Question

A \(0.3-\mathrm{m}^{3}\) rigid tank is filled with saturated liquid water at \(200^{\circ} \mathrm{C}\). A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water such that the temperature in the tank remains constant. Determine the amount of heat that must be transferred by the time one-half of the total mass has been withdrawn.

Short answer

Question: Determine the amount of heat required to keep the temperature constant when half of the water is withdrawn from a rigid tank initially filled with 0.3 m³ of saturated liquid water at 200°C. Solution: Step 1: Calculate the initial mass of water in the tank: Since \(v_{f1} = 0.001061 m^3/kg\), we can find the initial mass of water: \(m_1 = \frac{V_1}{v_{f1}} = \frac{0.3}{0.001061} = 282.93 kg\) Step 2: Calculate the enthalpy of the initial state of water: The initial enthalpy is \(H_1 = m_1 \cdot h_{f1} = 282.93 kg × 908.79 kJ/kg = 257150.66 kJ\) Step 3: Determine the final mass of water in the tank: The final mass of water in the tank is \(m_2 = \frac{1}{2}m_1 = 0.5 × 282.93 kg = 141.46 kg\) Step 4: Calculate the enthalpy of the final state of water: The final enthalpy is \(H_2 = m_2 \cdot h_{f2} = 141.46 kg × 908.79 kJ/kg = 128575.33 kJ\) Step 5: Use the energy balance to find the heat transfer: The heat transfer is \(Q = H_2 - H_1 = 128575.33 kJ - 257150.66 kJ = -128575.33 kJ\) Therefore, the amount of heat required to keep the temperature constant during the process is -128,575.33 kJ. Since the heat transfer is negative, it indicates that heat is being removed from the system during the process.

Step-by-step solution

Determine the initial mass of water in the tank

We are given the initial volume of the tank, \(V_1 = 0.3 m^3\), and the temperature of saturated liquid water in the tank, \(T_1 = 200^\circ C\). We can look up the specific volume of saturated liquid water at this temperature in a thermodynamic property table for water. At \(200^\circ C\), the specific volume of saturated liquid water, \(v_{f1}\), is approximately \(0.001061 m^3/kg\). Using this, we can find the initial mass of water in the tank: $$m_1=\frac{V_1}{v_{f1}}$$

Calculate the enthalpy of the initial state of water

We can look up the specific enthalpy of saturated liquid water at \(200^\circ C\) in the thermodynamic property table for water. The specific enthalpy, \(h_{f1}\) is approximately \(908.79 kJ/kg\). Knowing the initial mass of water, \(m_1\), we can calculate the total enthalpy of the initial state of water: $$H_1 = m_1 \cdot h_{f1}$$

Determine the final mass of water in the tank

The problem states that half of the water mass is withdrawn from the tank. Therefore, the final mass of water in the tank, \(m_2\), will be half of the initial mass, \(m_1\): $$m_2 = \frac{1}{2}m_1$$

Calculate the enthalpy of the final state of water

Since the temperature does not change during the process, it remains \(200^\circ C\). Thus, the specific enthalpy of saturated liquid water at the final state, \(h_{f2}\), is also \(908.79 kJ/kg\). Knowing the final mass of water, \(m_2\), we can calculate the total enthalpy of the final state of water: $$H_2 = m_2 \cdot h_{f2}$$

Use the energy balance to find the heat transfer during the process

Since the tank is rigid, work done is zero. We can use the energy balance equation for the closed system to find the heat transfer: $$Q = \Delta{H} = H_2 - H_1$$ Substitute the expressions for \(H_1\) and \(H_2\) from steps 2 and 4: $$Q = m_2 \cdot h_{f2} - m_1 \cdot h_{f1}$$ Substitute the expression for \(m_2\) from step 3 and solve for the amount of heat transfer, \(Q\).