Question

Air enters a compressor steadily at the ambient conditions of \(100 \mathrm{kPa}\) and \(22^{\circ} \mathrm{C}\) and leaves at \(800 \mathrm{kPa}\). Heat is lost from the compressor in the amount of \(120 \mathrm{kJ} / \mathrm{kg}\) and the air experiences an entropy decrease of \(0.40 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} .\) Using constant specific heats, determine ( \(a\) ) the exit temperature of the air, \((b)\) the work input to the compressor, and \((c)\) the entropy generation during this process.

Short answer

Question: Determine the exit temperature, work input to the compressor, and the entropy generation during the compression process given the following information: air enters the compressor at \(22^\circ C\) and \(100 \mathrm{kPa}\) and leaves at \(800 \mathrm{kPa}\). The compressor loses \(120 \mathrm{kJ/kg}\) of heat to the surroundings. During the process, the entropy of the air decreases by \(0.40 \mathrm{kJ/kg.K}\). Answer: The exit temperature is \(2361.2 \mathrm{K}\), the work input to the compressor is \(2196.43 \mathrm{kJ/kg}\), and the entropy generation during the process is \(0.40 \mathrm{kJ/kg.K}\).

Step-by-step solution

Find the enthalpy change

First, we need to find the change in enthalpy during the compression process. We are given that heat is lost from the compressor in the amount of \(120 \mathrm{kJ} / \mathrm{kg}\). Using constant specific heats, we know that the enthalpy change \(Δh = c_p(ΔT)\). Since we know the temperature at the entrance, we'll find the change in temperature and then calculate the enthalpy change.

Calculate the change in temperature

We will use the ideal gas equation of state for the process, knowing that \(PV=nRT\) which allows us to find the change in temperature during the compression: $$\frac{P_2}{P_1}=\frac{T_2}{T_1}$$ $$T_2 = T_1 \frac{P_2}{P_1}$$ Using the given values, we have: $$T_2 = (22+273.15) \frac{800}{100}$$ $$T_2 = 295.15 \times 8$$ $$T_2 = 2361.2 \mathrm{K}$$ Now we can find the change in temperature: $$ΔT = T_2 - T_1 = 2361.2 - 295.15 = 2066.05 \mathrm{K}$$

Calculate the enthalpy change

Using the specific heat for air at constant pressure \(c_p=1.005 \mathrm{kJ/kg.K}\), we can find the enthalpy change: $$Δh = c_p(ΔT)$$ $$Δh = 1.005 \times 2066.05 = 2076.43 \mathrm{kJ/kg}$$

Apply conservation of energy principle

Now we will apply the conservation of energy principle to the compression process. The energy balance equation can be written as: $$Q = Δh - W$$ We are given that heat is lost from the compressor in the amount of \(120 \mathrm{kJ/kg}\), which means \(Q = -120 \mathrm{kJ/kg}\). Now, we can solve for the work input: $$W = Δh - Q$$ $$W = 2076.43 + 120$$ $$W = 2196.43 \mathrm{kJ/kg}$$

Calculate the entropy generation

Finally, let's find the entropy generation during the process. We know that: $$ΔS_{total} = ΔS_{air} + S_{generation}$$ The entropy decreases by \(0.40 \mathrm{kJ/kg.K}\), thus \(ΔS_{air} = -0.40 \mathrm{kJ/kg.K}\). Then, we can solve for the entropy generation: $$S_{generation} = ΔS_{total} - ΔS_{air}$$ $$S_{generation} = 0 - (-0.40)$$ $$S_{generation} = 0.40 \mathrm{kJ/kg.K}$$ Now we can summarize the results: a) The exit temperature is \(2361.2 \mathrm{K}\) b) The work input to the compressor is \(2196.43 \mathrm{kJ/kg}\) c) The entropy generation during the process is \(0.40 \mathrm{kJ/kg.K}\)