- #1
Goldberg
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1.Question
I need to check if i solved it right can someone please check it , thank you. Its worth a lot of points
http://postimg.org/image/vbictz4mt/
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
http://file:///C:/Users/Saad%20khan/AppData/Local/Packages/oice_15_974fa576_32c1d314_2427/AC/Temp/msohtmlclip1/01/clip_image001.png
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