7- 1

CHAPTER SEVEN

7.1 080 35 10 0 30 2 33 2 34. .

. . L kJ . kJ work 1 h 1 kW

h L 1 kJ heat 3600 s 1 k J s kW kW

=

21

312 33.33 kW 10 W 1.341 10 hp

kW 1 W hp .1 hp

3 =

-

.

7.2 All kinetic energy dissipated by friction

(a) Emu

k =

=

=

-

2

2 2 2 2 4

2 2

25500 lb 5280 1 9 486 10

715

m2 2

f2

2 2m

2f

55 miles ft 1 h lb Btu2 h 1 mile 3600 s 32.174 lb ft / s 0.7376 ft lb Btu

.

(b)

8

4 6

3 10 brakings 715 Btu 1 day 1 h 1 W 1 MW2617 MW

day braking 24 h 3600 s 9.486 10 Btu/s 1 0 W

3000 MW

-

=

7.3 (a) Emissions:

Paper1000 sacks oz 1 lb

sack 16 oz lbm m

+=

( . . ).

0 0510 0 05166 41

Plastic2000 sacks oz 1 lb

sack 16 oz lbm m

+=

( . . ).

0 0045 0 01462 39

Energy:

Paper1000 sacks Btu

sack Btu

+=

( ).

724 905163 106

Plastic2000 sacks Btu

sack Btu

+=

( ).

185 464130 106

(b) For paper (double for plastic)

Raw MaterialsAcquisition and

Production

SackProduction and

UseDisposal

Materialsfor 400 sacks

1000 sacks 400 sacks

7- 2

7.3 (contd) Emissions:

Paper

400 sacks .0510 oz 1 lb sack 16 oz

1000 sacks .0516 oz 1 lb sack 16 oz

lb

reduction

m mm + =

0 04 5

30%

.

Plastic

800 sacks .0045 oz 1 lb sack 16 oz

2000 sacks .0146 oz 1 lb sack 16 oz

lb

reduction

m mm + =

0 02 05

14%

.

Energy:

Paper400 sacks Btu

sack

1000 sacks Btu

sack Btu; 27% reduction + =

724 905119 106.

Plastic800 sacks Btu

sack

2000 sacks Btu

sack Btu; 17% reduction + =

185 464108 106.

(c) .3 10 persons 1 sack 1 day 1 h 649 Btu 1 J 1 MW

person - day h 3600 s 1 sack 9.486 10 Btu J / s MW

8

-4

=24 10

2 375

6

,

Savings for recycling: 017 2 375. ( , MW) = 404 MW (d) Cost, toxicity, biodegradability, depletion of nonrenewable resources.

7.4 (a) Mass flow rate: gal 1 ft (0.792)(62.43) lb 1 min

min 7.4805 gal 1 ft 60 s lb s

3m

3 m& . .m = =3 00 0 330

Stream velocity: gal 1728 in 1 1 ft min

min 7.4805 gal 0.5 in 12 in 60 s ft s

3

2u = =300 1

12252.

.Pb g

Kinetic energy: .330 lb ft 1 1 lb

s s 32.174 lb ft / s

ft lbs

ft lb s hp

ft lb shp

m2

f2

m2

f

ff

Emu

k = =

=

=

FHG

IKJ =

-

--

-

2 23

33

5

20 1225

27 70 10

7 70 101341 1007376

140 10

..

. /.. /

.

b g

d i

(b) Heat losses in electrical circuits, friction in pump bearings.

7- 3

7.5 (a) Mass flow rate:

( )

( )

2 3

3

42.0 m 0.07 m 10 L 673 K 130 kPa 1 mol 29 g127.9 g s

s 4 1 m 273 K 101.3 kPa 22.4 L STP molm

p= =&

& & .E muk = =

=2 2

21 42 0

113127.9 g kg m 1 N 1 J2 s 1000 g s 1 kg m / s N m

J s2

2 2

(b)

( ) 3

3 2 2

127.9 g 1 mol 22.4 L STP 673 K 101.3 kPa 1 m 449.32 m s

s 29 g 1 mol 273 K 130 kPa 10 L (0.07) mp=

& & . .

& & ) & )

Emu

E E E

k

k k k

= =

=

2 2

21 49 32

1558127.9 g kg m 1 N 1 J2 s 1000 g s 1 kg m / s N m

J / s

= (400 C - (300 C = (155.8 - 113) J / s = 42.8 J / s 43 J / s

2

2 2

D o o

(c) Some of the heat added goes to raise T (and hence U) of the air 7.6 (a)

D DE mg zp = =-

=-

1 gal 1 ft 62.43 lb ft ft lb7.4805 gal 1 ft s 32.174 lb ft / s

ft lb3

m f3 2 m 2

f32174 10 1

834.

.

(b) E Emu

mg z u g zk p= - = - = - =FHG

IKJ

LNM

OQP =D D D

21 2

1 2

22 2 32174 10 254b g b g b g. .ft

s ft

fts2

(c) False 7.7 (a) D &E positivek When the pressure decreases, the volumetric flow rate increases, and

hence the velocity increases.

D &E negativep The gas exits at a level below the entrance level.

(b) &

.

.

m =

=

5 m 1.5 cm 1 m 273 K 10 bars 1 kmol 16.0 kg CH s 10 cm 303 K bars 22.4 m STP 1 kmol

kg s

2 34

4 2 3

p b gb g

2

101325

0 0225

( )

2out out out outin in

2in in in out in out

inout in

out

(m/s) A(m) (m/s) A(m)

10 bar 5 m s 5.555 m s

9 bar

P V V uP PnRTP V nRT V P u P

Pu u

P

= = =

= = =

& &&& &&

2 2 22 2

2 2

1

2

0.5(0.0225) kg (5.555 5.000 )m 1 N 1 W( )

s s 1 kg m/s 1 N m/s

0.0659 W

0.0225 kg 9.8066 m -200 m 1 N 1 W( )

s

k out in

p out in

E m u u

E mg z z

-D = - =

=

D = - =

& &

& & 2 s kg m/s 1 N m/s 44.1 W

= -

7- 4

7.8 D D& & .

.

E mg zp = =-

= -

-10 m 10 L kg H O m m N 1 J 2.778 10 kW h h 1 m L s 1 kg m/ s 1 N m 1 J

kW h h

5 3 32

3 2 2

1 981 75 11

204 10

7

4

The maximum energy to be gained equals the potential energy lost by the water, or

2.04 10 kW h 24 h 7 days h 1 day 1 week

kW h week (more than sufficient)4

= 3 43 106.

7.9 (b) Q W U E Ek p- = + +D D D

D

D

E

Ek

p

=

=

0

0

system is stationary

no height change

b gb g

Q W U Q W- = < >D , ,0 0

(c) Q W U E Ek p- = + +D D D

Q W

EE

k

p

= ===

0 000

adiabatic , no moving parts or generated currents system is stationary no height change

b g b gb gb g

DD

DU = 0 (d). Q W U E Ek p- = + +D D D

W

EE

k

p

===

000

no moving parts or generated currents system is stationary no height change

b gb gb g

DD

Q U Q= Q U W T0 7 65 30D . L bar > 0, Cfinal

7- 5

7.11 A = = 2

-p 3 cm m cm

m2 2

22b g 1

102 83 10

43.

(a) Downward force on piston:

F P A m gd = +

=

+

=-

atm piston+weight

5 2 2

2 2

1 atm 1.01325 10 N / m matm

24.50 kg 9.81 m 1 Ns kg m / s

N283 10

1527

3.

Upward force on piston: F AP Pu g= = -

gas2 2 m N m283 10 3.d i d i

Equilibrium condition:

F F P Pu d= = = = -

2 83 10 527 186 10 186 103 0 05 5. . .m N m Pa2 2

V nRTP0 0

10677= =

=1.40 g N mol N 303 K 1.01325 10 Pa 0.08206 L atm

28.02 g 1.86 10 Pa 1 atm mol K L2 2

5

5.

(b) For any step, D D D D

DD

U E E Q W U Q Wk pEE

kp

+ + = - = -==

00

Step 1: Q U W = -0 D

Step 2: DU Q W= - As the gas temperature changes, the pressure remains constant, so

that V nRT Pg= must vary. This implies that the piston moves, so that W is not zero.

Overall: T T U Q Winitial final= = - =D 0 0

In step 1, the gas expands > < W U T0 0D decreases

(c) Downward force Fd = + =-100 101325 10 2 83 10 4 50 9 81 1 3315 3. . . . .b gd id i b gb gb g N (units

as in Part (a))

Final gas pressure P FAf

= =

= -

33110

116 103

5 N2.83 m

N m2

2.

Since T T f0 30= = C , P V P V V VPPf f f f

= = =

=0 0 00

5

50 677

186 10116 10

108...

