewise, we define states 3 and 4 at these temperatures on a 100-kpa line and;states 5 and 6 at these temperatures ona 30,000-kPa line.. Thencp1 = (h2 - h1) / (T2-T1) ; at 10 kPa, 400 kPacp2 = (h4 - h3) / (T4 - T3) ; at 100 kPa, 400 kPacp3 = (h6 - h5 ) / (T6 - T5) ; at 30000 kPa, 400 kPa

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and ;GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STM8SI.tkw

Variable Sheet




h1 3280 kJ/kg Enthalpys1 9.6 kJ/(kg*K) Entropyv1 31 m^3/kg Specific Volumex1 'mless Qualityphase1 'SH Phase


h2 3280 kJ/kg Enthalpys2 9.61 kJ/(kg*K) Entropyv2 31.2 m^3/kg Specific Volumex2 'mless Qualityphase2 'SH Phase


h3 3270 kJ/kg Enthalpys3 8.54 kJ/(kg*K) Entropyv3 3.09 m^3/kg Specific Volumex3 'mnless Qualityphase3 'SH Phase



398 T5 C Temperature

44

30000 p5 kPa Pressureh5 2120 kJ/kg Enthalpys5 4.43 kJ/(kg*K) Entropy

v5 0.00269 m^3/kg Specific Volume

x5 'mless Qualityphase5 'SH Phase



*THUNITS.TKW Units for thermocp1 2.07 kJ/(kg*K) Approx cp at 10 kPa, 400 Ccp2 2.07 kJ/(kg*K) Approx cp at 100 kPa, 400 Ccp3 13.9 kJ/(kg*K) Approx cp at 30000 kPa, 400 C

4.26 a)1373.9 1219.3

0.386 Btu/lbm- R400v

uc

T

b)1373.7 1218.6

0.388 Btu/lbm- R400v

uc

T

c)1156.5 960.7

1.96 Btu/lbm- R100v

uc

T

4.27 a)117.6 126.4

0.22 Btu/lbm- R120 80p

hc

T

115.5 107.60.20 Btu/lbm- R

120 80vu

cT


; cp = the slope of a constant-pressure line on an hT diagram or approximately:; cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 98 F and T2 = 102 F; at 30 psia. Then cp = (h2 - h1) / (T2-T1) ; at 30 psia, 100 F; cv = the slope of a constant-volume line on a uT diagram or approximately;cv = (u4 - u3) / (T4 - T3) where states 3 and 4 are close together, say T3 = 98 F and T4 = 102 F atthe specific volume of the state of 30 psia, 100 F: 1.89 ft^3/lbm. Thencv = (u4 - u3) / (T4 - T3);*R1348e.tkw, R134a, 1-8 States, English units; R134a tables based on 'Thermodynamic Properties of HFC-134a' DuPont Technical Information,which is based upon the Modified Benedict-Webb-Rubin equation of state. R1348E.tkw

45

Variable Sheet


*R1348e.tkw, R134a, 1-8 States, EnglishP4-27.tkw Problem 4.27





98 T3 F Temperaturep3 29.9 psi Pressureh3 122 B/lbm Enthalpys3 0.257 B/(lbm*R) Entropy

1.89 v3 ft^3/lbm Specific Volumex3 'mngless Qualityphase3 'SH Phaseu3 111.98 B/lbm

102 T4 F Temperaturep4 30.1 psi Pressureh4 123 B/lbm Enthalpys4 0.258 B/(lbm*R) Entropy

1.89 v4 ft^3/lbm Specific Volumex4 'mngless Qualityphase4 'SH Phaseu4 112.74 B/lbm

*THUNITS.TKW Units for thermocp 0.213 B/(lbm*R) Constant-pressure specific heatcv 0.189 B/(lbm*R) Constant-volume specific heat

4.28 a) 2 1( ) 1.006(700 300) 402.4 kJ/kgph c T T

b)700

3 6 2

300(0.946 0.213 10 0.031 10 ) 417.7 kJ/kgh T T dT

c) Using the gas table F.1 we find 2 1 713.3 300.2 413.1 kJ/kgh h h

46

d) The error in (a) is 2.6% and that in (b) is 1.1% assuming that of (c) is thebest answer. All three methods are acceptable for the temperature range ofthis problem.

4.29 a) 2 1( ) 2 .006 (600 400) 402 kJpH mc T T

b)3 6

2 2 3 3.213 10 .031 102 .946 200 (600 400 ) (600 400 ) 418 kJ

2 3H

c) 2 1( ) 2(607 401) 412 kJH m h h

4.30 water 2 4.18(60 10) 418 kJpH mc T

ice 2 1.86[ 10 ( 60)] 186 kJpH mc T

where we averaged the ice( )pc between 10 C and 60 C in Table B.4.

4.31 Assume that all of the ice melts. The ice warms up to 0 C , melts at 0 C , thenwarms up to the final temperature 2 .T First, let’s find the mass of theice:

iVm

616 8 10 0.1174 kg0.00109v

where we found v in Table C.5. Energy is conserved so that the heat gained by theice equals the heat lost by the water:

( ) ( ) ( ) ( ) ( ) ( )i p i i if p w iw w p w wm c T h c T m c T

32 20.1174 2.1 10 330 4.18( 0) (1000 10 ) 4.18(20 )T T

2 9.07 CT

4.32 a) Assume the ice melts and then the water is heated:

2 2

( ) ( ) ( ) ( )

2000 2.3 1.89 60 330 4.18 . 102 C

p i i if p w wQ m c T mh m c T

T T

b) Assume the ice melts and then the water is heated:

2 2

( ) ( )

2000 2.3 330 4.18 . 129.1 C

if p w wQ mh m c T

T T

This is in the superheat region so the above temperature is too high. The watermust first vaporize at 2 120.2 CT , so that’s the final temperature.

