THERMOPHYSICS

THERMOPHYSICS

a) Produce a schematic diagram of the system, together with an appropriate labeled T-s diagram.

b) Using steam tables only, calculate:

(i) The isentropic efficiencies of the three turbines

An isentropic compressor requires the minimum power input.

We can determine the isentropic work by applying the 1st Law to an isentropic turbine that takes in the same feed and yields as effluent at the same pressure.

For a steady-state, single-inlet, single outlet system with negligible heat transfer, kinetic and potential energy changes, the 1st Law is:

Eqn 1

The entropy change for this process can determined using Gibbs 1st Eqn in terms of the Ideal Gas Entropy Function.

Eqn 2

We can also apply Eqn 2 to our hypothetical, isentropic turbine:

Eqn 3

We can solve Eqn 3 forthe unknown SoT2S :

Eqn 4

We can evaluate SoT1 using theIdeal Gas Property Tables:

SoT1

0.00617

kJ/kg-K

 

We can get HoT1 while we are at because wewill need it later when we evaluate Eqn 1.

H1

87.41

kJ/kg

Now, we can plug values into Eqn 4 :

SoT2S

0.2974

kJ/kg-K

 

 

 

R

8.314

kJ/kmol-K

MW

29

kg/kmol

Now, we can use SoT2S and the Ideal Gas Property Tables to determine T2S and H2S by interpolation.

T (K)

H (kJ/kg)

So(kJ/kg-K)

400

188.40

0.2965

T2S

H2S

0.2974

T2S

400.37

K

410

198.63

0.3217

H2S

188.78

kJ/kg

Now, we can plug values back into Eqn 1 :

(WS)min

-182.46

kW

We can determine the actual power input for the compressor by applying the 1st Law to the real compressor, just as we did in Eqn 1 for the isentropic compressor.

 

Eqn 5

We can evaluate HoT2 using theIdeal Gas Property Tables:

T2, part (b)

420

K

H2

208.88

kJ/kg

Now, we can evaluate WS,act using Eqn 5:

(WS)act

-218.65

kW

 

Since we determined the isentropic work in part (a) and the actual work in part (b), we are ready to plug numbers into Eqn 6 and wrap up this problem.

hS, comp

83.45%

 

Verify :

Check the ideal gas assumption:

V1 =

23.75

L/mole

V2 =

12.04

L/mole

(ii) The cycle efficiencies

T) TableA-22 h) =300.19~'Pr) = 1.386 (Unchanged)

P2 _ P r2 - *10 -1386 TableA-22 h = 57987 kJ

Pr2 - Prj = 2s' kg

Prj

». h) =h«; [h) =h«; 1 kJ llc = (h2a = h) - = 673.1 kg' T2 = 662.46K

Wa h)-h2a llc

T3 =1400K TableA-22 h; =1515.42~:'Pr3 = 450.5 (Unchanged)

P4 Pr4

-=-Pr4 = Pr3 /10=45.05

P3 Pr3

n., = 808.51;

llT = ;: = ~3 =~4a ::::}h4a = h3 +llT (h4s -h3) =914.54 ~: ,T4 = 883.56K

s 3 4s

Now we can calculate the Tis and the Q..

w.r = h3 - h., = 600.88 ~; w.c = h; - h2 = -372.91~

m m

Qln h h 84232 k.J Qout = h - h = -614.35 k.J

-.=3-2= . kg; . 14 kg

m m

Checking for 1 st Law satisfaction:

W~et = 600.88 ~ - 372.91 ~ = 227.97~m

WNet a: -.-=-.-V

m m

QNet =842.32 k.J -614.35 k.J = 227.97 k.J

The Back Work RatIo =62.0%

Wr h3 - h4 600.88

Now, we determine the Net Power developed in the system. = (Wr - ~. c J

WNet m.

m m

V RT (0.887 k ~K X300K)3

Til = pV=_1 ,VI =_1 = g=0.861t-

VI r; (100 ~ ) g

5~

m = s 3 = 5.81 k; ~ w: = 5.81 k;* (227.97 ~)= 1325 ": = 1.32MW

0.861 ~g

The following table summarizes the solution:

TI

T2

T3

T4

PI

P2

P3

P4

K

K

K

K

kPa

kPa

kPa

kPa

300

662.46

1400

883.56

100

1000

1000

100

Wc/m

WT/m

Qln/m

bwr

Wnetfm

11

Net Power ...