. L Pa Pa

Lb g

Distance traversed by piston = =-

=

-

DVA

1.08 L 1 m L 2.83 10 m

m3

2

067710

01423 3

..b g

W Fd= = = =331 0142 47 47 N m N m Jb gb g. Since work is done by the gas on its surroundings, W Q

Q W= + = +

- =47 47

0 J J

(heat transferred to gas)

7.12 $ .V = =32.00 g 4.684 cm 10 L mol g 10 cm

L mol3 3

6 301499

$ $ $H U PV= + = +

=1706 2338 J mol41.64 atm 0.1499 L 8.314 J / (mol K)

mol 0.08206 L atm / (mol K) J mol

7- 6

0

7.13 (a) Ref state $U = 0d i liquid Bromine @ 300 K, 0.310 bar

(b) D $ $ $ . . .U U U= - = - = -final initial kJ mol0 000 2824 2824 D D D D D$ $ $ $ $H U PV U P V= + = +d i (Pressure Constant)

( )3

0.310 bar 0.0516 79.94 L 8.314 J 1 kJ 28.24 kJ mol 30.7 kJ mol

mol 0.08314 L bar 10 JH

-D = - + = -

D DH n H= = - = - -$ . . .500 30 7 15358 154 mol kJ / mol kJ kJb gb g (c) $U independent of P U U = =$ , . $ , . .300 0 205 300 0310 28 24 K bar K bar kJ molb g b g $ , $ , .U P Uf340 340 1 29 62 K K .33 bar kJ mold i b g= =

D

D

$ $ $$ . . .

U U U

U

= -

= - =E final initial

kJ mol29 62 28 24 1380

$ $ $ $ $$

..

V changes with pressure. At constant temperature PV = P' V' V'= PV / P'

V' (T = 300K, P = 0.205 bar) =(0.310 bar)(79.94 L / mol)

bar L / mol

=0 205

120 88

n = =500

00414.

. L 1 mol

120.88 L mol

D DU n U= = =$ . .0 0414 138 0 mol kJ / mol .0571 kJb gb g D D DU E E Q Wk p+ + = - =Q 0 0571. kJ

(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is being neglected; internal energy is not completely independent of pressure. 7.14 (a) By definition $ $ $H U PV= + ; ideal gas PV RT H U RT$ $ $= = + $ , $ $ , $ $U T P U T H T P U T RT H Tb g b g b g b g b g= = + = independent of P

(b) D D D$ $ .H U R T= + = +

=35001987

3599calmol

cal 50 K

mol K cal mol

D DH n H= = = $ .25 3599 8998 mol cal / mol cal 9.0 10 cal3b g b g 7.15 D D DU E E Q Wsk p+ + = -

DD

D

E m uE

W P V

k

p

s

==

=

00 no change in and no elevation change

since energy is transferred from the system to the surroundings

b gb gb g

D D D D D D DU Q W U Q P V Q U P V U PV H= - = - = + = + =( )

0 0

7- 7

7.16. (a) DD

DD

E u uE

PW P V

k

p

s

= = ==

==

0 00

0

1 2 no elevation change

(the pressure is constant since restraining force is constant, and area is constrant) the only work done is expansion work

b gb gb g

.

$ .

..

$ $

H T

T

= +

=

=

=

34980 355125 10 1

83140 0295

480

3

2

(J / mol), V = 785 cm , T = 400 K, P =125 kPa, Q =83.8 J

n =PVRT

Pa 785 cm m m Pa / mol K 400 K 10 cm

mol

Q = H = n(H - H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol)

83.8 J = 0.0295 35.5T - 35.5(400) K

13

1

3 3

3 6 3

2 1 2

2

D

i VnRT

P

ii W P V

iii Q U P V U Q PV

). .

) .

) . . .

mol m Pa cm K

Pa mol K m cm

N (941- 785)cm m

m cm J

J J J

3 3

33

3 3

2 3

= =

=

= =

=

= + = - = - =

0 0295 8 314 10 480125 10 1

941

125 10 110

19 5

838 19 5 64 3

6

5

5

6D

D D D D

(b) DEp = 0

7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could

occur, the temperature would drop during these periods.) (b) D D D D DU E E Q t W tp R+ + = -& &

D DE E W U t

Q

p k= = = = =

=

=

0 0 0 0 0

0 90 11 26

, , & , $ ( )

& . .

1.4 W J s

1 W J s

U t( ) .J = 126

Moles in tank: 1 atm 2.10 L 1 mol K

K L atm moln PV RT= =

+

=25 273 0 08206

0 0859b g . . $ . .U U

nt

t= = =126 14 67 (J)0.0859 mol

Thermocouple calibration: C mV

T aE b T ET ET E

= + = += =-= =

0 0.249100 5

181 451,

, .27

. .b g b g

$ .

. .U tT E

== +

14 67 0 440 880 1320181 451 25 45 65 85

(c) To keep the temperature uniform throughout the chamber.

(d) Power losses in electrical lines, heat absorbed by chamber walls.

(e) In a closed container, the pressure will increase with increasing temperature. However, at the low pressures of the experiment, the gas is probably close to ideal =$U f Tb g only. Ideality could be tested by repeating experiment at several initial pressures same results.

7- 8

7.18 (b) D D D& & & & &H E E Q Wk p s+ + = - (The system is the liquid stream.)

DD

&&

&

Ek m uE p

Ws

==

=

00

0

no change in and no elevation change

no moving parts or generated currents

c hc h

c h

D & & , &H Q Q= > 0 (c) D D D& & & & &H E E Q Wk p s+ + = - (The system is the water)

DD

D

& ~&

&

H T a PEk m u

Q T

==

=

00

0

nd constant no change in and

no between system and surroundings

c hc h

c h

D & & , &E W Wp s s= - > for water system0 b g (d) D D D& & & & &H E E Q Wk p s+ + = - (The system is the oil) D &Ek =0 no velocity changec h D D& & & &H E Q Wp s+ = - &Q < 0 (friction loss); &Ws < 0 (pump work). (e) D D D& & & & &H E E Q Wk p s+ + = - (The system is the reaction mixture) D D

D

& &&

Ek E pWs

= =

=

0

0

given

no moving parts or generated currentc h

c h

D & & , &H Q Q= pos. or neg. depends on reaction

7.19 (a) molar flow: m 273 K kPa 1 mol L

423 K 101.3 kPa 22.4 L STP m mol min

3125 122 101

4343

3

.min

.b g = D D D& & & & &H E E Q Wk p s+ + = -

D D& &

&Ek E p

Ws

= =

=

0

0

given

no moving partsc h

c h

& & & $ . .Q H n H= = = =D D 4337 2 63 mol 1 min 3640 J kW min 60s mol 10 J / s

kW3

(b) More information would be needed. The change in kinetic energy would depend on the cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet and outlet pipes would be needed to answer this question.

7- 9

7.20 (a) $ . $H T H= -104 25C in kJ kgb g $ . .Hout 9.36 kJ kg= - =104 34 0 25 $ . . .Hin 5 kJ kg= - =104 30 0 25 20

D $ . . .H = - =9 36 520 16 4 kJ kg D D D& & & & &H E E Q Wk p s+ + = -

assumed no moving parts

D D& &&Ek E p

Ws

= =

=

0

0c h

c h

& & & $Q H n H= =D D

1.25 kW kg 1 kJ / s 10 g 1 mol

4.16 kJ kW 1 kg 28.02 g mol s

3

= = =&&& .n

QHD

10 7

= 10.7 mol 22.4 L STP 303 K kPa

s mol 273 K 110 kPa L / s L s = & . .V b g 1013 245 5 246

(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is assumed to depend linearly on temperature and to be independent of pressure, errors in measured temperature and in wattmeter reading.

7.21 (a) $ $ $ . . .$ . . .

$ . .H aT b aH HT T

b H aTH T

= + =--

=-

-=

= - = - = -

UV|W|

= -2 1

2 1

1 1

129 8 25850 30

52

258 5 2 30 130252 1302

b gb gb g b gkJ kg C

$ ..

H T= = = 0 130252

25ref C

Table B.1 = = = -S G V. . . $ .b g b gC H l3

36 14

m659 kg

m kg06591

152 10 3

$ $ $ . .U H PV TkJ kg kJ / kg

1 atm 1.0132 10 N / m 1.52 10 m J kJ

1 atm 1 kg 1 N m 10 J

5 2 3

3

b g b g= - = --

-

52 1302

1 13

= -$ . .U TkJ kgb g 52 1304

(b) D D

DE E W

Q Uk p, ,

Energy balance: 20 kg [(5.2 20 - 130.4) - (5.2 80 -130.4)] kJ

1 kg kJ

== =

= -

06240

Average rate of heat removal kJ min

5 min 60 s kW= =

6240 1208.

&n (mol/s) N2 30 oC

34 oC

&Q=1.25 kW

P=110 kPa

7- 10

7.22

0

&m(kg/s)260C, 7 barsH = 2974 kJ/kg$u = 0

(kg/s)200C, 4 barsH = 2860 kJ/kg$u (m/s)

&m

D D D

D D

D

& & & & &

& & & & $ $

$ $ ( ) .