47

4.33 First, the temperature of the ice is raised, then the ice is melted, then thetemperature is raised to the boiling point, then the water is vaporized, then it issuperheated. This requires the following:

2( ) ( ) ( )p i p w fg gQ mc T m h mc T mh m h h

10 0.486 32 143 1.0(250 32) 945 (1334 1164)

=14,900 Btu

where for ice 0.47 0.001pc T and we used an average

temperature of 16 C .

4.34 The heat gained by the ice cubes is lost by the cola. The mass of the ice is64 2 2 5 10 917 0.0734 kgcm

a) Assume all of the ice melts:( ) ( ) ( ) ( )i p i i i p w c p c cm c T m h m c T m c T

32 2 20.0734(2.0 20 330 4.18 ) 2 10 4.18 1000(20 ). 16.1 CT T T

b) Assume that all of the ice does not melt:( ) ( ) ( )i p i c p c cm c T x h m c T

30.0734 2.0 20 330 0.25 10 1000 4.18 20. 0.0561 kg0.0561

100 76.4%0.0734

x x

4.35 The heat lost by the copper is gained by the water:

( ) ( ) ( ) ( )c p c c w p w wm c T m c T

Use an average value of pc for copper from Table B.4:3

2 2 25 0.39 (300 ) 20 10 1000 4.18 ( 0). 6.84 CT T T

4.36 The heat lost by the copper is gained by the water:

( ) ( ) ( ) ( )c p c c w p w wm c T m c T

2 1( ) 4 0.717(400 100) 860 kJvu c T T

4.37 The heat lost by the iron is gained by the copper:

( ) ( ) ( ) ( )c p c c i p i im c T m c T

2 2 250 0.39 100 0.45(200 ). 139.5 CT T T

4.38 2 1( ) 4 1.00(400 100) 1200 kJph c T T

2 1( ) 4 0.717(400 100) 860 kJvu c T T

T

v1

2

48

4.39 a) 0, 60 kJ,U H W Q 1 100 C,T V 32

0.4 0.287 3730.856 m ,

50

0.85660 0.4 0.287 373ln

V

1. V 3

1 1 20.211 m ,V

p p 2

V 1

0.85650 203 kPa

0.211

b) If V 1 1const, 0. 0.4 1.00(300 ) 60. 150 CW H T T

0.4 0.717(300 150) 43 kJ, 43 kJ,U Q U

11 2

2

423200 147.6 kPa,

573T

p p VT

2 V 31

0.4 0.287 5730.329 m

200

c) 2 2 2. 100 kJ 0.4 1.00( 200). 450 C, 500 kPa,pQ mc T H T T p

V 32

0.4 0.287 7230.166 m ,

500V

1 V 31

2

2

4730.166 0.109 m

723TT

(W p V 2 V 1) 500(0.166 0.109) 28.5 kJ Also, 0.4 0.717(450 200) 71.7 kJU

d) 2 1V

T T 1

V

1 0.4

2

0.48323 605 K or 332 C,

0.1

k

0.4 0.717(332 50) 80.9 kJ, 80.9 kJ,U W U

10.4 0.287 323

0.4 1.00(332 50) 113 kJ, 77.3 kPa,0.48

H p

20.4 0.287 605

695 kPa0.1

p

4.40 a)1

3.42

1( ). 60 0.4(2512 )v

Q m u m pdv u pdv . Integrate graphically selecting

various initial states. We find 49.4 kJ, 10.2 kJ, 11.8 kJ,W U H

1 1100 C, 100 kPa,T p V 32 1.37 m , V 3

1 0.662 m

b) To locate state 1 using the steam tables is very difficult. So, let’s assume that the steam acts almost as an ideal gas and use the constants from Table B.2:If V 1 1const, 0. 0.4 1.872(300 ) 60. 220 CW H T T .

0.4 1.411(300 220) 45 kJ, 45 kJ,U Q U

11 2

2

493200 172 kPa,

573T

p p VT

2 V 31

0.4 0.462 5730.529 m

200

c) 2 1 2 2 2( ). 100 kJ 0.4( 2855). 3105, 0.5 MPa.Q H m h h H h h p

2Interpolate and find 320 C. ThenT V 32 0.4 0.542 0.217 m and

V 31 0.4 0.4249 0.170 m and 500(0.217 0.170) 23.5 kJW

Also, 0.4(2835 2643) 76.8 kJU

49

4.41 Q U 1

2

20ln 2 (53.5 / 778) 560ln 176.7 Btu

200p

W mRTp

4.42p V

m 22 1

1

200 2 8000.596 kg. 323 1292 K

2.077 323 200p

T TRT p

Q W 0.596 3.116(1292 323) 1800 kJvmc T

4.43 2 1( )Q m h h

a) 22 2 2

2

18321000 2.0( 1332). 1832 kJ/kg. 1551 F

60 psiah

h h Tp

b) 22 2 2

2

19391000 2.0( 1439). 1939 kJ/kg. 1741 F

60 psiah

h h Tp


;*P4-43.tkw Problem 4.43;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or, for part (a),Q = m * (h2 - h1);Similarly, for part (b):Q = m * (h4 - h3)

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;andGeorge S. Kell, Hemisphere Publishing Corp., 1984. ; STM8E.tkw


` Thermal Sciences, Potter & Scott*Stm8e.tkw Steam, 1-8 States, EnglishP4-43.tkw Problem 4.43


h1 1330 B/lbm Enthalpys1 1.82 B/(lbm*R) Entropyv1 10.4 ft^3/lbm Specific Volumeu1 1220 B/lbm Internal Energyx1 'mngless Qualityphase1 'SH PhaseT2 F Final temperature, part (a)

60 p2 psi Pressureh2 1830 B/lbm Enthalpys2 B/(lbm*R) Entropyv2 ft^3/lbm Specific Volumeu2 B/lbm Internal Energy

