Question

An inventor claims to have invented an adiabatic steady-flow device with a single inlet-outlet that produces \(230 \mathrm{kW}\) when expanding \(1 \mathrm{kg} / \mathrm{s}\) of air from \(1200 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C}\) to \(100 \mathrm{kPa} .\) Is this claim valid?

Short answer

A. The inventor's claim is valid. B. The inventor's claim is not valid. C. The calculated power output is 230 kW. D. The calculated power output is 254.7 kW. Answer: B. The inventor's claim is not valid.

Step-by-step solution

Understand the adiabatic steady-flow process

In an adiabatic steady-flow process, the system doesn't exchange heat with the surroundings. The mass flow rate through the system remains constant, and the change in internal energy and kinetic energy can be neglected compared to the changes in enthalpy. The equation for this process can be simplified as follows: \(W_{out} = h_{in} - h_{out}\), where \(W_{out}\) is the work output, and \(h_{in}\) and \(h_{out}\) are the specific enthalpies at the entrance and exit, respectively.

Calculate the specific enthalpy at the entrance (\(h_{in}\))

We are given that the air enters the adiabatic steady-flow device at a pressure of 1200 kPa (or \(12 \times 10^5 \mathrm{Pa}\)) and temperature of 300°C (or 573.15 K). We can use the ideal gas equation (\(h = C_pT\)) to calculate the specific enthalpy at the entrance, where \(C_p\) is the specific heat capacity at constant pressure for air (\(C_p = 1005 \mathrm{J/kg \cdot K}), T_{in} = 573.15 \mathrm{K}\) \(h_{in} = C_pT_{in} = 1005 \mathrm{J/kg\cdot K} \cdot 573.15 \mathrm{K} = 576280.575 \mathrm{J/kg}\)

Calculate the specific enthalpy at the exit (\(h_{out}\))

We are given that the air exits the adiabatic steady-flow device at a pressure of 100 kPa (or \(10^5 \mathrm{Pa}\)). Since it is an adiabatic process, we can use the isentropic relation to calculate the final temperature \(T_{out}\): \(\frac{T_{out}}{T_{in}} = \left(\frac{P_{out}}{P_{in}}\right)^{\frac{k-1}{k}}\), where \(k\) is the specific heat ratio for air (\(k = 1.4\)), \(P_{in} = 12 \times 10^5 \mathrm{Pa}\), and \(P_{out} = 10^5 \mathrm{Pa}\). \(\frac{T_{out}}{573.15} = \left(\frac{10^5}{12 \times 10^5}\right)^{\frac{1.4-1}{1.4}}\) \(T_{out} = 319.75 \mathrm{K}\) Now, we can use the ideal gas equation to calculate the specific enthalpy at the exit as \(h_{out} = C_pT_{out} = 1005 \mathrm{J/kg \cdot K} \cdot 319.75 \mathrm{K} = 321573.75 \mathrm{J/kg}\)

Calculate the work output (\(W_{out}\))

Now, calculate the work output using the equation \(W_{out} = h_{in} - h_{out}\). \(W_{out} = 576280.575 \mathrm{J/kg} - 321573.75 \mathrm{J/kg} = 254706.825 \mathrm{J/kg}\) Since the mass flow rate is given to be \(1 \mathrm{kg/s}\), the power output can be calculated as \(P_{out} = \dot{m}\times W_{out} = 1 \mathrm{kg/s} \cdot 254706.825 \mathrm{J/kg} = 254706.825 \mathrm{W}\) or \(254.7 \mathrm{kW}\)

Compare the calculated work output with the inventor's claim

We have calculated the work output to be 254.7 kW, which is different from the inventor's claim of 230 kW. Therefore, based on our analysis, the inventor's claim is not valid.