& & &H E E Q W

E Hmu

m H H

u H H u

k p s

k

E p Q Ws

+ + = -

=- =- -

= - = - = =

= = = 0

2

2 5

2

22 2974 2860 1

228 10 477

kJ 10 N m kg m / skg 1 kJ 1 N

ms

m / s

out in

in out

3 2 2

2

d i

d i b g

7.23 (a)

in

5 L/min0 mm Hg (gauge)

Q

100 mm Hg (gauge)

outQ

5 L/min

Since there is only one inlet stream and one outlet stream, and & & &m m min out= ,

Eq. (7.4-12) may be written

& $ & $ & & & &$

& $ & $ &

&& & &

m U m PVm

u mg z Q Ws

U

m PV mV P P V P

u

z

W

Q Q Q

s

D D D D

D

D D

D

D

+ + + = -

( )

( )

( )

=

= - =

=

=

=

= -

d i d i

a f2

2

0

0

0

0

2

given

assume for incompressible fluid

all energy other than flow work included in heat terms

out in

in out

& & &V P Q QD = -in out

(b) Flow work: 5 L mm Hg 1 atm 8.314 J

min 760 mm Hg 0.08206 liter atm J min& .V PD = -

=

100 0667

b g

2in2

5 ml O 20.2 JHeat input: 101 J min

min 1 ml OQ = =&

in

66.7 J minEfficiency: 100% 66%

101 J minV PQD

= =&&

7.24 (a) D D D& & & & &H E E Q Wk p s+ + = - ; D &E k , D &E p , & & &W H Qs = =0 D $H 400 3278 =C, 1 atm kJ kgb g (Table B.7) $H 100 2676 =C, sat' d 1 atm kJ kgb g (Table B.5)

7- 11

7.24 (contd)

100 kg H 100 kg H2 2o o

O(v) / s O(v) / s

100 C, saturated 400 C, 1 atm

(kW)&Q

& . .Q = - = 100 kg kJ 10 J s kg 1 kJ

J s33278 2676 0

6 02 107b g

(b) D D DU E E Q Wk p+ + = - ; DE k , DE p , W U Q= =0 D

( ) ( ) ( )3

final

kJ m Table B.5 100 C, 1 atm 2507 , 100 C, 1 atm 1.673 400 C, kg kg

U V V P = = =

Interpolate in Table B.7 to find P at which V =1.673 at 400oC, and then interpolate again to find U at 400oC and that pressure:

3 ofinal

3.11 1.673 1.673 m /g 1.0 4.0 3.3 bar , (400 C, 3.3 bar) = 2966 kJ/kg3.11 0.617

V P U-

= = + =-

( )[ ]( )3 7 100 kg 2966 2507 kJ kg 10 J kJ 4.59 10 JQ U m U = D = D = - = The difference is the net energy needed to move the fluid through the system (flow work). (The

energy change associated with the pressure change in Part (b) is insignif icant.)

7.25 $ , .H lH O C kJ kg2 b gc h20 83 9 = (Table B.5) $ .H steam, sat'd kJ kg20 bars, 2797 2b g = (Table B.6)

& &

& . ( )

m m

Q

[kg H O(l) / h] [kg H O(v) / h]

20 C 20 bar (sat'd)

= kW kW

2 2

o

0 65 813 528=

(a) D D D& & & & &H E E Q Wk p s+ + = - ; D &E k , D &E p , & & &W H Qs = =0 D D D& & $H m H=

&&$ . .m

QH

= =-

=D

5282797 2 83 9

701 kW kg 1 kJ / s 3600 s

kJ 1 kW 1 h kg hb g

(b) & . .V = =A

701 0 0995 69 7 kg h m kg m h sat'd steam @ 20 bar

Table B.6

3 3b gd i

(c) & & .V nRTP

= =

=

70178 5

kg / h 10 g / kg 485.4 K 0.08314 L bar 1 m

18.02 g / mol 20 bar mol K 10 L m / h

3 3

33

The calculation in (b) is more accurate because the steam tables account for the effect of pressure on specific enthalpy (nonideal gas behavior).

(d) Most energy released goes to raise the temperature of the combustion products, some is transferred to the boiler tubes and walls, and some is lost to the surroundings.

7- 12

7.26 $ , .H lH O C, 10 bar kJ kg2 b gc h24 100 6 = (Table B.5 for saturated liquid at 24oC; assume $H independent of P).

$ .H 10 bar, 27762 sat'd steam kJ kgb g = (Table B.6) D $ . . .H = - =2776 2 100 6 26756 kJ kg

2 2

o 3

[kg H O(l)/h] [kg HO(v)/h]

24 C, 10 bar 15,000 m /h @10 bar (sat'd)

(kW)

m m

Q

& &

&

& . .m = = A

1500001943 7 72 10

4 m kg

h m kg h

3

3

Table 8.6b g

Energy balance D D D& , & : & & &E W H E Qp s k= + =0d i

D D& & & & &&

E E E E Ek k k k kfinal initial finalEkinitial

= - =0

D & & .

.

Emu

A D

kf= =

=

=

A

2

2 3

2 2 3

5

2

1 1 1

4

015 4 2 1

596 10

2

7.72 10 kg 15,000 m h h J

h m 3600 s kg m / s

J / s

4 3

2 3 2 2

d i

p

p

& & $ & . .Q m H Ek= + =

+

= =

D D7 72 10 596 10

57973 5

4 5 kg 2675.6 kJ 1 h h kg 3600 s

J 1 kJ s 10 J

kJ s .80 10 kW

3

4

7.27 (a) 228 g/min 228 g/min 25oC T(oC)

& (Q kW)

D D& , & , &Ex E p Ws

Energy balance:=0

& & &Q H Q= =D Wb g 228 g min Jmin 60 s g

1 ( $ $ )H Hout in-

=$ . &H Q Wout J gb g b g0 263

T

H Q W

=

C

J gb g

b g b g25 264 27 8 29 0 32 4

0 263 0 4 47 9 28 134 24 8

. . . .$ . & . . . .

(b) $ $ .H b T b H T Tii i ii= - = - - = 25 25 25 3342b g b g b g Fit to data by least squares (App. A.1)

$ .H TJ g Cb g b g= -334 25 7.27 (contd)

=0

7- 13

(c) & &Q H= = - =D 350 kg 10 g 1 min 3.34 40 J kW smin kg 60 s g 10 J

kW3

3

20390

b g heat input to liquid (d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads.

7.28

& [ & [

& [ & [

m m

m m

w 2 w 2o

e 2 6 e 2 6o o

kg H O(v) / min] kg H O(l) / min] 3 bar, sat' d 27 C

kg C H / min] kg C H / min] 16 C, 2.5 bar 93 C, 2.5 bar

(a) C H mass flow: m L 2.50 bar 1 K - mol 30.01 g kg

min m 289 K 0.08314 L - bar mol 1000 g kg min

2 6

3

3&

.

me =

=

795 10 1

2 487 10

3

3

$ , $H Hei ef= =941 1073 kJ kg kJ kg

Energy Balance on C H :

kg

minkJkg

kJ min min 60 s

kW

2 6

3

D D D& , & , & & &

& . . .

E W E Q H

Q

p s k= @ =

= -LNM

OQP =

=

0 0

2 487 10 1073 9412 487 10 1

547 1033b g

(b) $ .H s1 3 2724 7.00 bar, sat' d vapor kJ kgb g = (Table B.6) $ .H s2 1131liquid, 27 C kJ kg =b g (Table B.5) Assume that heat losses to the surroundings are negligible, so that the heat given up by the

condensing steam equals the heat transferred to the ethane 547 103. kWd i Energy balance on H O: 2 & & & $ $Q H m H Hs s= = -D 2 1d i =

-=

- -

=&&

$ $.

. ..m

QH Hs s2 1

547 101131 2724 7

2 093 kJ kg

s kJ kg s steamb g

= =A

& . .Vs 2 09 0 606 1 kg / s m kg .27 m sTable B.6

3 3b gd i Too low. Extra flow would make up for the heat losses to surroundings. (c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it

would require heat flow from the ethane to the steam over some portion of the exchanger. (Observe the two outlet temperatures)

& ( )Q kW

7- 14

7.29 250 kg H O( )/minv240 bar, 500CH1$ (kJ/kg)

Turbine

W s =1500 kW

250 kg/min5 bar, T 2 (C), H2$ (kJ/kg)

Heatexchanger

Q(kW)

250 kg/min5 bar, 500CH3$ (kJ/kg)

H O 40 bar, 500 C kJ kg2 v H, : $ =b g 1 3445 (Table B.7) H O 5 bar, 500 C kJ kg2 v H, : $ =b g 3 3484 (Table B.7) (a) Energy balance on turbine: D D& , & , &E Q Ep k= = @0 0 0

D & & & $ $ & $ $ & &H W m H H W H H W ms s s= - - = - = -

= - =

2 1 2 1

3445 15003085

d i kJ

kg kJ min 60 s

s 250 kg 1 min kJ kg

$H P= = 3085 5 kJ kg and bars T = 310 C (Table B.7) (b) Energy balance on heat exchanger: sD D& , & , &E W Ep k= = @0 0 0

& & & $ $Q H m H H= = - = - =D 3 2250 3484 3085

1663d i b g kg kJ 1 min 1 kWmin kg 60 s 1 kJ / s

kW

(c) Overall energy balance: D D& , &E Ep k= @0 0

D & & & & $ $ & &H Q W m H H Q Ws s s= - - = -3 1d i

& & &Q H Ws= + =

-+

=

D D250 3484 3445 1500

1663

kg kJ 1 min 1 kWmin kg 60 s 1 kJ / s

kJ 1 kWs 1 kJ / s

kW

b g

(d) H O 40 bar, 500 C m kg2

3v V, : $ . =b g 1 0 0864 (Table B.7) H O 5 bar, 310 C m kg2

3v V, : $ . =b g 2 05318 (Table B.7) u1

4183= =

250 kg 1 min 0.0864 m 1min 60 s kg 0.5 m

m s3

2 2p.

u24

113= =250 kg min 0.5318 m 1

min 60 s kg 0.5 m m s

3

2 2p.

D& & . .

.