50

x2 Qualityphase2 Phase


h3 1440 B/lbm Enthalpys3 1.91 B/(lbm*R) Entropyv3 12.6 ft^3/lbm Specific Volumeu3 1300 B/lbm Internal Energyx3 'mngless Qualityphase3 'SH PhaseT4 1740 F Final temperature, part (b)

60 p4 psi Pressureh4 1940 B/lbm Enthalpys4 2.2 B/(lbm*R) Entropyv4 21.7 ft^3/lbm Specific Volumeu4 1700 B/lbm Internal Energyx4 'mngless Qualityphase4 'SH Phase

*THUNITS.TKW Units for thermo1000 Q B Heat transfer2 m lbm Mass of steam

4.44 2 1( )pQ mc T T , 1p Vm 1

1

400 0.21.021 kg

0.287 273RT

a) 2 250 1.021 1.0( 0). 49 CT T

b) 2 250 1.021 1.0( 200). 249 CT T

4.45 Q W 12 1( ).v

p Vmc T T m

61

1

800 8000 100.05978 kg

0.287 373RT

a) 0.05978 0.717(93.2 373) 11.99 kJQ where 22 1

1

200373 93.2 K

800p

T Tp

b) 0.05978 0.717(1399 373) 43.98 kJQ where 22 1

1

3000373 1399 K

800p

T Tp

4.46 We assume a quasiequilibrium process with q = 0 so that1/ 0.4 /1.4

22 1

1

10000373 1390 K

100

k kp

T Tp

q 2 1. ( ) 0.717(1390 373) 729 kJ/kgvw u w c T T

51

4.47 For the polytropic process1/ 0.2 /1.2

22 1

1

100373 276.7 K

600

n np

T Tp

2 2 1 1 2 1( ) 0.287(276.7 373) 138.2 kJ/kg

1 1 1 1.2p v p v R T Tw

n n

0.745(276.7 373) 138.2 66.5 kJ/kgvq c T w

4.48 2 1100 (3 5 2.4)

( ) 1.00(27 10) 753 kJ0.287 283pQ mc T T

4.49 Q paddle 2 1. ( )vW m u pA h W mc T T

The pressure is found from a force balance on the piston:

atm 2

17514.7 18.18 psia

4W

p pA

The mass is found from the ideal-gas law:

1p Vm

21

1

(18.18 144) 4 10 /17280.0255 lbm

53.3 560RT

The temperature at state 2 is

22

p VT

22 (18.18 144) 4 15 /1728

840 R0.0255 53.5mR

Finally,2

paddle 18.18 4 5/12 0.0255 0.171 778 (840 560) 1331 ft-lbfW

4.50 a) 31 .0011 .5(.4625 .0011) 0.2318 m /kgv

1 604.3 .5(2553.6 604.3) 1579 kJ/kgu

Q W 2 1 2 20.15

( ). 800 ( 1579). 2815 kJ/kg0.2318

m u u u u

Now, search Table C.3 for the location of state 2. It takes a couple interpolations:

22 2

2

2815314 C and 1.14 MPa

0.2318u

T pv

We used:2

2 22

22 2

2

1.0248 C and 2707 kJ/kg

0.2318

1.5489 C and 3102 kJ/kg

0.2318

pT u

v

pT u

v

p

v

1

(a)

(b)

52

b) Use some values from part (a):

Q W 2 1 2 20.15

( ). 200 ( 1579). 1888 kJ/kg0.2318

m u u u u

Try several guesses for 2T . We tried 2 2150 C and 160 CT T . Finally

22 2

2

1888154 C and 0.533 MPa

0.2318u

T pv

4.51 a)60 144 3

0.9925 lbm.53.3 490

m Q W

2 1 2( ) 0.9925( 83.5)m u u u

2 2285 Btu/lbm and 1595 R or 1135 Fu T

b)600 144 3

3.80 lbm.53.3 1280

m Q W

2 1 2( ) 3.80( 244)m u u u

2 2297 Btu/lbm and 1655 R or 1195 Fu T

4.52 a) 5 1.00(313 20) 1465 kJpQ mc T

where we used 2 1V

T T 2

V 1293 2 586 K or 313 C .

b) Q W 5 0.717(313 20) 1050 kJvmc T

where we used 22 1

1

293 2 586 Kp

T Tp

or 313 C .

c) lnV

Q W mRT 2

V1

1 2

1ln 5 0.287 293ln 291 kJ

2p

mRTp

d) 5 1.00(586 293) 1465 kJpQ mc T

4.53 10.5

1.35 kg, 604.7 .8 2133.8 2312 kJ/kg.00108 .8(0.4625 .00108)

m h

a) 2 1( ) 1.35(3485 2312) 1584 kJQ m h h

b) 2 1( ) 1.35(3870 2312) 2104 kJQ m h h where 2h is found by interpolation.


This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) at constant pressure becomesQ = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) orQ = m * (h2 - h1)v1 = V1/m ; Definition of v1W = m * p1* (v2 - v1) ; from w = INT pdv for a constant pressure process

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and George S.Kell, Hemisphere Publishing Corp., 1984. ; STM8SI.tkw

53

Variable Sheet


*STEAM8SI.TKW Steam, 1-8 States, SI UnitsP4-53.tkw Problem 4.53(b)


h1 2310 kJ/kg Enthalpys1 5.87 kJ/(kg*K) Entropyv1 0.37 m^3/kg Specific Volume


0.5 V1 m^3 Initial volume675 T2 C Temperature400 p2 kPa Pressure


*THUNITS.TKW Units for thermoQ 2110 kJ Heat transferm 1.35 kg Mass of steamW 390 kJ Work

