Question

A gas-turbine power plant operates on the regenerative Brayton cycle between the pressure limits of 100 and \(700 \mathrm{kPa}\). Air enters the compressor at \(30^{\circ} \mathrm{C}\) at a rate of \(12.6 \mathrm{kg} / \mathrm{s}\) and leaves at \(260^{\circ} \mathrm{C}\). It is then heated in a regenerator to \(400^{\circ} \mathrm{C}\) by the hot combustion gases leaving the turbine. A diesel fuel with a heating value of \(42,000 \mathrm{kJ} / \mathrm{kg}\) is burned in the combustion chamber with a combustion efficiency of 97 percent. The combustion gases leave the combustion chamber at \(871^{\circ} \mathrm{C}\) and enter the turbine whose isentropic efficiency is 85 percent. Treating combustion gases as air and using constant specific heats at \(500^{\circ} \mathrm{C}\), determine (a) the isentropic efficiency of the compressor, ( \(b\) ) the effectiveness of the regenerator, \((c)\) the air-fuel ratio in the combustion chamber, \((d)\) the net power output and the back work ratio, \((e)\) the thermal efficiency, and \((f)\) the second-law efficiency of the plant. Also determine \((g)\) the second-law efficiencies of the compressor, the turbine, and the regenerator, and \((h)\) the rate of the energy flow with the combustion chamber with a combustion efficiency of 97 percent. The combustion gases leave the combustion chamber at \(871^{\circ} \mathrm{C}\) and enter the turbine whose isentropic efficiency is 85 percent. Treating combustion gases as air and using constant specific heats at \(500^{\circ} \mathrm{C}\), determine (a) the isentropic efficiency of the compressor, (b) the effectiveness of the regenerator, (c) the air-fuel ratio in the combustion chamber, \((d)\) the net power output and the back work ratio, \((e)\) the thermal efficiency, and \((f)\) the second-law efficiency of the plant. Also determine \((g)\) the second-law efficiencies of the compressor, the turbine, and the regenerator, and \((h)\) the rate of the energy flow with the combustion gases at the regenerator exit.

Short answer

a) To find the isentropic efficiency of the compressor, we first need to calculate the work input to the compressor and the isentropic work input. Using the properties found in step 1, we can calculate the work input to the compressor and then find the isentropic efficiency of the compressor using the given formula. b) The effectiveness of the regenerator can be found using the properties at the regenerator inlet and outlet from steps 1 and 4, and the formula provided in step 5. c) To find the air-fuel ratio in the combustion chamber, we must first calculate the fuel mass flow rate and the heat input from the combustion of fuel using the information given in step 6. Then, the air-fuel ratio can be found using the provided formula. d) The net power output can be calculated by finding the difference between the turbine power output and the compressor power input, using the already calculated compressor work input and the formula provided in step 7. The back work ratio can also be found in step 7 using the provided formula. e) The thermal efficiency of the cycle can be calculated using the formula provided in step 8. f) The second-law efficiency of the plant can be calculated using the relationships provided in step 9. This will require finding the ideal reversible efficiencies or effectiveness and substituting into the given formulas to find the second-law efficiencies of the compressor, turbine, and regenerator. g) The second-law efficiencies of the compressor, turbine, and regenerator can also be found using the relationships provided in step 9. h) The rate of the energy flow with the combustion gases at the regenerator exit can be calculated using the formula provided in step 10. To find this, we must know the mass flow rate of the combustion gases and the properties at the regenerator exit.

Step-by-step solution

(Step 1: Find the properties of air at compressor inlet and outlet)

(We are given the temperature and pressure at the compressor inlet (state 1), \(T_{1}=30^{\circ} \mathrm{C}\) and \(P_{1}=100 \,\mathrm{kPa}\). We need to find the specific volume \(v_{1}\), specific internal energy \(u_{1}\), and specific entropy \(s_{1}\). We also need to find the temperature \(T_{2}=260^{\circ} \mathrm{C}\) at the compressor outlet (state 2), and find \(v_{2}\), \(u_{2}\), and \(s_{2}\). We can use the air tables or constant-specific heat relationships to find these values.)

(Step 2: Find compressor work input and isentropic efficiency)

(Using the properties found in step 1, we can calculate the work input to the compressor: \(W_{c} = m(u_{2} - u_{1}) = m_{air}(h_{2} - h_{1})\) where \(m_{air}\) is the air mass flow rate. We can then find the isentropic efficiency of the compressor using \(\eta_{c}=\frac{W_{c(isentropic)}}{W_{c(actual)}}\), where \(W_{c(isentropic)} = m_{air}(h_{2isentropic}-h_{1})\). To find \(h_{2isentropic}\), use the isentropic relationship for pressure: \(P_{2isentropic} = P_{1}\left(\frac{T_{2isentropic}}{T_{1}}\right)^{\frac{k}{k-1}}\), and then find the new enthalpy, \(h_{2isentropic}\).)