Em

u uk = - =-

=

2250 113 183

026

22

12

2 2 2 kg 1 1 min m 1 N 1 kW s

min 2 60 s s 1 kg m / s 10 N m kW

7- 15

7.30 (a) D &E p , D &E k , & & & .W Q H hA T T h T Ts s o s o= = - - = - - =0 300 18 300D b g b g kJ h kJh (b) Clothed: 8

= .C

s

hT

To= = 34 2

134.

Nude, immersed: 64= .

Cs

hT

To= = 34 2

316. (Assuming Ts remains 34.2C)

(c) The wind raises the effective heat transfer coefficient. (Stagnant air acts as a thermal insulator i.e., in the absence of wind, h is low.) For a given To , the skin temperature must drop to satisfy the energy balance equation: when Ts drops, you feel cold.

7.31 Basis: 1 kg of 30C stream

1 kg H2O(l)@30oC

3 kg H2O(l)@Tf(oC)

2 kg H2O(l)@90oC

(a) Tf = + =13

3023

90 70o C C Cb g b go o (b) Internal Energy of feeds: C, liq. kJ kg

C, liq. kJ kg

$ .$ .

UU

30 125790 376 9

= =

UV|W|b gb g

(Table B.5 - neglecting effect of P on $H ) Energy Balance: - = + +

= = = Q W U E E Up k

Q W E p EkD D D DD D =

=0

0

- - =3 1 1257 2 376 9 0$ ( . ( .U f kg) kJ / kg kg) kJ / kgb g b g = = $ . .U Tf f2932 70 05 kJ kg C (Table B.5)

Diff.= - =70 05 70 0070 05

100% 007%. .

.. (Any answer of this magnitude is acceptable).

7.32 .

.

.

.Q (kW)

52.5 m3 H2O(v)/hm(kg/h)5 bar, T(oC)

m(kg/h)0.85 kg H O( )/kgv20.15 kg H O( )/kgl25 bar, saturated, T(oC)

(a) Table B.6 C barsP

T=

= 5

1518. , $ . $ .H HL V kJ kg , kJ kg= =6401 27475

$ & .V(5 bar, sat'd) = 0.375 m / kg m m 1 kg h 0.375 m

kg h33

3 = =52 5 140

(b) H O evaporated kg h kg h2 = =015 140 21.b gb g (Q = energy needed to vaporize this much water)

( )Energy balance:

21 kg 2747.5 640.1 ] kJ 1 h 1 kW 12 kW

h kg 3600 s 1 kJ s

Q H= D

-= =

& &

7-16

7.33 (a) P T o= =5 bar CTable B.6 saturation 1518. . At 75C the discharge is all liquid

(b) Inlet: T=350C, P=40 bar Table B.7 in = 3095 kJ / kg$H , $Vin 3= 0.0665 m / kg

Outlet: T=75C, P=5 bar Table B.7 out = 314.3 kJ / kg$H , $Vout -3 3= 1.03 10 m / kg

uVA

uVA

inin

in

3

2 2

outout

out

3

2 2

kg 1 min m / kg min 60 s 0.075) / 4 m

m / s

kg 1 min 0.00103 m / kg

min 60 s 0.05) / 4 m m / s

= = =

= = =

& .(

.

&(

.

200 0 06655018

200175

p

p

Energy balance: & & & & & ( $ $ ) & ( )Q W H E m H H m u us k- + = - + -D D 2 1 22 122

2 2 2

2

200 kg 1 min (314-3095) kJ 200 kg 1 min (1.75 -50.18 ) m min 60 s kg 2 min 60 s s

13,460 kW ( 13,460 kW transferred from the turbi

sQ W- = +

= -

& &

ne)

7.34 (a) Assume all heat from stream transferred to oil

1.00 10 kJ 1 min min 60 s

kJ s4&Q = = 167

25 bars, sat'd

100 kg oil/min

&m (kg H2O(v)/s)25 bars, sat'd

135C100 kg oil/min185C &m (kg H2O(l)/s)

Energy balance on H O: 2 out in

& & & $ $& , & , &

Q H m H H

E E Wp k s

= = -

=

D

D D

d i0

$ , .H l 25 bar, sat' d kJ kg( ) = 962 0 , $ , .H v 25 bar, sat' d kJ kg( ) = 2800 9 (Table B.6)

&&

$ $ . . .mQ

H H=

-=

-- =

out in

kJ kgs kJ kg s

167962 0 2800 9 0 091b g

Time between discharges: g 1 s 1 kg

discharge 0.091 kg 10 g s discharge3

120013=

(b) Unit Cost of Steam: $1 kJ 0.9486 Btu

10 Btu kg kJ/ kg6

2800 9 8396 10 3

. .$2.

-= -b g

Yearly cost:

1000 traps 0.091 kg stream 0.10 kg last 2.6 10 $ 3600 s 24 h 360 daytrap s kg stream kg lost h day year

year

=

-3

54 10$7. /

7-17

7.35 Basis: Given feed rate

200 kg H2O(v)/h 10 bar, satd, $ .H = 2776 2 kJ / kg & [n3 kg H O(v) / h]2 10 bar, 250oC, $H = 2943 kJ / kg

& [

$n

H2

3052

kg H O(v) / h]

10 bar, 300 C, kJ / kg2

o =

&Q(kJ / h) $H from Table B.6 (saturated steam) or Table B.7 (superheated steam) Mass balance: 200 2 3+ =& &n n (1)

D DD

& , & , && & & . & , &

E E WK p

Q H n n Q

Energy balance: in kJ h=

= = - -0

3 22943 200 2776 2 3052b g b g b g (2) (a) &n3 300= kg h ( ) &1 1002n = kg h ( ) & .2 2 25 104Q = kJ h

(b) &Q = 0 ( ),( ) &1 2 3062n = kg h , &n3 506= kg h

7.36 (a) Tsaturation @ 1.0 bar = 99.6 C =Tf 99 6.o C

H O (1.0 bar, sat'd) kJ / kg, kJ / kg

H O (60 bar, 250 C) kJ / kg2

2

= =

=

$ . $ ..

H Hl v417 5 2675 4

1085 8o

Mass balance: kgEnergy balance:

kg)(1085.8 kJ / kg) = 0 (2)

,

m m

H

m H m H m H m H m H

v l

E Q E W

v v l l v v l l

K p

+ =

=

+ - = + -

=

100 1

0

100

0

1 1

( )&

$ $ $ $ $ (& , & & , &D D

D

( , ) . .12 704 29 6m ml v= = kg, kg yv = =29 6

0 296.

. kg vapor

100 kgkg vapor

kg

(b) is unchanged.T The temperature will still be the saturation temperature at the given final

pressure. The system undergoes expansion, so assuming the same pipe diameter, 0.kED >& would be less (less water evaporates) because some of the energy that would have

vaporized water instead is converted to kinetic energy.

vy

(c) Pf = 398. bar (pressure at which the water is still liquid, but has the same enthalpy as the feed)

7-18

7.36 (contd)

(d) Since enthalpy does not change, then when Pf 39 8. bar the temperature cannot increase,

because a higher temperature would increase the enthalpy. Also, when Pf 39 8. bar , the product is only liquid no evaporation occurs.

0

0.1

0.2

0.3

0.4

0 20 40 60 80

Pf (bar)

y

050

100150200250300

1 5 10 15 20 25 30 36 39.8 60

Pf (bar)

Tf (

C)

7.37 10 m3, n moles of steam(v), 275C, 15 bar 10 m3, n moles of water (v+l), 1.2 bar

Q 1.2 bar, saturated

10.0 m3 H2O (v)

min (kg)275oC, 1.5 bar

10.0 m3

mv [kg H2O (v)]ml [kg H2O (l)]

(a) P=1.2 bar, saturated, CTable B.6 T2 104 8= .o

(b) Total mass of water:min3

3=10 m kg

0.1818 m kg

155=

Mass Balance:

Volume additivity: m m kg) m kg)

kg, kg condensed

3 3 3

m m

V V m m

m m

v l

v l v l

v l

+ =

+ = = +

= =

55 0

10 0 1428 0 001048

7 0 480

.

. ( . / ( . /

. .

(c) Table B.7 = 2739.2 kJ / kg; = 0.1818 m / kg

Table B.6 = 439.2 kJ / kg; = 0.001048 m / kg= 2512.1 kJ / kg; = 1.428 m / kg

in in3

3

3

RS|T|

$ $$ $$ $

U V

U VU V

l l

v v

D DD

E E Wv v l l

p k

Q U m U m U m U, ,

$ $ $

[( . ) ( . )( .

in in

5

Energy balance: = =

(2512.1 kJ / kg) + ) - kg (2739.2)] kJ

= 1.12 10 kJ

=+ -

=

-

0

7 0 48 0 439 2 55

7.38 (a) Assume both liquid and vapor are present in the valve effluent.