4.54 10.005 1.495 1.688

4.713 kg, 1.688 kg. 0.26370.001061 0.8857 1.688 4.713f gm m x

3

1 2 0.001061 0.2637(0.8857 0.001961) 0.234 m / kgv v

1 504.5 0.2637(2025) 1038 kJ/kgu

a) Find 2u in Table C.2: 2 1( ) 6.401(2578 1038) 9850 kJQ m u u

b) 22 2 1

2

400 C2953 and ( ) 6.401(2953 1038) 12 260 kJ

0.234T

u Q m u uv

c) 22 2 1

2

0.8 MPa2530 and ( ) 6.401(2530 1038) 9550 kJ

0.234p

u Q m u uv

4.55 . 10 4 0.171 . 14.62 F and 4 0.24 14.62 14.04 BtupQ mc T T T H

4.56 11

2

lnp

Q W mRT p Vp

611

2

200ln 200 (8000 10 ) ln 2.22 kJ

800pp

54

4.57 a)1/ 0.4 /1.4

22 1

1

15000473 1332 K or 1059 C

400

k kp

T Tp

2 0.717(1059 200) 1230 kJvW mc T

b)1/ 0.4 /1.4

22 1

1

15000623 1755 K or 1482 C

400

k kp

T Tp

2 0.717(1482 200) 1620 kJvW mc T

4.58 11

p VT

1/21 0.28572

2 11

100 0.3 0.8394 K and 394 50 1205 K

0.2 0.287

k kp

T TmR p

2 1( ) 0.2 0.717(1205 394) 116 kJvW mc T T

4.59 a)15 100 10 75 150

. 400 1000 0.24 . 49.4 F60 53.3 530pQ mc T T T

b)15 100 10 75 150

. 400 1000 0.171 . 69.4 F60 53.3 530vQ mc T T T

c) Probably the constant pressure assumption since there are air passages under the doorsand through the air vents so the volume is not constant. The passages allow the pressureto be essentially constant.

4.60 2 2200 2

. 200 0.717( 100). 174.7 C0.287 373vQ W mc T T T

2mRT

pV

2

3.737 0.287 447.7240 kPa

2

where m = 3.737 kg.

4.61 Maximum increase would be for an insulated container so that Q = 0:

Q 3. ( 2 10 9.81) 10 10 1000 . 4.69 CpW mc T T T

4.62 paddle( ) pQ W mc T

a)400 0.1

1.00(1092 273) 10 100 45 /1000 373.1 kJ0.287 273

Q

where we used 2 1V

T T 2

V 1273 4 1092 K . The factor of 1000 in the above

equation changed J to kJ.

55

b)400 0.1

1.00(2292 573) 10 100 45/1000 373.1 kJ0.287 573

Q

where we used 2 1V

T T 2

V 1573 4 2292 K

4.63 V 31

0.8 53.3 12606.218 ft

60 144

a) net 1-2 2-3W W W 3-1 1 (W p V 2 V 1 ) lnV

mRT 1

V 3

6.21860 144(10 6.218) 0.8 53.3 1260ln 7150 kJ

10

net net7150

9.19 Btu778

Q W

b) net 1-2 2-3W W W 3-1 1 (W p V 2 V 1 1 3) ( )vmc T T

60 144(10 6.218) 0.8 0.171(1042 1260) 9480 kJ

net net9480

12.2 Btu778

Q W

where we used 3 1V

T T 1

V

1 0.4

3

6.2181260 1042 K.

10

k

4.64 The temperatures and V 3 are

1 2 3100 0.08 800 0.08

278.7 K. 2230 K0.1 0.287 0.1 0.287

T T T

V 3 V 322

3

8000.8 0.64 m

100pp

net 1-2W W 22-3 3-1 1

3

ln (p

W W mRT p Vp

1 V 3 )

8000.1 0.287 2230 ln 100(0.08 0.64) 77.1 kJ

100

For a cycle net net 77.1 kJQ W

4.65 a) net 1-2 2-3W W W 3-4 4-1W W 1 2 1 3 4 3( ) ( )mp v v mp v v

10.04 4000(0.04978 0.00125) 47.39(0.00125 0.04978) 1930 kJ

where we usedVm 2

2

0.5 10.04 kg.0.04978v

net net 1930 kJQ W

56

b) TK solutionRule Sheet

;For this cycle of a closed system, assume that all processes are frictionless. ThenQnet = Wnet ; = W12 + W23 + W34 + W 41 =m * (INT pdv + 0 + INT pdv + 0) andWnet = m * (p1 * (v2 - v1) + p3 * (v4 - v3)) ; for the two constant pressure processesm = V2 / v2v3 = v2v4 = v1; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. STM8si.tkw


` Thermal Sciences, Potter & Scott*Stm8si.tkw, Steam, 1-8 States, SI UnitsP4-65.tkw Problem 4.65


v1 0.00125 m^3/kg Specific Volume0 x1 Quality

phase1 'SAT PhaseT2 250 C Temperature (starting guess needed)

4000 p2 kPa Pressurev2 0.0498 m^3/kg Specific Volume (transfer value to input)

1 x2 Qualityphase2 'SAT Phase

*THUNITS.tkw Units for thermo80 T3 C

p3 47.4 kPa0.0498 v3 m^3/kg Specific volume (transfer value to input)

phase3 'SAT80 T4 C

p4 47.4 kPa0.00125 v4 m^3/kg Specific volume (transfer value to input)

phase4 'SAT0.5 V2 m^3 Volume at state 2

m 10 kg Mass of steamQnet 1930 kJ Net heat transfer for the cycleWnet 1930 kJ Net work for the cycle

4.66 3 1 3100 0.02

232.3 K.0.03 0.287

VT T T

1

V

1 0.4

3

0.02232.3 583.5 K

0.002

k

net 1-2 2-3W W W 3-1 lnV

W mRT 2

V 1 33

( )vmc T T

0.020.03 0.287 583.5ln 0.717(583.5 232.3) 4.014 kJ

0.002

57

4.672

12 1 2

2

5100 25 m/s

10A

V VA

4.68 1 1 1 2 2 2 20.02

. 0.0025 m/s1000 (800 0.01 0.001)

m AV A V V

0.60240 s or 4 min

0.0025L

tV

4.692

12 1

2

1150 37.5 fps

2 4 0.5A

V VA

4.70 41 1 1

4000(10 10 ) 150 3.65 kg/s

0.287 573m AV

242 2

3.65195 m/s

400(50 10 )

0.287 373

mV

A

4.71 a) The continuity equation with one inlet is. .

