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A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The air enters the turbine at 120 psia and \(2000 \mathrm{R}\) and leaves at 15 psia and \(1200 \mathrm{R} .\) Heat is rejected to the surroundings at a rate of 6400 \(\mathrm{Btu} / \mathrm{s},\) and air flows through the cycle at a rate of \(40 \mathrm{lbm} / \mathrm{s}\) Assuming the turbine to be isentropic and the compresssor to have an isentropic efficiency of 80 percent, determine the net power output of the plant. Account for the variation of specific heats with temperature.

Short Answer

Expert verified
Answer: To find the net power output of the Brayton cycle, follow these steps: 1. Calculate the isentropic work done in the turbine (\(W_{T,s}\)) using ideal gas relations and specific heats of air. 2. Determine the actual work done in the turbine (\(W_{T}\)). 3. Calculate the work done by the compressor (\(W_{C}\)) using isentropic efficiency. 4. Find the net work output of the plant by subtracting the compressor work input from the turbine work done, i.e., \(W_{net} = W_{T} - W_{C}\).

Step by step solution

01

Calculate the Isentropic Work done in the Turbine (\(W_{T,s}\))

We'll use the relation between specific heat at constant pressure (\(C_p\)), temperature and work output:\(W_{T,s} = m*C_p*(T_{1} - T_{2s})\) where \(T_{2s}\) is the temperature at the outlet of the turbine under the isentropic process. As we assume the turbine to be isentropic and account for the variation of specific heat with temperature, we can use the specific heat data tables for air to determine \(C_p\) at both conditions \(T_{1} = 2000\, \text{R}\) and \(T_{2} = 1200\, \text{R}\). After that, we can average them to get the average specific heat, and hence compute the isentropic work done by the turbine.
02

Calculate the Actual Work done in the Turbine (\(W_{T}\))

Since the turbine is isentropic, the actual work done in the turbine is equal to the isentropic work done i.e., \(W_{T} = W_{T,s}\)
03

Calculate Work done by the Compressor (\(W_{C}\))

Using the isentropic efficiency, we can express the work done by the compressor as: \(W_{C} = \frac{W_{T}}{\eta_{c}}\) where \(\eta_{c}\) is the isentropic efficiency of the compressor (80%). Hence, we can compute the work done by the compressor.
04

Calculate the Net Work Output of the Plant

The net work output of the plant is equal to the difference between the turbine work done and the compressor work input i.e., \(W_{net} = W_{T} - W_{C}\).

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Most popular questions from this chapter

Consider an ideal Brayton cycle executed between the pressure limits of 1200 and \(100 \mathrm{kPa}\) and temperature limits of 20 and \(1000^{\circ} \mathrm{C}\) with argon as the working fluid. The net work output of the cycle is \((a) 68 \mathrm{kJ} / \mathrm{kg}\) \((b) 93 \mathrm{kJ} / \mathrm{kg}\) \((c) 158 \mathrm{kJ} / \mathrm{kg}\) \((d) 186 \mathrm{kJ} / \mathrm{kg}\) \((e) 310 \mathrm{kJ} / \mathrm{kg}\)

Develop an expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100 percent. Use constant specific heats at room temperature.

A four-cylinder two-stroke 2.4 -L diesel engine that operates on an ideal Diesel cycle has a compression ratio of 22 and a cutoff ratio of \(1.8 .\) Air is at \(70^{\circ} \mathrm{C}\) and \(97 \mathrm{kPa}\) at the beginning of the compression process. Using the cold-airstandard assumptions, determine how much power the engine will deliver at \(3500 \mathrm{rpm}\)

Consider an aircraft powered by a turbojet engine that has a pressure ratio of \(9 .\) The aircraft is stationary on the ground, held in position by its brakes. The ambient air is at \(7^{\circ} \mathrm{C}\) and \(95 \mathrm{kPa}\) and enters the engine at a rate of \(20 \mathrm{kg} / \mathrm{s}\) The jet fuel has a heating value of \(42,700 \mathrm{kJ} / \mathrm{kg},\) and it is burned completely at a rate of \(0.5 \mathrm{kg} / \mathrm{s}\). Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components, determine the force that must be applied on the brakes to hold the plane stationary.

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.

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