Answer:
a) T_5 = 782.8 K
b) W_cyc = 108.04 KJ/kg
c) n_th = 22.47 %
d) X_dest = 289.924 KJ/kg , X_exhaust = 126.6768 KJ/kg
Explanation:
Given:
- P_2 / P_1 = 7
- T_4 = 1150 K
- T_1 = 310 K
- n_s,comp = 0.75
- n_s,turb = 0.82
- R_air = 0.287 KJ/kg
Find:
- T_5 - Temperature of air at turbine exit ?
- W_cycle ?
- n_th ?
- X_dest , X_exhaust ?
Solution:
Assumptions:
1) The cycle operates at steady-state.
2) Air is the working fluid and it behaves as an ideal gas.
3) The Brayton Cycle is modeled as as a closed cycle.
4) The combustor is replaced by a HEX. (External Combustion)
5) The compressor and turbine are not internally reversible.
6) Changes in kinetic and potential energies are negligible.
7) Air has variable specific heats.
8) The compressor and turbine are adiabatic.
Analysis:
- The efficiency of turbine is given by:
n_s,turb = (H_4 - H_5) / (H_4 - H_5,s)
- For H_4 and S_ 4 we have T_4 = 1150 K, use the ideal gas air property table:
T_4 = 1150 K -------------> S_4 = 3.129 KJ/kgK , H_4 = 1219.25 KJ/kg
- For H_5s, the enthalpy of the effluent from a hypothetical isentropic turbine. We can do this because we know the values of two intensive variables: P_5 and S_5 = S_4. The key to using this information is the 2nd Gibbs equation:
S_5,s,o = S_4,s,o + R_air*Ln(P_5 / P_4)
S_5,s,o = 3.129 + 0.287*Ln(1 / 7) = 2.57054 KJ/kgK
- Now use the value S_5,s,o and ideal gas air property table and evaluate:
S_5,s,o = 2.57054 KJ/kgK ---------> T_5,s = 698.6 K ,
H_5s = 711.72 KJ/kg
- Now use the efficiency relation for turbine:
H_5 = H_4 - n_s,turb*(H_4 - H_5,s)
H_5 = 1219.25 - 0.82*(1219.25-711.72)
H_5 = 803.08 KJ/kg
- Using H_5 and ideal gas air property table and evaluate:
H_5 = 803.08 KJ/kg -----------> T_5 = 782.8 K , S_5 = 2.6940 KJ/kg-K
- In order to determine the specific shaft work for the cycle, we need to determine the specific shaft work for the compressor and for the turbine
W_cyc = W_comp + W_turb
W_cyc = H_1 - H_2 + H_4 - H_5
- The efficiency of compressor is given by:
n_s,comp = (H_1 - H_2,s) / (H_1 - H_2)
- For H_1 and S_ 1 we have T_1 = 310 K, use the ideal gas air property table:
T_1 = 310 K -------------> S_1 = 1.73498 KJ/kgK , H_1 = 310.24 KJ/kg
- For H_2s, the enthalpy of the effluent from a hypothetical isentropic compressor. We can do this because we know the values of two intensive variables: P_2 and S_2 = S_1. The key to using this information is the 2nd Gibbs equation:
S_2,s,o = S_1,s,o + R_air*Ln(P_2 / P_1)
S_2,s,o = 1.73498 + 0.287*Ln(7) = 2.29343 KJ/kgK
- Now use the value S_2,s,o and ideal gas air property table and evaluate:
S_2,s,o = 2.29343 KJ/kgK ---------> T_2,s = 537.1 K ,
H_2s = 541.34 KJ/kg
- Now use the efficiency relation for compressor:
H_2 = H_1 - (H_1 - H_2,s)/n_comp
H_2 = 310.24 - (310.24-541.34)/0.72
H_2 = 618.37 KJ/kg
Hence,
The work out for the cycle is:
W_cyc = 310.24 - 618.37 + 1219.25 - 803.08
W_cyc = 108.04 KJ/kg
- The thermal efficiency of a cycle is:
n_th = W_cyc / Q_H
Q_H = H_4 - H_3
- The effectiveness of re-generator is e:
e = (H_3 - H_2) / (H_5 - H_2)
H_3 = (H_5 - H_2)*e + H_2
H_3 = (803.08 - 618.37)*0.7 + 618.37 = 738.43 KJ/kg
Hence,
Q_H = 1219.25 - 738.43 = 480.81 KJ/kg
Finally,
n_th = 108.04 / 480.81 = 22.47 %
- The amount of heat loss is given by:
Q_L = H_6 - H_1
H_6 = H_5 + H_2 - H_3 = 803.08 + 618.37 - 738.43 = 682.97 KJ/kg
Q_L = 682.97 - 310.24 = 372.73 KJ/kg
- The amount of exergy destroyed for whole cycle:
X_dest = T_L * ( Q_L / T_L - Q_H / T_H)
X_dest = 310 * (372.73 / 310 - 480.81 / 1800)
X_dest = 289.924 KJ/kg
- The amount of exergy of exhaust gasses:
X_exhaust = H_6 - H_0 - T_L*(S_6 - S_o )
X_exhaust = 682.97 - 310.24 - 310*(2.52861 - 1.73489 )
X_exhaust = 126.6768 KJ/kg