1 1 1c vdm

AV Vdt

ddt

where V is the volume of the tank. Then, using 1 11/ ,v

1 1d A Vdt V

23

1

0.05 20 0.0409 kg/m s10 0.3843v


; State 1: State of entering steam in the pipe; State 2: State of steam in the tank. A mass;balance on the tank is: Rate of mass flow in = rate of mass increase in the tank. This ismdot1 = dm2%dt ; where dm2%dt = dm2/dtmdot1 = A1* V1/ v1A1 = pi() * d1^2/4drho%dt = dm2%dt/Volume2

; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw

Variable Sheet

Input Name Output Unit Comment` Engineering Thermo, Jones & Dugan

*Stm8si.tkw Steam, 1-8 States, SI UnitsP4-71.tkw Problem 4.71

58

400 T1 C Temperature in inlet pipe800 p1 kPa Pressure in inlet pipe

v1 0.384 m^3/kg Specific Volume in inlet pipex1 'mngless Qualityphase1 'SH Phase

T2 C Temperaturep2 kPa Pressurev2 m^3/kg Specific Volumex2 Qualityphase2 Phasemdot1 0.409 kg/s Mass flow rate inA1 0.00785 m^2 Cross-sectinal area of pipe

20 V1 m/s Velocity in pipe10 d1 cm Diamter of pipe10 Volume2 m^3 Volume of tank

dm2%dt 0.409 kg/s Rate of iuncreas of mass in tankdrho%dt 0.0409 kg/(m^3*s) Rate of increase of density in tank

4.72 2 21 2 3 2

2000 150. .04 125 .05 40

.287 623 .287 423m m m m

2 22

6.646.64 kg/s, 255 m/s

4500.05

0.287 473

m V

4.73 avg15

1.442 fps62.4 4 1/ 24

mV

A

2 3

max max max2 2

4 1/ 4815 1 4 62.4 4 62.4 4

33

hh

hh

m V dA

y yV dy V y V

h h

max 2.16 fpsV

4.74 a) 21 1 2 2

1

1 1 4.950.25 30 4.95 kg/s, 109 m/s0.5951 2 0.125 / 1.08

m A V Vv


;For steady flow, mdot1 = 2 * mdot2, with state 1 in large pipe, state 2 in each smaller one.A1 * V1 / v1 = 2 * A2 * V2 / v2A1 = pi() * d1^2/4A2 = pi() * d2^2 /4mdot2 = A2 * V2 / v2; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw

59

Variable Sheet


*Stm8si.tkw Steam, 1-8 States, SI Units250 T1 C Temperature400 p1 kPa Pressure


30 V1 m/s Velocity in large pipe50 d1 cm Diameter of large pipe200 T2 C Temperature200 p2 kPa Pressure

h2 2870 kJ/kg Enthalpys2 7.51 kJ/(kg*K) Entropyv2 1.08 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH PhaseV2 109 m/s Velocity in each small pipe

25 d2 cm Diameter of each small pipe*THUNITS.TKW Units for thermo

A1 0.196 m^2 Area of large pipeA2 0.0491 m^2 Area of each small pipemdot2 4.95 kg/s Flow rate in each small pipe

4.75 1 1 11

1 1 10 2.18800 2.18 lbm/sec, 182 fps25.43 144 (2 / 144) / 1.1583

m A V Vv

4.76 a) V 2 21 1 1 2 2 2

1 1. 10 0.04 20 0.06 10

0.1996 0.3066d d

AV A Vdt dt

30.01348 kg/m s

ddt


; State 1 is defined as the state of steam in the entrance.pipe. State 2 is defined as the; state of steam in the exit pipe. m is the mass of steam in the tank at a specified instant.; A mass balance on the tank is therefore dm/dt = mdot1 - mdot2 ordm%dt = A1* V1 /v1 - A2* V2 / v2A1 = pi() * d1^2/4A2 = pi() * d2^2/4drho%dt = dm%dt/Volumet

60


Variable Sheet


*Stm8si.tkw Steam, 1-8 States, SI UnitsP4-76.tkw Problem 4.76

600 T1 C Temperature in inlet pipe2000 p1 kPa Pressure in inlet pipe

v1 0.2 m^3/kg Specific Volume in inlet pipex1 'mngless Qualityphase1 'SH Phase

400 T2 C Temperature in exit pipe1000 p2 kPa Pressure in exit pipe

v2 0.307 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH PhaseA1 0.00503 m^2 Cross-sectinal area of pipe

20 V1 m/s Velocity in pipe8 d1 cm Diamter of pipe10 Volumet m^3 Volume of tank

A2 0.0113 m^210 V2 m/s12 d2 cm

dm%dt 0.135 kg/s Rate of iuncreas of mass in tankdrho%dt 0.0135 kg/(m^3*s) Rate of increase of density in tank

