Question

The idea of using gas turbines to power automobiles was conceived in the 1930 s, and considerable research was done in the \(1940 \mathrm{s}\) and \(1950 \mathrm{s}\) to develop automotive gas turbines by major automobile manufacturers such as the Chrysler and Ford corporations in the United States and Rover in the United Kingdom. The world's first gasturbine-powered automobile, the 200 -hp Rover Jet \(1,\) was built in 1950 in the United Kingdom. This was followed by the production of the Plymouth Sport Coupe by Chrysler in 1954 under the leadership of G. J. Huebner. Several hundred gas- turbine-powered Plymouth cars were built in the early 1960 s for demonstration purposes and were loaned to a select group of people to gather field experience. The users had no complaints other than slow acceleration. But the cars were never mass-produced because of the high production (especially material) costs and the failure to satisfy the provisions of the 1966 Clean Air Act. A gas-turbine-powered Plymouth car built in 1960 had a turbine inlet temperature of \(1700^{\circ} \mathrm{F}\), a pressure ratio of \(4,\) and a regenerator effectiveness of \(0.9 .\) Using isentropic efficiencies of 80 percent for both the compressor and the turbine, determine the thermal efficiency of this car. Also, determine the mass flow rate of air for a net power output of 130 hp. Assume the ambient air to be at \(510 \mathrm{R}\) and 14.5 psia.

Short answer

Based on the provided information on a gas-turbine-powered Plymouth car from 1960 and our calculations, we found the thermal efficiency to be approximately 39.2%, and the mass flow rate of air when producing 130 hp to be approximately 0.149 lbm/s (pound mass per second).

Step-by-step solution

Calculate the temperature and pressure at the end of the compressor

The pressure and temperature of the ambient air are given as 14.5 psia and 510 R, respectively. The pressure ratio is given as 4. We will use this pressure ratio to calculate the pressure at the end of the compressor. Let \(P_1\) be the pressure at the beginning of the cycle, and \(P_2\) be the pressure at the end of the compressor. We have: \(P_2 = P_1 \times \text{pressure ratio} = 14.5 \times 4 = 58 \, \text{psia}\) Now we need to find the temperature at the end of the compressor. The isentropic efficiency of the compressor is given to be 80%. We can use this efficiency to find the actual temperature at the end of the compressor. For this, we should first find the isentropic temperature (\(T_{2s}\)). Assuming an ideal gas with constant specific heat and using the relation for isentropic processes, we have: \(T_{2s} = T_1 \times (\text{pressure ratio})^{(\gamma - 1)/\gamma}\) For air, \(\gamma = 1.4\). So the relation becomes: \(T_{2s} = 510 \times (4)^{(1.4 - 1)/1.4} = 510 \times (4)^{0.286} \approx 688.0 \,\text{R}\) Now we can use the isentropic efficiency of the compressor to find the actual temperature (\(T_2\)): \(\text{Isentropic efficiency}_{\text{compressor}} = \frac{T_{2s} - T_1}{T_2 - T_1}\) Solving for \(T_2\), we have: \(T_2 - T_1 = \frac{T_{2s} - T_1}{\text{Isentropic efficiency}_{\text{compressor}}} = \frac{688.0 - 510}{0.8} \Longrightarrow T_2 \approx 733.5\, \text{R}\)

Calculate the temperature at the end of the regenerator

The regenerator effectiveness is given to be 0.9. We can use this effectiveness to calculate the temperature at the end of the regenerator, \(T_3\): \(T_3 = T_2 + \text{effectiveness} \times (T_1 - T_2) = 733.5 + 0.9 \times (510 - 733.5) \approx 522.4\, \text{R}\)

Calculate the temperature and pressure at the end of the turbine

The temperature at the turbine inlet, \(T_4\), is given as \(1700^{\circ} \mathrm{F}\), which equals 1700 + 460 = 2160 R. The pressure ratio is 4, so the pressure at the end of the turbine, \(P_5\), can be calculated as: \(P_5 = \frac{P_4}{\text{pressure ratio}} = \frac{58}{4} = 14.5 \,\text{psia}\) Now we need to find the temperature at the end of the turbine. First, we find the isentropic temperature (\(T_{5s}\)): \(T_{5s} = T_4 \times (\frac{P_5}{P_4})^{(\gamma - 1)/\gamma} = 2160 \times (\frac{14.5}{58})^{0.286} \approx 1461.6\, \text{R}\) Now we can use the isentropic efficiency of the turbine to find the actual temperature (\(T_5\)): \(\text{Isentropic efficiency}_{\text{turbine}} = \frac{ T_4 - T_5 }{T_4 - T_{5s}}\) Solving for \(T_5\), we get: \(T_5 - T_4 = -\frac{T_4 - T_{5s}}{\text{Isentropic efficiency}_{\text{turbine}}} = -\frac{2160 - 1461.6}{0.8} \Longrightarrow T_5 \approx 1279.4\, \text{R}\)

Calculate the thermal efficiency

Now that we have the temperatures and pressures at all points in the cycle, we can calculate the thermal efficiency. Thermal efficiency is given by the relation: \(\text{Thermal efficiency} = \frac{\text{Net work output}}{\text{Heat input}}\) The net work output is the work done by the turbine minus the work required by the compressor: \(\text{Net work output} = \text{Work}_{\text{turbine}} - \text{Work}_{\text{compressor}}\) For the work done by the turbine and the compressor, we use the relation: \(\text{Work} = \text{mass flow rate} (\text{specific heat} \times \Delta \text{temperature})\) We can rewrite the thermal efficiency as: \(\text{Thermal efficiency} = \frac{ (\text{specific heat} \times (T_4 - T_5)) - (\text{specific heat} \times (T_2 - T_1)) }{(\text{specific heat} \times (T_4 - T_3))}\) Since specific heat is the same for both numerator and denominator, it cancels out: \(\text{Thermal efficiency} = \frac{(T_4 - T_5) - (T_2 - T_1)}{(T_4 - T_3)}\) Plugging in the temperatures we have calculated, we get the thermal efficiency: \(\text{Thermal efficiency} \approx \frac{(2160 - 1279.4) - (733.5 - 510)}{(2160 - 522.4)} \approx 0.392\) So, the thermal efficiency of the gas-turbine powered car is about 39.2%.

Calculate the mass flow rate of air

Now we can use the net power output and the work done by the turbine to find the mass flow rate of air. The net power output is given as 130 hp. We should convert hp to Btu/s: \(\text{Net power output} = 130\, \text{hp} \times \frac{2544}{33000}\, \frac{\text{Btu}}{\text{hp} \cdot \text{s}} \approx 10.193\, \text{Btu/s}\) We have the relation for the net work output: \(\text{Net work output} = \text{mass flow rate} \times (\text{specific heat} \times (T_4 - T_5)) - \text{mass flow rate} \times (\text{specific heat} \times (T_2 - T_1))\) We can rewrite this equation for the mass flow rate: \(\text{mass flow rate} = \frac{\text{Net work output}}{(\text{specific heat} \times (T_4 - T_5)) - (\text{specific heat} \times (T_2 - T_1))}\) For air, the specific heat at constant pressure, \(c_p\), is about 0.240 Btu/lbm·R. Plugging in the values we have calculated, we can find the mass flow rate of air: \(\text{mass flow rate} \approx \frac{10.193}{(0.240 \times (2160 - 1279.4)) - (0.240 \times (733.5 - 510))} \approx 0.149\, \text{lbm/s}\) So, the mass flow rate of air for the gas-turbine powered car when producing 130 hp is about 0.149 lbm/s.