Question

A 100 -L well-insulated rigid tank is initially filled with nitrogen at \(1000 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\). Now a valve is opened and one-half of nitrogen's mass is allowed to escape. Determine the change in the exergy content of the tank.

Short answer

#tag_title#Question#tag_content#Estimate the change in the exergy content of a well-insulated rigid tank initially filled with nitrogen at 1000 kPa and 20°C. Half of the nitrogen is allowed to escape through a valve. Assume the reference temperature and pressure are 25°C and 100 kPa, and the initial volume of the tank is 100 L. #tag_title#Answer#tag_content#To estimate the change in exergy content of the tank, first determine the initial and final state properties of the nitrogen gas using the ideal gas law. Then, calculate the specific exergy change as Δe = Δu - T_0Δs - P_0Δv, using ideal gas behavior equations for Δu, Δs, and Δv. Finally, calculate the total change in exergy content by multiplying the specific exergy change by the mass of nitrogen in the final state: ΔE = m_2 Δe.

Step-by-step solution

Determine initial state properties

Initially, the tank has nitrogen at \(1000 kPa\) and \(20^\circ C\). We can calculate the initial volume (\(V_1\)) of the nitrogen in the tank using the ideal gas law: \(PV=nRT\) Where: \(P=1000 kPa = 1000 \times 10^{3} Pa\) \(V=100 L = 100 \times 10^{-3} m^3\) \(R= 296.8 J/kg \cdot K\) (gas constant for Nitrogen) \(T = 20^\circ C = 293.15 K\) Now, we solve for the number of moles (n) in the tank initially: \(n = \frac{PV}{RT}\)

Determine final state properties

Now, let's find the properties in the final state (after half of the nitrogen's mass has escaped). Only half of the Nitrogen remains in the tank, so the number of moles in the final state is: \(n_2 = \frac{1}{2} n\) We know that the volume of the tank (\(V_2\)) remains constant, as it is a rigid tank, and the mass of nitrogen has decreased by half. Therefore, we can determine the final pressure (\(P_2\)) and temperature (\(T_2\)) using the ideal gas law: \(P_2V_2=n_2RT_2295.15 K\) Solving for \(P_2\): \(P_2 = \frac{n_2RT_2}{V_2}\)

Calculate specific exergy change

Now that we have the properties of both initial and final states, we can calculate the specific exergy change. The specific exergy change can be defined as: \(\Delta e = e_2 - e_1 = (u_2 - u_1) - T_0(s_2 - s_1) - P_0(v_2 - v_1)\) Where \(u\), \(s\), and \(v\) are specific internal energy, specific entropy, and specific volume, respectively. \(T_0\) and \(P_0\) are the reference temperature and pressure. Since the nitrogen is an ideal gas, change in exergy simplifies to: \(\Delta e=\Delta u - T_0\Delta s - P_0\Delta v\) To find \(\Delta u\), \(\Delta s\), and \(\Delta v\), we can use the ideal gas tables for nitrogen or the equations for ideal gas behavior: \(\Delta u = C_v(T_2 - T_1)\) \(\Delta s = C_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1}\) \(\Delta v = \frac{RT_2}{P_2} - \frac{RT_1}{P_1}\) Where \(C_v\) and \(C_p\) are the specific heat capacities at constant volume and constant pressure, respectively.

Calculate the change in exergy content

Now, let's substitute the values of \(\Delta u\), \(\Delta s\), and \(\Delta v\) into the specific exergy change equation: \(\Delta e=\Delta u - T_0\Delta s - P_0\Delta v\) To find the total change in exergy content inside the tank, multiply the specific exergy change by the mass of nitrogen in the final state: \(\Delta E = m_2 \Delta e\) Finally, substituting values and simplifying the terms will yield the change in exergy content of the tank.