4.77 Use Eq. 4.58 with 1 2 : 20.006

21 1 2 2 max max2

0. 0.006 0.8 1 2 . 1.6 m/s

0.006r

AV A V V r dr V

21 1 1 1000 0.006 0.8 0.0905 kg/sm AV

4.781

2 4 4

0

1 12(1 )2 10 1000 4 10 0.314 kg/s

2 4m r r dr

22 20.314 1000 0.0025 . 16 m/sV V

4.792

1 1 160 144 2

100 2.526 lbm/sec53.3 560 144

m AV

2

2 2 2 2 270 144 2

. 116.3 fps.53.3 760 144

A V V V

61

If we neglect KE , 2 1( ) 2.526 0.24(300 100) 121.2 Btu/sec pQ mc T T

Note:2 2 2 2

2 1 116.3 1002.526 4453 ft-lbf/sec or 5.72 Btu/sec

2 2V V

KE m

4.80 a) Across a valve 0h . From Table D.2 at 0.8 MPa we find 1 93.42h :

2 1 2 293.42 3.46 (221.3). 0.4065h h x x

2 2 ( ) 3.41 0.4065(206.1 3.41) 85.8 kJ/kgf g fu u x u u b) TK solution:

Rule Sheet;This is a throttling process. w = 0. Assume q = 0 and change in ke is negligible. Therefore,h2 = h1 ; from the first law.

; R134a tables based on 'Thermodynamic Properties of HFC-134a'; DuPont Technical Information, which is based upon the Modified; Benedict-Webb-Rubin equation of state. R1348si.tkwh2 = u2 + p2 * v2

Variable Sheet


*R1348si.tkw, R134a, 1-8 States, SI unitsP4-80.tkw Problem 4.80


h1 93.602 kJ/kg Enthalpys1 0.34761 kJ/(kg*K) Entropyv1 0.00084403 m^3/kg Specific Volumex1 'mngless Qualityphase1 'CLQ Phase

T2 -36.91 C Temperature60 p2 kPa Pressure93.602 h2 kJ/kg Enthalpy (transfer value to input)

s2 0.39624 kJ/(kg*K) Entropyv2 0.12501 m^3/kg Specific Volumex2 0.40047 Qualityphase2 'SAT Phaseu2 86.101 kJ/kg Internal energy

*THUNITS.TKW Units for thermo

4.81 Neglecting kinetic energy changes and using 1 1344fh h kJ/kg (from Table C.1),

2 2 1 2 20.6 MPa. 1344 670.6 (2086.3). 0.3228p h h x x

2 2 ( ) 669.9 0.3228(2567.4 669.9) 1282 kJ/kgf g fu u x u u

62

4.82 Neglecting kinetic energy changes and using 1 3634h kJ/kg (from Table C.3),

a) 22

2

0.6 MPa569 C

3634 kJ/kgp

Th

b) With 2 1V V and 2 1m m we have

2 21

0.9695A V 1 1

10.04341

A V 2

1

. 22.3AA

c) For air 2 10. 600 Ch T T

4.83 Using the energy equation in the form of Eq. 4.65 we have, with 0SQ W and

2 1 2 1andV V z z ,

2 12 1 2 1 2

2 1

(450 30) 144or . 38.09 39.34 Btu/lbm

62.4 778p p

h h u u u

The factor 778 converts ft-lbf to Btu.

4.84 The energy equation in the form of Eq. 4.68 is used. First, the velocities are

1 22 21

200 2006.366 m/s. 17.68 m/s

1000 0.1 1000 0.06m

V VA

2 22 1

2P

V V pW m g z

2 2 400000017.68 6.366200 827000 W or 1110 hp

2 1000

4.85 Q 2 1( ). ( 5) 1.00(530.7 293). 0.021 kg/sS pW mc T T m m

where we used1/ 0.4 /1.4

22 1

1

400293 530.7 K

50

k kp

T Tp

.

4.86 First, let’s find the mass flux:2

2 2 2500

0.05 100 2.72 kg/s0.287 503

m A V

where0.4 /1.4

2500

293 503 K80

T

.

The energy equation gives (the kinetic energy change is negligible)

Q 2 1( ) ( ). 2.72 1.00(503 293) 558 kWC p CW mc T T W

4.87 a) 2 1( ) ( ). 0.01 (2839 2592) 6 3.53 kJ/sCQ W m h h Q The negative sign indicates a heat loss.

63


Qdot = mdot * (h2 - h1) + Wdot ; First law for steady flow through compressor with negligible; changes in kinetic and potential energy.


Variable Sheet


*Stm8si.tkw Steam, 1-8 States, SI Units50 T1 C Temperature

p1 12.3 kPa Pressureh1 2590 kJ/kg Enthalpy

1 x1 Qualityphase1 'SAT Phase


h2 2840 kJ/kg Enthalpyx2 'mngless Qualityphase2 'SH Phase

*THUNITS.TKW Units for thermoQdot -3.52 kJ/s Rate of heat addition, i.e., heat is removed

0.01 mdot kg/s Mass flow rate-6 Wdot kW Power output

4.88 We use the energy equation in the form of Eq. 4.68:

2 1 2000 (2000 2) 144( ) 2594 ft-lbf/sec or 4.72 hp1/ 3600 1/ 0.01623P

p pW m

v

4.89 Use Eq. 4.58 with 2 1 : 3

2 2 1 1 0.1 m /sA V AV

1 22 2

0.1 0.150.93 m/s and 12.73 m/s

0.025 0.05V V

The volume flow rate is the product .AV Assuming a 100% efficient pump forminimum power,

2 22 1 2 1( )

2P

V V p pW m

2 212.73 50.93 4000 2000.1 1000 258.4 kW or 346 hp

2 1000 1000

The constant “1000” in the denominator of the kinetic energy term converts W to kW.

64

4.90 Assume a 100% efficient turbine:2 2

2 11 1 2T

V VW AV

2 1 0 300

1000 20 6000 kW1000

p p

We neglected the kinetic energy change because information was not given andbecause it is most often negligible. Since the pressure is in kPa the work rate is in kW.