1 kg H O( ) / s

15 bar, T C2

sato

v

+ 150 & [& [

m lm v

l

v

2

2

kg H O ( ) / s]kg H O( ) / s]

1.0 bar, saturated

, 15 bar

7-19

7.38 (contd) (b) Table B.6 T bar) =198.3 C T C

Table B.7 C, 15 bar) 3149 kJ / kg Table B.6 1.0 bar, sat'd) = 417.5 kJ / kg; 1.0 bar, sat'd) = 2675.4 kJ / kg

sat'no

ino

in

= =

( .$ $ ( .$ ( $ (

15 3483348 3H H

H Hl v

o

D DD

& , & , & , && & $ & $ & $

& $ & $ & $ & ( . ) ( & )( . )& &E E Q W

l l v v

l l v v l l

p k s

H m H m H m H

m H m H m H m mm mv l

in in

in in

Energy balance:

kJ / kg

== + - =

= + = + -+0

0 0

3149 417 5 1 2675 4

There is no value of &ml between 0 and 1 that would satisfy this equation. (For any value in this range, the right-hand side would be between 417.5 and 2675.4). The two-phase assumption is therefore incorrect; the effluent must be pure vapor.

(c) Energy balance = = =

& $ & $ & & $ ( )m H m H m m HT

out out in inin out

out

out

3149 kJ / kg = bar, T

337 CTable B.7

11

o

(This answer is only approximate, since D &E k is not zero in this process). 7.39 Basis: 40 lb min circulationm (a) Expansion valve

R = Refrigerant 12

40 lbm R(l)/min

H$ = 27.8 Btu/lbm93.3 psig, 86F

40

1

77 8 9 6

lb min lb R lb

lb R( / lb

Btu / lb Btu / lb

m

m m

m m

m m

/( ) /

( ) )$ . , $ .

x vx l

H H

v

v

v l

-

= = Energy balance: neglect

out in

D D D& , & , & , & & & $ & $E W Q E H n H n Hp s k i i i i= = - = 0 0

40 778 40 1 9 40 27.8 Btu 0X R v X R lv v lb Btu min lb

lb .6 Btu min lb

lb min lb

m

m

m

m

m

m

b g b g b g. + - - = X v =

E0 267. 26.7% evaporatesb g

(b) Evaporator coil

40 lb /minm

H$v = 77.8 Btu/lb ,11.8 psig, 5F

m

40

H$ = 77.8 Btu/lb11.8 psig, 5F

m

0.267 R( )v0.733 R( )l

H$l = 9.6 Btu/lbm

lb R( )/minm v

Energy balance: neglect D D D& , & , & & &E W E Q Hp s k= =0

& . .Q R v R l= - -

=

40 778 40 0267 77.8 Btu 9

2000

lb Btu min lb

lb min lb

40 0.733 lb .6 Btu min lb

Btu min

m

m

m

m

m

m

b gb g b g b gb g b g

7-20

7.39 (contd)

(c) We may analyze the overall process in several ways, each of which leads to the same

result. Let us first note that the net rate of heat input to the system is

& & &Q Q Q= - = - = -evaporator condenser Btu min2000 2500 500

and the compressor work Wc represents the total work done on the system. The system is

closed (no mass flow in or out). Consider a time interval Dt minb g . Since the system is at steady state, the changes DU , DE k and DE p over this time interval all equal zero. The

total heat input is &Q tD , the work input is &W tc D , and (Eq. 8.3-4) yields & & & & .

..Q t W t W Qc cD D- = = =

-

=-

-0500 1341 10

9 486 10118

3

4 Btu 1 min hp

min 60 s Btu s hp

7.40 Basis: Given feed rates

&

.

n1

080

(mol / h)

0.2 C H C HC, 1.1 atm

3 84 10

o

&&nn

C H 3 8

C H 4 10o

3 8

4 10

(mol C H / h) (mol C H / h)

227 C

&Q (kJ / h)&n2

3 84 10

o

(mol / h)0.40 C H.60 C H

25 C, 1.1 atm0

Molar flow rates of feed streams:

300 L 1.1 atm 1 mol hr 1 atm 22.4 L STP mol h& .n1 14 7= =b g

& .n2 9 00= =200 L 273 K 1.1 atm 1 mol hr 298 K 1 atm 22.4 L STP mol hb g

Propane balance 14.7 mol 0.20 mol C H

h mol9.00 mol 0.40 mol C H

h mol mol C H h

C H3 8 3 8

3 8

3 8 = +

=

&.

n

654

Total mole balance: mol mol C H h C H hC H 4 20 4 204 10& ( . . . ) .n = + - =14 7 9 00 654 1716 Energy balance: neglect D D D& , & , & & &E W E Q Hp s k= =0

& & & $ & $Q H N H N Hi i i i= = - = +

- - =

Dout in

3 8 4 10

3 8 4 10

6.54 mol C H 20.685 kJ h mol

17.16 mol C H 27.442 kJ h mol

0.40 9.00 mol C H 1.772 kJ h mol

0.60 9.00 mol C H 2.394 kJ h mol

kJ hb g b g 587

( $Hi = 0 for components of 1st feed stream)

7-21

7.41 Basis: 510 m 273 K L 1 mol 1 kmol min 291 K m 22.4 L STP 10 mol kmol min3

3 310 214

3

b g = . (a)

38C, ( mol H O(v)/ mol ) 2

18C, sat'd y

( mol dry air/ mol ) x (1 )

21.4 kmol /min

( mol dry air)

n 0 ( kmol /min) h r = 97%

0 0

( kmol H O( )/ mol ) 2 n 2 l 18C

y (1 )

1 1 y

( mol H O(v)/ mol) 2

.

.

( ) ( )2H O

2

38 C 0.97 49.692 mm HgInlet condition: 0.0634 mol H O mol

760 mm Hgr

o

h Py

P

* = = =

( )

2H O1 2

18 C 15.477 mm HgOutlet condition: 0.0204 mol H O mol

760 mm Hg

Py

P

* = = =

Dry air balance: kmol min1 0 0634 1 0 0204 214 22 4- = - =. & . . & .b g b gn no o

Water balance: 0.98 kmol min

kmol kg kmol 18 kg / min H O condenses2

00634 22 4 0 0204 214

098 18022 2. . & . . &

. .min

b g b g= + ==

n n

(b). Enthaphies: C kJ molair$ . .H 38 0 0291 38 25 0 3783 = - =b g b g $ . .H air C kJ mol18 0 0291 18 25 0204 = - = -b g b g

$ , . .

$ , . .$ , .

H v

H v

H l

H O 3

H O 3

H O 3

2

2

2

38 C kJ 1 kg 18.02 g

kg 10 g mol kJ mol

18 C kJ 1 kg 18.02 g

kg 10 g mol kJ mol

18 C 1 kg 18.02 g

kg 10 g mol kJ mol

Table B.5

= =

= =

= =

U

V|||

W|||

b gb gb g

2570 846 33

253454567

75.5 kJ136

D D

D

& & &, ,& & & $ & $ & . . .

. . . . . . . .

. . . .

E W Ep s kQ H n H n H Qi i i i

Energy balance:

kJ min

out in

= @

= = - = - -

+ + - -

- = -

0 0

1 00204 214 10 0204

0 0204 214 10 4567 0 98 10 136 1 0 0634 22 4 10 0 3783

0 0634 22 4 10 46 33 567 10

3

3 3 3

3 4

b gd ib gb gd ib g d ib g b gd ib gb gd ib g

=567 10 270 tons o4. kJ 60 min 0.9486 Btu 1 ton cooling

min h kJ 12000 Btuf cooling

(kJ/min)Q&

7-22

7.42 Basis: 100 mol feed

n

0.65 A( ) l

2 (mol), 63.0C

100 mol, 67.5C

0.35 B( ) l

0.98 A( ) v 0.02 B( ) v

n 2 (mol) 0.98 A( ) l 0.02 B( ) l

0.5

n 5 (mol), 98.7C 0.544 A( ) v 0.456 B( ) v

56.8C

Q c (cal)

n 2 (mol) 0.98 A( ) l 0.02 B( ) l

0.5

Q r (cal)

n 5 (mol), 98.7C 0.155 A( ) l 0.845 B( ) l

A - Acetone B - Acetic Acid

(a) Overall balances:

Total moles: 100

0.65 100 mol

mol= +

= +UVW

==

050 98 05 0155

12040

2 5

2 5

2

5

.: . . .

n nA n n

nnb g b g

Product flow rates: Overhead 0.5 120 mol

0.5 120 mol b gb g

0 98 5880 02 12. .. .

==

AB

Bottoms 0 40 mol

0 40 mol . .. .155 6 2845 338

b gb g

==

AB

D D

DE W Ep x

Q H n H n Hi i i i, ,

$ $

Overall energy balance: out in2 0 0= @

= = -

( ) ( ) ( ) ( ) ( ) ( )interpolate in table interpolate in table

458.8 0 1.2 0 6.2 1385 33.8 1312 65 354 35 335 1.82 10 calQ

= + + + - - = (b) Flow through condenser: mols 2 58 8 117 6. .b g = A 2 12 2 4. .b g = mols B

D DD

E W Ep kQ Hc

, ,

Energy balance on condenser: 3 0 0= @

=

Qc = - + - = - 117 6 0 7322 2 4 0 6807 8 77 105. . .b g b g cal heat removed from condenser

Assume negligible heat transfer between system & surroundings other than Qc & Qr

( )4 5 51.82 10 8.77 10 8.95 10 calr cQ Q Q= - = - - = heat added to reboiler 7.43

Q= 0

1.96 kg, P1= 10.0 bar, T1

1.00 kg, P2= 7.0 bar, T

2

2.96 kg, P3= 7.0 bar, T

3=250oC

7-23

7.43 (contd) (a) T T P2 7 0= =( . bar, sat'd steam) =165.0 C

o

$ ( ( ),$ ( ( ), (

H vH v

3

2

29542760

H O P = 7.0 bar, T = 250 C) kJ kg (Table B.7)H O P = 7.0 bar, sat'd) kJ kg Table B.6)

2o

2

==

D D

D

E Q W Ep s k

H H H H H

H

, ,,

. $ . $ . $ . $ .$ ( . )

Energy balance

kg(2954 kJ / kg) -1.0 kg(2760 kJ / kg)

bar, T kJ / kg T C1 1

@

= = - - =

= @

0

0 2 96 196 10 196 2 96

10 0 3053 3003 1 2 1

1o

(b) The estimate is too low. If heat is being lost the entering steam temperature would have to be higher for the exiting steam to be at the given temperature.

