Question

A \(2-m^{3}\) rigid insulated tank initially containing saturated water vapor at \(1 \mathrm{MPa}\) is connected through a valve to a supply line that carries steam at \(400^{\circ} \mathrm{C}\). Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be \(300^{\circ} \mathrm{C}\). Determine the mass of the steam that has entered and the pressure of the steam in the supply line.

Short answer

To find the mass of the steam that entered the tank (ms) and the pressure of the steam in the supply line (Ps), we need to solve the conservation of energy equation and the isentropic process assumption equation: 1. 10.3 kg × 2584.7 kJ/kg + ms × us = (10.3 kg + ms) × 2967 kJ/kg 2. Ps × vs^k = 2 MPa × (0.1943 m^3 / kg)^k By solving these equations, we can find ms and Ps, which are the mass of steam that entered the tank during the process and the pressure of the steam in the supply line, respectively.

Step-by-step solution

Find the initial properties of the steam inside the tank

Since the tank initially contains saturated water vapor at 1 MPa, we can determine the initial specific volume (\(v_1\)), initial internal energy (\(u_1\)), and initial temperature (\(T_1\)) from the saturated steam tables. Using the steam tables, we can find the initial specific volume \(v_1 = 0.1943 \, m^3/kg\) and initial internal energy \(u_1 = 2584.7 \, kJ/kg\) for saturated water vapor at 1 MPa.

Calculate the final properties of the steam inside the tank

The final pressure inside the tank is given as 2 MPa, and the final temperature is measured to be \(300^{\circ}\mathrm{C}\). To find the final specific volume (\(v_2\)), we need to use the fact that in a rigid tank, the specific volume remains constant since the volume is fixed. Thus, \(v_2 = v_1 = 0.1943 \, m^3/kg\). Now, we look up the properties of steam at \(P_2 = 2 \, MPa\) and \(T_2 = 300^{\circ}\mathrm{C}\) in the superheated steam tables. We can find the final internal energy of the steam inside the tank: \(u_2 = 2967 \, kJ/kg\).

Analyze the process and find the mass of the steam, pressure, and temperature in the supply line

Since the tank is insulated, no heat is transferred, and the process is adiabatic. We can apply the first law of thermodynamics which states that the change in internal energy of a closed system is equal to the work done, but in our case, there is no work done on the system, so the specific internal energy remains constant. Initially, the tank has a mass \(m_1 = V / v_1 = (2 \, m^3) / (0.1943 \, m^3/kg) = 10.3 \, kg\) of steam. Let \(m_s\) be the mass of the steam that entered the tank from the supply line. Then, the total mass of the steam in the final state is \(m_2 = m_1 + m_s\). Now we apply the conservation of energy equation: \(m_1 u_1 + m_s u_s = m_2 u_2\), where \(u_s\) is the internal energy of the steam in the supply line. Substituting the values and solving we get: \(10.3 \, kg \times 2584.7\, kJ/kg + m_s \times u_s = (10.3 \, kg + m_s) \times 2967\, kJ/kg\) To find the pressure of the steam in the supply line (\(P_s\)), we'll use the isentropic process assumption for the flow of steam from the supply line. We have: \(P_s \times v_s^k = 2\, MPa \times (0.1943\, m^3 / kg)^{k}\) Where \(v_s\) is the specific volume of the steam in the supply line, and \(k = c_p/ c_v\) is the ratio of specific heats. Once we find \(P_s\), we can determine the mass of the steam that entered the tank (\(m_s\)) and the pressure (\(P_{supply}\)) in the supply line.