4.91 Assume a 100% efficient turbine with the pressure of zero gage on the surface of thebackwater and at the turbine outlet:

2 22 1

1 1 2T

V VW AV

2 1p p

2 1( )g z z

100 9.81 ( 40) 39 240 W or 39.24 kW

4.92 Assume a 100% efficient turbine with the pressure of zero gage on the surface of thebackwater and at the turbine outlet:

2 22 1

1 1 2T

V VW AV

2 1p p

2 1( )g z z

1000 (0.6 1.2) 1.5 9.81 ( 2) 21200 W or 21.2 kW

4.93 a) 2 13414

( ). 10000 30(1131 1512).3600TQ W m h h Q

1954 Btu/secQ b) TK solution:

Rule SheetWdot = mdot * (h1 - h2) + Qdot ; First law for steady flow with no changes in potential or; kinetic energy

;*Stm8e.tkw Steam, 1-8 States, English units; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,; and Kell, Hemisphere Publishing Corp., 1984. See Comment sheet.


` Thermal Sciences, Potter & Scott*Stm8e.tkw Steam, 1-8 States, English units


h1 1510 B/lbm Enthalpyv1 1.05 ft^3/lbm Specific Volumex1 'mngless Qualityphase1 'SH PhaseT2 162 F Temperature

5 p2 psi Pressureh2 1130 B/lbm Enthalpy

65

v2 73.5 ft^3/lbm Specific Volume1 x2 Quality

phase2 'SAT Phase*THUNITS.TKW Units for thermo

10 Wdot MW Power output30 mdot lbm/s Mass flow rate

Qdot -1950 B/s Heat is removed at a rate of 1950 B/s

4.94 2 11000

( ) (1116.1 1623.8) 8462 Btu/sec or 11,970 hp60TW m h h

2

1000 / 60230 fps

2 /173.75m

VA

where we used 1/ .v

4.95 1 2 23445.2 kJ/kg, 251.4 0.9(2358.3) 2374 kJ/kgf fgh h h x h

2 1( ) 6(2374 3445) 6430 kWTW m h h

2 2

682.2 m/s

0.4 /[0.001 0.9(7.649 0.001)]m

VA

4.96

2

2 2 2 1 1 1 22

6000.05 100

0.287 373. 18.17 m/s140

0.20.287 253

A V AV V

2

1 1 1600

0.05 100 4.402 kg/s0.287 373

m AV

2 2 2 2

2 12 1

18.17 100( ) 4.402 1.00( 20 100) 539 kW

2 2 1000T p

V VW m c T T

The factor “1000” in the denominator of the kinetic energy term converts W to kW.

4.97 21 1 1

8000.05 20 0.9257 kg/s

0.287 473m AV

2 1( ) 0.9257 1.0(200 20) ( 400) 233 kJ/sp SQ mc T T W The negative sign indicates a heat loss, as expected.

4.98 22 1

450( ) .06 150 1.0 (20 100) 500 70.5 kJ/s

.287 373p SQ mc T T W

66

4.99 First, let’s calculate the velocities:

1 22 21 1

30 305.509 fps, 137.7 fps

62.4 2 /144 62.4 0.4 /144

m

V VA

Q SW2 2

2 1 2 1

2V V p p

m

2 21

1137.7 5.509 (14.7 ) 144

0 30 . 142 psia2 32.2 62.4

Pp

The factor 32.2 above converts slugs to lbm.

4.100 The energy equation: 2 2 22 1 1 2 22 ( ) 20 2 1.0(293 )pV V c T T T

Continuity equation: 11 1 1 2 2 2 1 1 2 2 2

1

.p

AV A V AV A VRT

22 2 2

80 2 420 5.384 kg/m s

0.287 293 3V

Assume an adiabatic expansion using 1/ v :k-1

2 2 20.4 0.4

1 1 2

293or 298.9

[80 / 0.287 293]T TT

The above three equations include three unknowns 2 2 2, , andV T . They are solvedby trial-and-error to give

32 2 23.47 kg/m , 1.55 m/s and 219 CV T

4.101 a) Assume an adiabatic process:/ 1 0.4 /1.4

22 1

1

585468 812 K or 539 C

85

k kp

T Tp

b)2 2 2 2

2 1 22 1 2

1000 ( ) 1.0(812 468) . 835 m/s

2 2 1000p

V V Vc T T V

c)2

2 2 22 12 2 1 1 2 2

2 1

585 0.1 100 812( / 4) . . 0.239 m

185 468 835 / 4

p p

d V r V d dRT RT

4.102 a) 2800.05 200 1.672 kg/s

0.297 253

pm AV

RT

b)2 2 2 2

2 12 1 2 2

15 2000 ( ). 0 1.042( 20). 0.91 C

2 2 1000p

V Vc T T T T

4.1032 2 2 2

2 12 1 2

50 6000 . 0 1154.

2 2 1000V V

h h h

2 1161 Btu/lbmh

22

2

20 psia238 F

1161P

Th

67

4.104 The heat transfer rate for the steam, assuming no pressure drop through the condenser, is

2 1( ) 600 (94.02 1116.1) 613, 200 Btu/mins s s sQ m h h This equals the energy gined by the water:

2 1( ). 613, 200 1.00 (15). 40,900 lbm/minw w p w w w wQ m c T T m m

4.105 a) The heat lost by the air is gained by the water:

2 1 2 1( ) ( ). 5 1.0 200 4.18 10. 23.9 kg/sa pa a a w pw w w w wm c T T m c T T m m b) 23.9 4.18 200 1000 kJ/sw w pwQ m c T

