Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentropic processes. Consider such a cycle when the compression ratio is 8, P1 = 95 kPa, T1 = 15°C, and the maximum cycle temperature is 900°C. Determine the heat transferred to and rejected from this cycle, as well as the cycle’s thermal efficiency. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

- 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