AME 50531Intermediate ThermodynamicsExamination 1: SOLUTIONProf. J. M. Powers2 October 2009

1. (20) Superheated water vapor enters a valve at 500 lbf/in2, 500 F and exits ata pressure of 80 lbf/in2. The expansion is a throttling process. Determine theexergy change per unit mass. Let T

o= 77 F , P

o= 1 atm.

Solution

The exergy at the inlet i is

ψi =

(

hi − ho +1

2Vi · Vi + g(Zi − Zo)

)

− To(si − so).

The exergy at the exit e is

ψe =

(

he − ho +1

2Ve ·Ve + g(Ze − Zo)

)

− To(se − so).

Now for a valve hi = he, Vi = Ve, and Zi = Ze. So the exergy change per unit of massflowing is simply

ψe − ψi = To(si − se).

Interpolating Table F.7.2 for the inlet at 500 lbf/in2, 500 F , we find

si =(1.5282 + 1.4592)

2= 1.4937

Btu

lbm R.

hi =1245.17 + 1216.21

2= 1230.69

Btu

lbm.

At the exit, we have

he = hi = 1230.69Btu

lbm.

along with Pe = 80 lbf/in2. Table F.7.2 is so close to having a proper value at this statethat we will not interpolate. We find that at this pressure and enthalpy that Te = 400 Fand se = 1.6790 Btu/lbm/R. The absolute To = 77 + 459.67 = 536.7 R. So the exergychange per unit mass is

ψe − ψi = (536.7)(1.5937− 1.6790) = −99.445Btu

lbm.

2. (40) Steam is the working fluid in an ideal Rankine cycle with superheat andreheat. Steam enters the first-stage turbine at 8.0 MPa, 480 C, and expandsto 0.7 MPa. It is then reheated to 440 C before entering the second stageturbine, where it expands to the condenser pressure of 0.008 MPa. The net

power output is 100 MW . Determine the thermal efficiency of the cycle.

Solution

We will assign the following states:

• 1. first turbine inlet/combustor exit

• 2. combustor inlet/first turbine exit

• 3. second turbine inlet/combustor exit

• 4. condenser inlet/second turbine exit

• 5. pump inlet/condenser exit

• 6. combustor inlet/pump exit

We have P1 = 8 MPa, T1 = 480 C. We can interpolate Table B.1.3 to find h1 and s1.Leaving out the details of the interpolation, we get

h1 = 3348.4kJ

kg, s1 = 6.6586

kJ

kg K.

At state 2, we have P2 = 0.7 MPa = 700 kPa, and s2 = s1 = 6.6586 kJ/kg/K. This isa two-phase mixture. Using Table B.1.2, we get the quality at state 2:

x2 =s2 − sf

sfg

=6.6586− 1.9922

4.7158= 0.9895.

Then for h2, we get

h2 = hf + x2hfg = 697.20 + (0.9895)(2066.30) = 2741.85kJ

kg.

State 3 has T3 = 440 C, P3 = 700 kPa.

This requires a complicated double interpolation of Table B.1.3, which yields

h3 = 3353.3kJ

kg, s3 = 7.7571

kJ

kg K.

At state 4, we have P4 = 8 kPa and s4 = s3 = 7.7571 kJ/kg/K. We have to interpolateTable B.1.2 at this pressure. We find here that sf = 0.5926, sfg = 7.6361 kJ/kg/K,hf = 173.88 kJ/kg, hfg = 2403.1 kJ/kg. So

x4 =s4 − sf

sfg

=7.7571− 0.5926

7.6361= 0.9382.

Thus at state 4, we have

h4 = hf + x4hfg = 173.88 + (0.9382)(2403.1) = 2428.57kJ

kg.

At state 5, the water is saturated liquid at 8 kPa, which gives, after interpolation,

h5 = 173.88kJ

kg.

And the pump exit has

h6 = h5 + v(P6 − P5) = 173.88 + (0.0010084)(8000− 8) = 181.94kJ

kg.

The net power per unit mass flow is

wnet = wturbine 1 + wturbine 2 − wpump,

wnet = (h1 − h2) + (h3 − h4) + (h6 − h5),

wnet = (3358.4 − 2741.8) + (3353.3− 2428.57)− (181.94 − 173.88),

wnet = 1523.22kJ

kg.

Then heat that is paid for is

qin = (h1 − h6) + (h3 − h2),

qin = (3348.4− 181.94) + (3353.3− 2741.85),

qin = 3777.91kJ

kg.

The thermal efficiency is

ηt =wnet

qin=

1523.22

3777.91= 0.403.

3. (40) An ideal Otto cycle has a compression ratio of 8. At the beginning ofthe compression process, the air is at 100 kPa and 17 C, and 800 kJ/kg ofheat is transferred to the air during the constant volume heat addition pro-cess. Accounting for the variation of the specific heats of air with temperature,determine the thermal efficiency.

Solution

State 1 has P1 = 100 kPa, T1 = 17 + 273 = 290 K. From Table A.7.1, we find that

u1 = 207.19kJ

kg.

Now we know thatv1v2

= 8.

From the ideal gas lawP2v2T2

=P1v1T1

.

Sov1v2

=P2

P1

T1

T2

= 8.

SoP2

P1

= 8T2

T1

.

We also know that s2 = s1. Now

0 = s2 − s1 = soT2

− soT1

−R ln

(

P2

P1

)

.

0 = soT2

− soT1

−R ln

(

8T2

T1

)

.

Using known quantities at state 1, we get

0 = soT2

− 6.83521− 0.287 ln

(

8T2

290

)

.

We guess values of T2 and use Table A.7.1 and interpolate. We find

T2 = 652.4 K, u2 = 475.11kJ

kg.

So

P2 = 8P1

T2

T1

= 8(100)

(

652.4

290

)

= 1799.7 kPa.

Now in the combustion stage, we have

u3 = u2 + qin = 475.11 + 800 = 1275.11kJ

kg.

We interpolate Table A.7.1 to then find that

T3 = 1575.1 K.

We use the ideal gas law to get P3:

P3v3T3

=P2v2T2

.

P3 = P2

T3

T2

v2v3

= (1799.7)1575.1

652.4(1) = 4245.05 kPa.

Now for the isentropic power stroke, we have

0 = s4 − s3 = soT4

− soT3

−R ln

(

P4

P3

)

.

We also have from the ideal gas law that

v3v4

=P4

P3

T3

T4

=1

8.

SoP4

P3

=1

8

T4

T3

.

Thus,

0 = soT4

− soT3

−R ln

(

1

8

T4

T3

)

.

Plugging in numbers we know, and interpolating Table A.7.1 to get soT3

at T3 = 1575.1K,we get

0 = soT4

− 8.67118− 0.287 ln

(

1

8

T4

1575.1

)

.

Using a trial and error process coupled with interpolation of Table A.7.1, we find

T4 = 795.6 K, u4 = 588.74kJ

kg.

For the thermal efficiency, we calculate

wnet = qnet = qin − qout = qin − (u4 − u1) = 800 − (588.74− 207.19) = 418.45kJ

kg.

So the thermal efficiency is

ηt =wnet

qin=

418.45

800= 0.523.