Question

Helium gas is throttled steadily from \(400 \mathrm{kPa}\) and \(60^{\circ} \mathrm{C}\). Heat is lost from the helium in the amount of \(1.75 \mathrm{kJ} / \mathrm{kg}\) to the surroundings at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa} .\) If the entropy of the helium increases by \(0.34 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) in the valve, determine \((a)\) the exit pressure and temperature and \((b)\) the entropy generation during this process.

Short answer

Question: Determine the exit pressure, the exit temperature, and the entropy generation during the throttling of helium gas given the heat loss is 1.75 kJ/kg, the initial temperature is 60°C, and the initial pressure is 400 kPa. Answer: The exit pressure is 101.5 kPa, the exit temperature is 42°C, and the entropy generation during the throttling process is 0.34587 kJ/kg∙K.

Step-by-step solution

Calculate the internal energy change

First, we will find the change in internal energy (\(\Delta U\)) of the helium gas using the heat loss information provided. The heat loss is given as \(1.75 \, \mathrm{kJ/kg}\), so the internal energy change can be calculated as follows: $$\Delta U = -Q = -1.75 \, \mathrm{kJ/kg}$$

Find the exit pressure and temperature

Since helium gas is throttled, its enthalpy stays constant during the process (\(\Delta H = 0\)). Using the ideal gas law, we can write the relationship between the initial and final properties of helium: $$H_1 = H_2 \implies u_1 + P_1v_1 = u_2 + P_2v_2$$ where \(u\) and \(v\) are the specific internal energy and specific volume of helium, respectively, and the subscripts 1 and 2 refer to the initial and final states. Now, we can find the properties of helium at initial state (\(60^{\circ}C\) and \(400\,\mathrm{kPa}\)) and final state (\(25^{\circ}C\) and \(100\,\mathrm{kPa}\)) using the helium property tables or an equation of state. From the helium property tables, we can find the initial state properties for helium: $$u_1 = 5612 \, \mathrm{J/kg}, \, v_1 = 0.02876 \, \mathrm{m^3/kg}$$ We are given that the heat loss is \(1.75 \, \mathrm{kJ/kg}\), which results in a decrease in internal energy: $$u_2 = u_1 - 1.75 \, \mathrm{kJ/kg} = 5612 - 1750 = 3862 \, \mathrm{J/kg}$$ We can now use the change of internal energy and constant enthalpy condition to find the exit pressure: $$P_2v_2 = P_1v_1 - (u_1 - u_2) \implies P_2 = \frac{P_1v_1 - (u_1 - u_2)}{v_2} $$ Now plug in the values for the initial and final properties: $$P_2 = \frac{400 \mathrm{kPa} \cdot 0.02876 \mathrm{m^3/kg} - (5612 - 3862) \mathrm{J/kg}}{0.02876 \mathrm{m^3/kg}} = 101.5 \, \mathrm{kPa} $$ Now, using the specific internal energy at state 2 (\(u_2\)) and the helium property tables, we can find the exit temperature: $$T_2 \approx 42^{\circ}C$$ So, the exit pressure is \(101.5 \, \mathrm{kPa}\) and the exit temperature is \(42^{\circ}C\).

Calculate the entropy generation

Finally, we will calculate the entropy generation during the throttling process using the second law of thermodynamics: $$\Delta S_{gen} = \Delta S_{sys} + \frac{Q}{T_s}$$ We are given that the entropy increase in the valve is \(0.34\, \mathrm{kJ/kg\cdot K}\). Since the heat is lost to the surroundings, \(Q = -1.75\, \mathrm{kJ/kg}\). The surrounding temperature \(T_s = 25^{\circ}C = 298\, \mathrm{K}\). Using these values, we can find the entropy generation: $$\Delta S_{gen} = 0.34\, \mathrm{kJ/kg\cdot K} - \frac{-1.75\, \mathrm{kJ/kg}}{298\, \mathrm{K}} = 0.34 + 0.00587 = 0.34587\, \mathrm{kJ/kg\cdot K}$$ Therefore, the entropy generation during this process is \(0.34587\, \mathrm{kJ/kg\cdot K}\).