Solution manual for thermodynamics for engineers 1st edition by kroos final example

May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part... A For this constant pressure process, the heat transfer is 11.. May not be s

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1 A

2 D W = N ⋅ m/s = ( kg ⋅ m/s 2)m/s = kg ⋅ m 2 /s3

3 C Sum forces in the vertical direction (be sure and use Pa and not kPa):

∑ F = 0 PA − W − Kx = 0

x = 1 ( 600 000 × π × 0 042 − 100 × 9.81) = 0.254 m or 25.4 cm 8000

If absolute pressure is used, the atmospheric pressure acting on the top of the cylinder must be included

4 A

5 B Since the temperature is below the boiling point (120.2°C from Table C-2) of water at 0.2 MPa,refer to

Table C-1 and use h f = 461 kJ/kg

6 D The specific volume at 160°C isv g= 0.307 m3/kg (Table C-1) From Table C-3 we search at 800°Cand observe at 1.6 MPa that v = 0.309 m3/kg So, P2 = 1.59 MPa (No careful interpolation is needed.)

7 C The volume is assumed to be constant (it doesn’t blow up like a balloon!) The ideal gas law is

used: P1V1 = mRT1 and P2V2 = mRT2 so P1/ P2 = T1/ T2 Then, using absolute temperatures and pressures (assume atmosphere pressure of 100 kPa since it is not specified),

340 = 273 ∴ P = 539 kPa or 439 kPa gage

2

8 B If the pressure is constant, the work is mP(v2 – v1) The result is

W = mP ( v 2− v1)=8×800×(0.2608−0.2404)=131 kJ

Units: kg ⋅ kN

⋅ m3

= kN ⋅ m

= kJ m2 kg

9 A First, find the height H of the piston above the cylinder bottom:

m =ρ AH . 0.1= 400 × (π × 0.082) H ∴ H =1.688 m

0.287 ×473 The temperature when the piston hits the stops in this constant-pressure process is

T= T V2= 473×1.188A= 333 K or 59.9°C

1

1

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10 A For this constant pressure process, the heat transfer is

11 B The temperature at the final state must be known It is T = T P2 = 673×100= 168 K So

400

2 1 P

1

Q = mC v T =2×0.717×(673−168)=724 kJ

12 C For an isothermal process Q = W = mRT ln V2 / V1 so the heat transfer is

13 D The heat from the copper enters the water:

14 A Only A does not have a fluid flowing in and/or out

15 C The density is found from Table C-3 usingρ=1/v :

m =ρ AV =1AV = 1 × (π × 0 142 ) × 2 = 1.89 kg/s

0.06525

v

16 C The power required by a pump involving a liquid is

P2− P1 6000 −20

W = m = 5 × = 29.9 kW or 40.1 hp

17 D Equate the expression for the efficiency of a Carnot engine to the efficiency in general:

10

η = 1− = ∴ Qin =

= 29.6 kJ/s

− 10 = 19.6 kJ/s

18 A The maximum possible efficiency would be ηmax=1− T L = 1 −283 = 0.237 Consequently,

= 0.237. ∴Wout, max = 0 237 × + 10 = 10 2 kW The engine is improbable.

60

Qin

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19 B If COP > 1, A is violated If COP = 1, C occurs The condition that W > Q L is very possible.

20 D The heat transfer from the high-temperature reservoir is Q H = Q L + W = 20 + 8 = 28 kJ/s The

8

efficiency of the engine is thus η = W = = 0.286 The Carnot efficiency yields

Q H 28

Obviously the temperatures must be absolute

21 C The entropy change of the copper is mC plnT2/T1 and that of the water is Q/T. Hence,

cu C

p ,cu T

S = m C

p ,cu

ln 2 +

= m C

p,cu

ln 2 + net cu T1 Twater cu T1 Twater

= 10 × 0.39ln293 +10 × 0.39 × (100 − 20)= 0.123 kJ/K

373

The copper loses entropy and the water gains entropy Make sure the heat transfer to the water

is positive

22 C The maximum work occurs with an isentropic process: s1 = s2 = 7.168 kJ/kg·K The enthalpy at the

turbine exit and the work are found as follows (use Tables C-3 and C-2):

s2= s1=7.168=0.6491+ x 2(7.502). ∴ x2=0.869

The copper loses entropy and the water gains entropy Make sure the heat transfer to the water

is positive

23 A The minimum work occurs with an isentropic process for which

24 D The maximum possible turbine work occurs if the entropy is constant, as in Problem 22 which is

1386 kJ/kg The actual work is w T = h1 − h2 = 3658 − 2585 = 1073 kJ/kg The efficiency is then

η =

w

T

=1073= 0.774 or 77.4%

T , s

25 D The heat that leaves the condenser enters the water:

( 2609.7 − 251) ∴ m w = 113 kg/s

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26 B Heat transfer across a large temperature difference, which occurs in the combustion process, ishighly

irreversible The losses in A, C¸ and D are relatively small.

27 B The combustion process is not reversible

28 A The rate at which heat is added to the boiler is

= m( h3− h2) = 2 × ( 3434 − 251) = 6366 kJ/s

29.C The turbine power output is

= m( h3− h4) = 2 × ( 3434 − 2609 7) = 1648 kW

30.D The rate of heat transfer from the condenser is

= m( h4− h1) = 2 × ( 2610 − 251) = 4718 kJ/s

31 B The required pump horsepower follows:

W P = m

k−1

32 C First, find the temperature at state 2: T2 = T1 V1

= 293× 80 4 = 673 K Then

33 A The efficiency of the Otto cycle is

r

8

34 A The work output can be found using the efficiency and the heat input:

w = ηq =1 − 1q =1 − 1 × 574= 250 kJ/kg

rk−1 80 4

4

35 D The heat input is provided by the combustor:

36.B The pressure ratio for the assumed isentropic process is

P T k /( k−1) 5133 5

2

= 2

=

= 7.1

37.A The back-work ratio is found to be

BWR =

wcomp

=C p(T2− T1) =240 − 20 = 0.44 or 44%

w C

p

(T − T ) 800 − 300

T 3 4

38.C The efficiency is the net work divided by the heat input:

η =

w

T − w

comp

=C p(T3 − T4) − C p(T2 − T1)

=800 − 300 − ( 240− 20) = 0.5

q

in

C p(T3− T2) 800 − 240

39.D The vapor pressure, using P g= 5.628 kPa from Table C-1, isP v = P g φ = 5.628× 0.9 = 5.065 kPa.

Then we have

P 5.065

ω = 0.622 v

= 0.622 ×

= 0.0332 kgw /kga

100 − 5.065

40 C Locate state 1 at 10°C and 60% humidity Move at constant ω (it is assumed that no moisture is

added If it were, the amount of water would be needed) to the right until T2 = 25°C There the chart

is read to provide the humidity as φ = 24%

5

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