7.44 (a) T T P

V P

V Pl

v

1 30

30

30

= =

=

=

( .

$ ( .$ ( .

bar, sat' d. ) =133.5 C

bar, sat' d. ) = 0.001074 m / kg

bar, sat'd.) = 0.606 m / kg

3

3

o

V

V

m

l

space

v

= =

=

= =

0001074 177 2

200 0

22.8 L 10 606

0 0376

. .

.

..

m 1000 L 165 kg kg m

L

L -177.2 L = 22.8 L

m 1 kg1000 L m

kg

3

3

3

3

m=165.0 kg

P=3 bar

V=200.0 LPmax=20 bar

Vapor

Liquid

(b) P P mtotal= = = + =max . . . .20 0 165 0 0 0376 165 04 bar; kg

T T P

V P V Pl v

1 20 0

20 0 200

= =

= =

( .

$ ( . $ ( . bar, sat'd. ) = 212.4 C

bar, sat' d. ) = 0.001177 m / kg; bar, sat'd.) = 0.0995 m / kg3 3

o

V m V m V m V m m V

m m

m m

total l l v v l l total l v

l l

l v

= + + -

=

= =

$ $ $ ( ) $. ( . / ) ( . /

. .

L m L

kg m kg) + (165.04 - kg m kg)

kg; kg

33 3200 0 1

10000 001177 0 0995

164 98 006

V V

m

l space

evaporated

= = =

= =

0 001177194 2 200 0

100020

.. ; .

(

m 1000 L 164.98 kg kg m

L L - 194.2 L = 5.8 L

0.06 - 0.04) kg g kg g

3

3

(c)

D DD

E W Ep s kU U P U P

, ,

( . ( .

Energy balance Q = bar, sat'd) bar, sat'd)@

= = - =0

20 0 3 0

$ ( . $ ( .$ ( . $ ( .

U P U P

U P U Pl v

l v

= =

= =

20 0 20 0

30 3 0

bar, sat'd.) = 906.2 kJ / kg; bar, sat'd. ) = 2598.2 kJ / kg

bar, sat'd. ) = 561.1 kJ / kg; bar, sat'd.) = 2543 kJ / kg

Q =

-

006. kg(2598.2 kJ / kg) +164.98 kg(906.2 kJ / kg) - 0.04 kg(2543 kJ / kg)

165.0 kg (561.1 kJ / kg) = 5.70 10 kJ4

Heat lost to the surroundings, energy needed to heat the walls of the tank

7-24

7.44 (contd)

(d) (i) The specific volume of liquid increases with the temperature, hence the same mass of liquid water will occupy more space; (ii) some liquid water vaporizes, and the lower density of vapor leads to a pressure increase; (iii) the head space is smaller as a result of the changes mentioned above.

(e) Using an automatic control system that interrupts the heating at a set value of pressure A safety valve for pressure overload.

Never leaving a tank under pressure unattended during operations that involve temperature and pressure changes.

7.45 Basis: 1 kg wet steam (a) H O21 kg 20 bars

0.97 kg H O(v)20.03 kg H O(l)2H$1 (kJ/kg)

H O,(v) 1 atm21 kg

H$2 (kJ/kg)H O21 kg

Tamb , 1 atm

Enthalpies: bars, sat'd kJ kg

bars, sat' d kJ kgTable B.7

$ , .$ , .

H v

H l

20 2797 2

20 908 6b gb g b g

=

=

UV|W|

D DD

E E Q Wp KH H H

H T

, , ,

$ $ . . . .

$ =

Energy balance on condenser:

= 2740 kJ / kg C Table B.7

o

3 00 0 97 2797 2 0 03 9086

132

2 1

2

= = = +

b g b g

(b) As the steam (which is transparent) moves away from the trap, it cools. When it reaches

its saturation temperature at 1 atm, it begins to condense, so that T = 100 C . The white plume is a mist formed by liquid droplets.

7.46 Basis: oz H O 1 quart 1 m 1000 kg

32 oz 1057 quarts m kg H O2

3

3 28

02365l

lb g b g= .

(For simplicity, we assume the beverage is water)

0.2365 kg H2O (l)18C

32F (0C)4Cm (kg H2O (s))

(m + 0.2365) (kg H2O (l))

Assume P = 1 atm Internal energies (from Table B.5):

$ . $ ) . $ )U l U l U sH O( ), 18 C kJ / kg; H O( , 4 C kJ / kg; H O( , 0 C = -348 kJ / kg 2 2 2 = = b g b g b g755 168

Energy balance closed system

out in

b g: $ $, , ,

= - = =

D

D D

U n U n Ui i i i

E E Q Wp k

0

0

+( . ) ( .m m02365 16 8kg kJ / kg) = 0.2365 kg(75.5 kJ / kg) + kg (-348 kJ / kg) =m 0 038 38. kg = g ice

Q=0 Q

7-25

7.47 (a) When T H= = =0 0 0o refoC, T C $ ,

(b) Energy Balance-Closed System: DU = 0 D DE E Q W

k p, , , = 0

(C)

25 g Fe, 175C

T20C

25 g Fe

f

1000 g H2O(l) 1000 g H2O

U T U T U Uf fFe H O Fe H O2 2C C, 1 atmd i d i b g b g+ - - =175 20 0 or D DU UFe H O2+ = 0 DU T Tf fFe

g 4.13 cal 4.184 J g cal

J=-

= -250 175

432 175. d i

Table B.5 1.0 L g J

1 L gJ

432

H OH O

H O

H O

22

2

2

= - = -

+ - = =

DU U T U T

T U T f T

ff

f f f

10 839 1000 839

1000 160 10 0

3

5

$ . $ .

$ .

d ie j d ie jd i d i

Tf T

Tf

ff

- + -

= C Interpolate

C30 40 35 34

21 10 2 5 10 1670 261234 64 4d i . . .

7-26

7.48

H O( )v

T0

I

2760 mm Hg100C

H O( ), 100 Cl2

H O( )v

Tf

II

2(760 + 50.1) mm Hg

H O( ), l2

Tf

Tf

1.08 bar sat'dTf = 101.8C (Table 8.5)

Energy balance - closed system: D DE E W Qp K, , , = 0

DU m U m U m U m U m U m Uv v lI

l b b v v l l b b

v

l

b

= = + + - - -0 II II I II II II I I I I I I-vapor

-liquid

-block

$ $ $ $ $ $

I IIVVUU

l

v

l

v

101 1081044 10461673 1576419 0 426 62506 5 2508 6

. .$ . .$$ . .$ . .

bar, 100 C bar, 101.8 CL kgL kgL kgL kg

b g b gb gb gb gb g

Initial vapor volume: 20.0 L L kg 1 L

8.92 kg L H OI 2V vv = - - =50

5014 4. . b g

Initial vapor mass: = 14.4 L 1673 L kg kg H OI 2m vv b g b g= -8 61 10 3. Initial liquid mass: 5 L 1.044 L kg kg H OI 2m ll = =. .0 4 79b g b g Final energy of bar: .36 101.8 kJ kgII$ .Ub = =0 36 6b g Assume negligible change in volume & liquid =Vv

II L14 4.

Final vapor mass: 14.4 L 1576 L kg kg H OII 2m vv = = -b g b g914 10 3.

Initial energy of the bar:

$

.. . . . . . . . . .

.

U bI

kg kJ kg

= + + - -

=

- -150

914 10 2508 6 4 79 4266 50 366 861 10 25065 4 79 419 0

441

3 3b g b g b g b g b gd i

(a) Oven Temperature: kJ / kg0.36 kJ / kg C

Co

To = = 441 122 5. .

H O kg - kg = kg = 0.53 g2 evaporatedII I= - = - - -m mv v 914 10 861 10 530 10

3 3 4. . .

(b) $ . . . .U bI kJ kg= + =44 1 8 3 50 458 To = = 458 036 127 2. . . C (c) Meshuggeneh forgot to turn the oven on ( To < 100 C )

7-27

7.49 (a) Pressure in cylinder = +weight of pistonarea of piston

atmospheri c pressure

P

Tsat

= + =

=

30.0 kg 9.807 N cm bar

400.0 cm kg 1 m 10 N m

atm 1.013 bar

atm bar

C

2 2 5 2

100 10 1108

1018

2

2

b gb g

..

.

