# Thermodynamics problem -- lunar-based power facility

1. Aug 5, 2015

### Goldberg

1.Question

I need to check if i solved it right can someone please check it , thank you. Its worth alot of points
http://postimg.org/image/vbictz4mt/
3. The attempt at a solution
GIven

P1 = 50 kpa

T1 = -20 C = 253 K

1 to 2 is isentropic process

Therefore we can use

PT^k/(1-k) = constant

k = 1.66

Given

P2/P1 = 6

Therefore

P2 = 6*50 kPa

P2 = 300 kPa

P2 = P3

P3 = 300 kPa

The

T3 relation is given in the figure given

P4 = P1 = 50 kPa

Let's divide the given graph into two sections

From 0-1 and 1-2 gas flow rate

From 0-1

Here we can write the linear equation

T3 = a*m+b

Here a and b are two constant

(0.200) and (1,150) are points on the line

Therefore

200 = b

150 = a*1+ 200

a = -50

The relation for 0-1 gas flow rate is

T3 = -50*m + 200

The relation for 1-2 gas flow rate is also linear equation and let it be

T3 = c*m + d

Here c and d are constants

(1,150) and (2,0) are points on the line

0 = c*2 + d

2c + d = 0 ---(1)

150 = c + d -- (2)

Solving (1) and (2)

c = -150

d = 300

The relation for 1-2 gas flow rate is

T3 = -150*m + 300

Since 3 to 4 is isentropic Expansion

We can use

P3T3^k/(1-k) = P4T4^k/(1-k)

T4 = T3*(P3/P4)^(1-k)/k

T4 = T3*(6)^-0.398

T4 = 0.49*T3

The Power developed by the

P = mCp*(T3-T4)

Cp = 0.520 kJ/kg/K

For 0-1 gas flow rate

P = m*0.520*(T3-0.49*T3)

P = 0.520*m*0.51*T3

P = 0.265*m*T3

P = 0.265*m*(-50m+200)

P = -13.26m^2 + 53.04*m

For 1-2 gas flow rate

P = 0.265*m*T3

P = 0.265*m*(-150*m + 300)

P = -39.78*m^2 + 79.56*m

Optimum at mass flow rate of 1 kg/s

Last edited by a moderator: May 7, 2017
2. Aug 10, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?