(Step 3: Find properties of air at the turbine inlet)

(We are given \(T_{4}=871^{\circ} \mathrm{C}\) and \(P_{4}=700 \,\mathrm{kPa}\) as the air temperature and pressure at the turbine inlet (state 4). We need to find the specific volume \(v_{4}\), specific internal energy \(u_{4}\), and specific entropy \(s_{4}\). We can use the air tables or constant-specific heat relationships to find these values.)

(Step 4: Find properties of air at the turbine outlet and regenerator outlet)

(We are given the temperature \(T_{5}=400^{\circ} \mathrm{C}\) at the regenerator outlet (state 5) and the isentropic efficiency of the turbine \(η_{t}=0.85\). We can use the isentropic process to find the specific entropy \(s_{5isentropic}=s_{4}\), and then we can use the isentropic efficiency of the turbine to find the actual specific entropy \(s_{5}=s_{4}+ \frac{s_{5isentropic}-s_{4}}{η_{t}}\). Then, we can find the specific volume \(v_{5}\) and internal energy \(u_{5}\) at the turbine outlet. To find the properties at the regenerator outlet, we can use the given temperature \(T_{3}=400^{\circ} \mathrm{C}\) and specified pressure \(P_{3}=100\,\mathrm{kPa}\).)

(Step 5: Find effectiveness of the regenerator)

(To find the effectiveness of the regenerator, we use the definition: \(ε=\frac{h_{3}-h_{2}}{h_{4}-h_{2}}\), where \(ε\) is the regenerator effectiveness.)

(Step 6: Find air-fuel ratio in the combustion chamber)

(The air-fuel ratio can be found using the equation: \(AFR=\frac{m_{air}}{m_{fuel}}\). We are given the fuel heating value (HV) and the combustion efficiency \(η_{c}=0.97\). We can calculate the fuel mass flow rate \(m_{fuel}= \frac{Q_{comb}}{η_{c} \cdot \text{HV}}\), where \(Q_{comb}\) is the heat input from the combustion of fuel and \(Q_{comb}=m_{air}(h_{4}-h_{3})\). Then, we can find the air-fuel ratio \(AFR= \frac{m_{air}}{m_{fuel}}\).)

(Step 7: Calculate the net power output and back work ratio)

(The net power output is the difference between the turbine power output and the compressor power input: \(W_{net}=W_{turbine}-W_{compressor}\). We already calculated \(W_{compressor}\) in step 2 and can calculate the turbine power output, \(W_{turbine}=m_{air}(h_{4}-h_{5})\). The back work ratio is the ratio of the compressor work to the turbine work: \(BWR= \frac{W_{compressor}}{W_{turbine}}\).)

(Step 8: Calculate the thermal efficiency)

(The thermal efficiency of the cycle can be calculated as: \(η_{th} = \frac{W_{net}}{Q_{comb}}\), where \(W_{net}\) is the net power output found in step 7, and \(Q_{comb}\) is the heat input from the combustion of fuel found in step 6.)

(Step 9: Calculate second-law efficiencies)

(We can calculate the second-law efficiencies of the plant and the compressor, turbine, and regenerator using the relationships: \(η_{II}^{plant}=\frac{η_{th}}{η_{r,ideal}}\), \(η_{II}^{compressor}=\frac{η_{c,actual}}{η_{c,ideal}}\), \(η_{II}^{turbine}=\frac{η_{t,actual}}{η_{t,ideal}}\), and \(η_{II}^{regenerator}=\frac{ε_{actual}}{ε_{ideal}}\), where \(η_{II}\) is the second-law efficiency, and \(η_{r,ideal}\), \(η_{c,ideal}\), \(η_{t,ideal}\), and \(ε_{ideal}\) are the ideal reversible efficiencies or effectiveness.)

(Step 10: Calculate the energy flow rate with combustion gases at regenerator exit)

(The rate of energy flow with the combustion gases at the regenerator exit can be calculated as: \(E_{exit}=m_{comb}(h_{5}-h_{1})\), where \(m_{comb}\) is the mass flow rate of the combustion gases, which can be found as the sum of the mass flow rates of air and fuel.) At the end of these steps, we will have determined all of the requested quantities (a) - (h).