Heat required to bring the water and block to the boiling point

Q U m U U m U T Uw wl wl Al Al sat Al= = - + -

=-

+-

=

D $ . $ , $ $. . ( . )

108 20

839 0 94 1018 202630

bar, sat'd l 20 C C

7.0 kg 426.6 kJ kg

3.0 kg [ ]kJ kg

kJ

b g b gd i b g b gd ib g

2630 kJ < 3310 kJ Sufficient heat for vaporization

(b) T Tf sat= = 1018. C . Table B.5 = == =

$ . , $ .$ , $ .V U

V Ul l

v v

1046 426 61576 2508 6

L kg kJ kg L kg kJ kg

T 101.8CP 1.08 bars

1576 L/kg, 2508.6 kJ/kg

1.046 L/kg, 426.6 kJ/kg

Q (kJ) W (kJ)

m vv (kg H O( ))2

m ll (kg H O( ))2

7.0 kg H O( )2 l

$ .$H = 426 6 kJ / kgV =1.046 L / kg

(Since the Al block stays at the same temperature in this stage of the process, we can

ignore it -i.e., $ $U Uin out= ) Water balance: 7 0. = +m ml v (1)

Work done by the piston:

bar L

8.314 J / mol K 1 kJ

0.08314 liter - bar / mol K 10 JkJ

piston atm

atm

3

W F z w P A z

wA

P A z P V W m m

m m

v l

v l

= = +

= +LNMOQP = = + -

= + -

D D

D Db g b g b gb g

b g

108 1576 1046 1 046 7 0

170 2 0 113 0 7908

. . . .

. . .

Energy balance: DD

U Q W

m m m mm m

v L

U Q

v L

W

v L

= -

+ - = - - + - + - =

25086 426 6 426 6 7 3310 2630 1702 0113 0 79082679 4267 3667 0

. . . ( ) ( . . . ).

b g6 744444 844444 6 744 844 6 744444 844444

(2)

Solving (1) and (2) simultaneously yields mv = 0 302. kg , ml = 6 698. kg

Liquid volume kg L kg L liquid= =6698 1046 7 01. . .b gb g Vapor volume .302 kg L kg 476 L vapor= =0 1576 b gb g Piston displacement:

L 10 cm 1

1 L 400 cm

3 3

2D

Dz

VA

= =+ -

=701 476 7 0 1046

1190 cm. . .b gb g

(c) Tupper All 3310 kJ go into the block before a measurable amount is transferred to the

water. ThenDU Q T TAL u u= - = = 30 094 20 3310 1194. . kg kJ kg Cb g b g if melting is neglected. In fact, the bar would melt at 660oC.

7-28

7.50 100. ), L H O( 25 C

m (kg)2

o

v1

v

4 00. ), L H O( 25 Cm (kg)

2o

L1

l

m [kg H O( ]= m m

v2 2v1 e

v)+

m [kg H O( ]= m m

L2 2L1 e

l)+

Q=2915 kJ

UVW =Assume not all the liquidis vaporized. Eq. at

kg H O vaporized.2T P mf f e, .

Initial conditions: Table B.5 kJ kg =$ .U L1 104 8 , $ .VL1 1003= L kg P = 00317. bar T = 25 C, sat' d $ .U v1 2409 9= kJ kg , $ ,VL1 43 400= L kg mv1

543400 2 304 10= = -1.00 l l kg kgb g b g . , mLI = =4 00 1 3988. . l .003 l kg kgb g b g

Energy balance:

DU Q m U T m U Te v f e L f= + + - - - -2 304 10 3988 2 304 10 2409 95 5. $ . $ . .d i d i b g d i d ib g

- =3988 1048 2915. ( . )b g kJ 2 304 10 3988 33335. $ . $ + + - =-

Em U T m U Te v f e v fd i d i b g d i

=- -

-

-

mU U

U Ue

v L

v L

3333 2 304 10 39885. $ . $$ $

d i (1)

V V V m V T m V T

mV V

V V

L v e L f e L f

ev L

v L

+ = +FHGG

IKJJ

+ - =

=- -

-

-

A A-

tan . $ . $ .

. . $ . $$ $

k

kg liters kg

L

2 304 10 3988 500

500 2 304 10 3988

5

5

d i b g d i

d i b g2

1 2

3333 2 304 10 3988

500 2 304 10 39880

5

5

b g b g d i d i d i d i

d i- =

- -

-

-- -

-=

-

-

f TU T U T

U U

V V

V V

fv f L f

v L

v L

v L

. $ . $$ $

. . $ . $$ $

Procedure: Assume Table 8.5

T U U V V f Tf v L v L f $ , $ , $ , $ d i Find T f such that f T fd i = 0

T U U V V f

T P

f v L v L

f f

$ $ $ $. . . . . .. . . . . .. . . . . .. . . . . . . .

201 4 25938 856 7 123 7 1159 512 10198 3 2592 4 842 9 1317 1154 193 10195 0 2590 8 828 5 140 7 1149 134 10196 4 25915 834 6 136 9 1151 4 03 10 196 4 14 4

2

2

2

4

- -

- @ =

-

-

-

- C, bars

or Eq 2

Eq 1

kg 2.6 g evaporatedb g

b gme =

-2 6 10 3.

7-29

7.51. Basis: 1 mol feed

(1 )

B = benzene

z

T = toluene

1 mol @ 130CB (mol B(l)/mol)

z B (mol T(l)/mol)

(1 )

nVy

(mol vapor)B(mol B(v)/mol)

yB (mol T(v)/mol)

(1 )

nLx

(mol liquid)B(mol B(l)/mol)

xB (mol T(l)/mol)

in equilibriumat T(C), P(mm Hg)

(a) 7 variables: ( , , , ,n y n x Q T PV B L B , , ) 2 equilibrium equations 2 material balances 1 energy balance 2 degrees of freedom. If T and P are fixed, we can calculate and . n y n x QV B L B, , , ,

(b) Mass balance: n n n nV L V+ = = -1 1 2 (1) Benzene balance: z n y n xB V B L B= + (2)

C H : 6 6 l T Hb g d i= =0 0, $ , T H H TBL= = =80 1085 01356, $ . $ . d i (3) C H : 6 6 v T Hb g d i= =80 4161, $ . , T H H TBV= = = +120 4579 01045 3325, $ . $ . . d i (4)

C H 7 8 l T Hb g d i: , $= =0 0 , T H H TTL= = =111 1858 01674, $ . $ . d i (5) C H 7 8 v T Hb g d i: , $ .= =89 4918 , T H H TTV= = = +111 52 05 01304 37 57, $ . $ . . d i (6) Energy balance: neglect D DE W Ep s k, ,= 0

Q H n y H n y H n x H n x H z H T

z H TV B BV V B TV L B BL L B TL B BL F

B TL F

= = + - + + - -

- -

D $ $ $ $ $$1 1 1

1 1

b g b g b g b gb gb g b g (7)

Raoult's Law:

(1 -

y P x p

y P x pB B B

B B T

=

= -

*

*) ( )1

(8)

(9)

Antoine Equation. For T= 90C and P=652 mmHg:

* [6.89272 1203.531/(90 219.888)]

* [6.95805 1346.773/(90 219.693)]

(90 C) 10 1021 mmHg

(90 C) 10 406.7 mmHg

oB

oT

p

p

- +

- +

= =

= =

Adding equations (8) and (9)

P x p x p x

P pp p

P pp p

l

yx p

Pv

B B B T BT

B T

T

B T

BB B

= - =--

=--

=-

=

= = =

* **

* *

*

* *

*

( ).

.

. ( ).

+ 1021 - 406.7

mol B( ) / mol

mmHg

mmHg mol B( ) / mol

1652 4067

0399

0399 1021652

0625

Solving (1) and (2) =--

=-

-=

= - = - =

mol vapor

mol liquid

nz xy x

n n

VB B

B B

L V

0 5 03990 625 0 399

0 446

1 1 0 446 0554

. .. .

.

. .

7-30

7.51 (contd)

Substituting (3), (4), (5), and (6) in (7)

Q

Q

= + + - ++ + - -- =

0446 0 625 01045 90 3325 0 446 1 0625 01304 90 37 570554 0399 01356 90 0554 1 0 399 01674 90 05 01356 13005 01674 130 814

. ( . )[ . ( ) . ] . ( . )[ . ( ) . ]. ( . )[ . ( )] . ( . )[ . ( )] . [ . ( )]. [ . ( )] .

kJ / mol

(c). If PPmax, all the output is liquid.