4.106 With 1 10 and 0 there results since . Across a valvei f fQ m u h m m the enthalpy

is constant so that 1 line 3674.4 kJ/kgh h . The final pressure is 4 MPa andfp

3674.4 kJ/kg.fu The temperature is interpolated from Table C.3 to be

3674.4 3650.150 800 813 C

3650.1 3555.5fT

The final specific volume is

3812.8 800 (0.1229 0.1169) 0.1229 0.1244 m /kg50fv

The mass of steam in the tank is then fVm

4 32.15 kg0.1244

f

fv

4.107 a)1

800 psia1587 F

1620

f

ff

pT

u h

b) Interpolate and find 31.512 m /kg.f fVv m

50 33.1 lbm1.512

f

fv

4.108 ii

p Vm

250 3 80 38.769 kg. 2.806 kg

0.287 298 0.287 298

if

i

mRT

2 2

2.806 0.717 298 8.769 0.717 298 (8.769 2.806) 1.0 298 503 kJf f i iQ m u m u m h

4.1091393790 5 800 5

5.351 kg.0.287 293 0.287i f

f f

m mT T

1 10 0.717 293 5.351 ( 5.351) 1.0 353f f i i f f fu m u m m h u m m

13937Then 0 0.717 1124 ( 5.351) 353. 30.48 kgf f f

f

T m mT

Finally, 113937

457 or 184 C. 30.48 5.351 25.1 kg30.48fT m

68

4.110 Since 0 and 0iQ m , 1 0.24 530 0.171 . 744 R or 284 Ff f fu h T T

f ff

p Vm

12 144 104.36 lbm

53.3 744

fRT

4.111 a)V f f fp RT

V

1/ 0.4 /1.495. 303 127 K or 146 C

2000

k kf f f

f ii i i i i i

p T pT T

p RT p T p

b) For V 22 1

1

303const 95 227 kPa

127

Tp p

T

4.112800 4

0.02 300 6 kg. 36.8 kg. 30.8 kg0.287 303i f i fm m m m

a)1 0.430.8

303 282 or 9 C36.8

kf

f ii

mT T

m

b) f ff

m RTp

V30.8 0.287 282

624 kPa4

f

c)1/ 1 2.5253

36.8 23.44 kg303

kf

f ii

Tm m

T

13.36 0.02 60. 11.13 minm t t

4.1131.8 0.2

1726 kg0.001043 1.694im . a) 158.9 CT

b)1 1

911.4 kg. 814.5 kg0.001101 0.3157fm m

c) 2 2 2 2 20.5 1.5

458.9 kg. ( )0.001101 0.3157f f

dm Q um h m u m h m

dt

2 2 210000

10000 ( ) . 1.878 kg/min2567 2757fu h m m

1.878 1725.9 458.9. 675 mint t

4.114 a) Assume steady state with the heat sink area equal to the transistor area:

1 2 3

4 4

44 8 4 4

4

412

( ) ( )1

0.5

3251 15 10 ( 325) 0.5 5.67 10 10 ( 300 )

10.5

15 102.835 10 0.003 2 0

jj j sur

jj j

j j

Q Q Q Q

T Th A T T A T T

h AT

T T

T T

69

By trial and error,

Tj = 568 K (295C) >> 125C

The transistor exceeds its maximum allowable temperature of 125C

b) 4 4 8 4 4 4( ) 0.5 5.67 10 10 (568 300 ) 0.27 W rad j surrQ A T T

0.27fraction 0.27

1

4.115 Assume the temperature to be uniform at any instant, negligible radiation, andconstant properties.

a) ch LBi

k

where Lc is the effective length and is defined as Volume/Surfce area. For thepancake,

2

2

(0.075) (0.003)4 0.0015

2 (0.075) 0.0075 0.0034

CL

m

Thus,8 0.0015

0.048 0.10.25

ch LBi

k

The lumped capacitance assumption is valid.

b) Energy balance:Eout = Estored

( ) sh A T T cV or

sh AdT dT

dt T T cVdt

Let = T T. Then,

sh Ad

cV. ln

s

i

h AT Tdt

T T cVt

cVt

500 3000 0.0015 25 20ln ln 647 s

8 80 20

s i

T Th A T T

7.5 cm

T= 20Ch = 8 W/m2·K

Air

3 cm

70

4.115 Assume steady-state, one-dimensional conduction, with negligible contactresistance and constant properties.

1 1 1 10.07 C/W, 0.21 C/W

20.8 0.6 8 0.6

0.0050.07 C/W

0.12 0.6

0.0150.015 C/W

0.17 0.6

0.20

oconv iconvo w i w

panelpanel

panel panel

plasterplaster

plaster plaster

studstud

stud stud

R Rh A h A

tR

k A

tR

k A

tR

k A0.2

312.5 C/W, 11.91 C/W.16 0.06 0.03 0.56

0.0250.443 C/W

0.094 0.6

insins

ins ins

sidingsidings

siding siding

tR

k A

tR

k A

stud instotal oconv panel plaster sidings iconv

stud ins

R RR R R R R R

R R

20 5 1516.67 W

312.5 11.91 0.90.07 0.07 0.015 0.443 0.21312.5 11.91

total

TQ

R

4.117 Neglect radiation.

a)2 1

2

200 109.98 kW

ln / ln 4 / 3.75 112 7 16.5 2 .04 7 302 2

inside air inside air

pipe air

T T T TQ

r rR Rl k r l h

b) 200 10536 W

ln 4 / 3.75 ln 9 / 4 12 7 16.5 2 7 0.055 2 0.09 7 30

inside air

pipe insulation air

T TQ

R R R

insulation 536% reduction 1 100 1 100 94%

9980

QQ

Tair, h

lpipe

D2, D1

Rpipe Rair

Tinside Tair

71

4.118 Assume the system to be at steady state.

21 2

1 2

22

.

1 1 1 14 4

1 1 1 10.025 0.035 8.23 W/m K

400 20 11 44500 4 0.035 1004

i amb i ambtot cond conv

cond conv tot

i amb

tot

i amb

tot

T T T TQ R R

R R Q

T Tk r r r hQ

r rk

T Tr hQ

1 cm

r2 = 3.5 cmr1 = 2.5 cm Q total = 500W

Ti = 400oC Tamb

Rcond Rconv

thermal science ch 04

Technology