Answer:
q_net,in = 585.8 KJ/kg
q_net,out = 304 KJ/kg
n = 0.481
Explanation:
Given:
- The compression ratio r = 8
- The pressure at state 1, P_1 = 95 KPa
- The minimum temperature at state 1, T_L = 15 C
- The maximum temperature T_H = 900 C
- Poly tropic index n = 1.3
Find:
a) Determine the heat transferred to and rejected from this cycle
b) cycle’s thermal efficiency
Solution:
- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:
W_out - Q_out = Δ u_1,2
- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:
c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2
- Using polytropic relation we will convert T_2 = T_L*r^(n-1):
c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2
- Hence, we have:
q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)
- Plug in the values:
q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)
q_1,2= 60 KJ/kg
- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:
Q_in = Δ u_2,3
q_2,3 = u_3 - u_2
q_2,3 = c_v*(T_H - T_2)
- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):
q_2,3 = c_v*(T_H - T_L*r^(n-1) )
q_2,3 = 0.718*(1173-288*8(1.3-1) )
q_2,3 = 456 KJ/kg
- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:
q_3,4 - w_in = Δ u_3,4
- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:
c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n + q_3,4
- Using polytropic relation we will convert T_4 = T_H*r^(1-n):
c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4
- Hence, we have:
q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)
- Plug in the values:
q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)
q_3,4= 129.8 KJ/kg
- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:
Q_out = Δ u_4,1
q_4,1 = u_4 - u_1
q_4,1 = c_v*(T_4 - T_L)
- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):
q_4,1 = c_v*(T_H*r^(1-n) - T_L )
q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )
q_4,1 = 244 KJ/kg
- The net gain in heat can be determined from process q_3,4 & q_2,3:
q_net,in = q_3,4+q_2,3
q_net,in = 129.8+456
q_net,in = 585.8 KJ/kg
- The net loss of heat can be determined from process q_1,2 & q_4,1:
q_net,out = q_4,1+q_1,2
q_net,out = 244+60
q_net,out = 304 KJ/kg
- The thermal Efficiency of a Otto Cycle can be calculated:
n = 1 - q_net,out / q_net,in
n = 1 - 304/585.8
n = 0.481
then you have a thin walled cylinder
then you have a thin walled cylinder