(d) At P=652 mmHg it is necessary to add heat to achieve the equilibrium and at P=714 mmHg, it is necessary to release heat to achieve the equilibrium. The higher the pressure, there is more liquid than vapor, and the liquid has a lower enthalpy than the equilibrium vapor: enthalpy out < enthalpy in.

zB T P pB pT xB yB nV nL Q 0.5 90 652 1021 406.7 0.399 0.625 0.446 0.554 8.14 0.5 90 714 1021 406.7 0.500 0.715 -0.001 1.001 -6.09 0.5 90 582 1021 406.7 0.285 0.500 0.998 0.002 26.20 0.5 90 590 1021 406.7 0.298 0.516 0.925 0.075 23.8 0.5 90 600 1021 406.7 0.315 0.535 0.840 0.160 21.0 0.5 90 610 1021 406.7 0.331 0.554 0.758 0.242 18.3 0.5 90 620 1021 406.7 0.347 0.572 0.680 0.320 15.8 0.5 90 630 1021 406.7 0.364 0.589 0.605 0.395 13.3 0.5 90 640 1021 406.7 0.380 0.606 0.532 0.468 10.9 0.5 90 650 1021 406.7 0.396 0.622 0.460 0.540 8.60 0.5 90 660 1021 406.7 0.412 0.638 0.389 0.611 6.31 0.5 90 670 1021 406.7 0.429 0.653 0.318 0.682 4.04 0.5 90 680 1021 406.7 0.445 0.668 0.247 0.753 1.78 0.5 90 690 1021 406.7 0.461 0.682 0.176 0.824 -0.50 0.5 90 700 1021 406.7 0.477 0.696 0.103 0.897 -2.80 0.5 90 710 1021 406.7 0.494 0.710 0.029 0.971 -5.14

(e). P = 714 mmHg, P = 582 mmHgmax min

nV vs. P

0

0.2

0.4

0.6

0.8

1

582 632 682 732

P (mm Hg)

nV

n PV = @05 640. @ mmHg

7-31

7.52 (a). Bernoulli equation: D D

DP u

g zr

+ + =2

20

DPr

= -

= -

-0 977 10 15 10467

5 5 2. ..d iPa 1 N / m m

Pa 1.12 10 kgms

3

3

2

2

g zD = =(9.8066 m / s m m s2 2 2) . /6 588b g Bernoulli m / s m / s2 2 2 2 = - = + -Du u u

2

22

12

2467 588 2 121. . .b g d i

= - = =5 00 2 121 0800 08942 2 2 2 2. / ( )( . ) / . / .b g m s m s m s m / s2 2 2 u (b). Since the fluid is incompressible, &V m s d u d u3 12 1 22 24 4d i = =p p

d duu1 2

2

1

60894500

2 54= = = cm m s

m s cmb g .

..

7.53 (a). &V A u A u u u AA

u uA A

m s m m s m m s 3 2 2d i d i b g d i b g= = = ==1 1 2 2 2 1 12

4

2 1

1 2

4

(b). Bernoulli equation ( Dz = 0)

D D D

P uP P P

u u

r

r+ = = - = -

-22 1

22

12

20

2

d i

Multiply both sides by

Substitute

Multiply top and bottom of right - hand side by

-

=

1

1622

12

12

u u

A

note &V A u2 12 12=

P PV

A1 2

2

12

152

- =r &

(c) P P ghV

AV

A gh1 2

2

12

2 1215

2

215

1- = - = = -FHG

IKJr r

r r

rHg H OH O Hg

H O2

2

2

d i&

&

& . .V 22 2 4

32 75 1 1955 10=-

= -pb g b gcm 1 m 9.8066 m 38 cm 1 m 13.6

15 10 cm s 10 cm ms

4

8 4 2 2

6

2

= =& .V 0 044 44 m s L s3

7-32

7.54 (a). Point 1- surface of fluid . P1 31= . bar , z1 7= + m , u1 0= m sb g Point 2 - discharge pipe outlet . P2 11== atm bar.013b g , z2 0= mb g , u2 = ? Dr

r=

-

= -1013 31

2635. .

.b gbar 10 N 1 m

m bar 0.792 10 kg m s

5 3

2 32 2

g zD =-

= -9.8066 m m s

68.6 m s22 27

Bernoulli equation m s m s2 2 2 2 = - - = + =D D

Du P

g z2

2263 5 68 6 3321

r. . .b g

Du u2 22 20= -

u u22

22 332 1 664 2 258= = =( . ) . . m s m s m / s2 2 2 2

& ( . )V = =p 100 2580 1222 cm cm 1 L 60 s

4 1 s 10 cm 1 min L / min

2

3 3

(b) The friction loss term of Eq. (7.7-2), which was dropped to derive the Bernoulli equation, becomes increasingly significant as the valve is closed.

7.55 Point 1- surface of lake . P1 1= atm , z1 0= , u1 0=

Point 2 - pipe outlet . P2 1= atm , z z2 = ftb g

uVA2

2 295

05 1049353= =

=

&. .

. gal 1 ft 1 144 in 1 min

min 7.4805 gal in 1 ft 60 s ft s

3 2

2p b g Pressure drop: D P P Pr = =0 1 2b g

LZ

z

F z z=

=FHG

IKJ

= =

sin

. . )

302

0 041 2 00822Friction loss: ft lb lb (ft lb lbf m f mb g

Shaft work:

-8 hp 0.7376 ft lb s 7.4805 gal 1 ft 60 s

1.341 10 hp gal 1 ft 62.4 lb 1 min

ft lb lb

f3

3m

f m

&&

/ minWm

s =

= -

-

1

95

333

3

Kinetic energy: ft 1 lb

2 s lb ft / s ft lb lb

2f

2m

2 f mD u2

2 2

235 3 0

3217419 4=

-

=

.

..b g

Potential energy: g ft ft lb

s lb ft / sft lb lbf2

m2 f mDz

zz=

=

32174 132174

..

b g b g

Eq. 7.7 - 2 ftb g + + + = - + + = =D D DP u g z F Wm

z z zsr

2

219 4 0 082 333 290

&& . .

7-33

7.56 Point 1 - surface of reservoir . P1 1= atm (assume), u1 0= , z1 60= m Point 2 - discharge pipe outlet . P2 1= atm (assume), u2 = ? , z2 0=

DP r = 0

Du u V A V

V

222

2 2 6 4

2 2 4

2

2 2 2

1

35 1

3376

= = =

=

& & / )

. &

d hb g

b g

(m s 10 cm 1 N

(2) cm 1 m kg m / s

N m kg

2 8

4 2p

g zD =-

= -

9.8066 m m N s 1 kg m / s

637 N m kg2 265 1

&&

.& &

Wm V

Vs =

= 080 10

8006 W 1 N m / s s 1 m

W m 1000 kgN m kg

3

3d i b g Mechanical energy balance: neglect Eq. 7.7 - 2F b g D D D

P ug z

Wm

VV

VsT E

r+ + =

- - = - = =

+22

23376 637

800 12776 2

&& .

& & &.

. m 60 ss 1 min

m min3

3

Include friction (add F > 0 to left side of equation) &V increases. 7.57 (a). Point 1: Surface at fluid in storage tank, P1 1= atm , u1 0= , z H1 = mb g Point 2 (just within pipe): Entrance to washing machine. P2 1= atm , z2 0=

u2 24 0 47 96= =

600 L 10 cm 1 min 1 m cm 1 L 60 s 100 cm

m s3 3

min ..

p b g

DPr

= 0 ; Du u2 2

2 2

2 27 96 1

2317= =

=. . m s J

1 kg m / s J / kg

2 2b g

g zH

HD =-

= -

9.807 m m J s 1 kg m / s

(J / kg)2 2 20 1

9 807b gc h

.

Bernoulli Equation: mD D

DP u

g z Hr

+ + = =2

20 323.

(b). Point 1: Fluid in washing machine. P1 1= atm , u1 0 , z1 0= Point 2: Entrance to storage tank (within pipe). P2 1= atm , u2 7 96= . m s , z2 323= . m

DPr

= 0 ; Du2

2317= .

Jkg

; g zD = - =9 807 323 0 317. . .b g Jkg

; F = 72 J

kg

Mechanical energy balance: & &W m P u g z Fs = - + + +LNM

OQP

D DD

r

2

2

= - = -& .min

.Ws600 L kg 1 min 31.7 + 31.7 + 72 J 1 kW

L 60 s kg J s1 kW

09610

303b g

(work applied to the system) Rated Power kW . 1.7 kW= =130 0 75.

7-34

7.58 Basis: 1000 liters of 95% solution . Assume volume additivity.

Eq. 6.1-1

Density of 95% solution l

kg kg liter

b g: .

...

. .1 0 95

1260 05100

0804 124r r

r= = + = = xii

Density of 35% solution lkg

kg liter:..

..

. .1 0 35

126065100

0 9278 108r

r= + = =

Mass of 95% solution: liters 1.24 kg

liter kg

10001240=

G = glycerol

m

W = water1240 kg (1000 L)

1 (kg)

0.95 G0.05 W

0.35 G0.65 W

23 m

m2 (kg)0.60 G0.40 W

5 cm I.D.

Mass balance: 1240Glycerol balance:

kg 35% solution kg 60% solution

+ =+ =

UVW==

m mm m

mm

1 2

1 2

1

2095 1240 0 35 06017402980. . .b gb g b gb g b gb g

Volume of 35% solution added kg 1 L

1.08 kg L= =

17401610

= + =Final solution volume L L1000 1610 2610b g Point 1. Surface of fluid in 35% solution storage tank. P1 1= atm , u1 0= , z1 0= Point 2. Exit from discharge pipe. P2 1= atm , z2 23= m

u2 21

2 51051= =

1610 L 1 m 1 min 10 cm13 min 10 L 60 s cm 1 m

m s3 4 2

3 2 2p ..b g

D P r = 0 , D Du u2 2

2 2

2 21051

1= =

= . /b g m s 1 N

(2) kg m / s0.552 N m kg

2 2

2

g zD =

= 9.8066 m 23 m 1 N s 1 kg m / s

225.6 N m kg2 2 , F = = 50 J kg 50 N m kg

Mass flow rate: kg 1 min

13 min 60 s kg s& .m = =1740 2 23

Mechanical energy balance Eq. 7.7 - 2b g

& & .

.

W mP u

g z Fs = - + + +LNM

OQP = -

= -

D DD

r

2

2

2 23

0 62

kg 0.552 + 225.6 + 50 N m 1 J 1 kW

s kg 1 N m 10 J s

kW 0.62 kW delivered to fluid by pump.

3

b g