# 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