Question

\(8-74 \quad\) A \(0.6-m^{3}\) rigid tank is filled with saturated liquid water at \(170^{\circ} \mathrm{C}\). A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of \(210^{\circ} \mathrm{C}\) so that the temperature in the tank remains constant. Determine \((a)\) the amount of heat transfer and \((b)\) the reversible work and exergy destruction for this process. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\).

Short answer

Question: Determine the amount of heat transfer, reversible work, and exergy destruction for a rigid tank filled with saturated liquid water at \(170^{\circ} \mathrm{C}\) when half of the mass is withdrawn from the tank and heat is transferred to keep the temperature constant inside the tank. Answer: To find the amount of heat transfer, reversible work, and exergy destruction for the given process, we need to perform energy and exergy balance calculations. The heat transfer can be calculated using a simplified energy balance for an open system, while the reversible work can be determined using an exergy balance for an open system. By obtaining the respective values for these parameters, we can then calculate the exergy destruction using the second-law efficiency equation.

Step-by-step solution

Find the initial state properties of water

To perform the energy balance, we need to find the properties of water at its initial state, which means saturated liquid water at \(170^{\circ} \mathrm{C}\). Using the steam tables, we find the specific volume \(v_1\), internal energy \(u_1\), and the enthalpy \(h_1\).

Determine mass and mass flow rate

We are given that the rigid tank has a volume of \(0.6 \, m^3\), and we can now determine the mass of the water initially inside the tank: \(m_1 = \frac{V_1}{v_1}\) Since half of the mass of the water is withdrawn from the tank, we can determine the mass flow rate (\(\dot{m}\)) of the withdrawn water: \(\dot{m} = \frac{m_1}{2}\)

Perform the energy balance for heat transfer

As mentioned, we will perform a simplified energy balance for an open system. Since the tank is rigid and there is no work interaction other than heat transfer, the energy balance can be written as: \(\dot{Q} - \dot{W} = \dot{m}(\, h_{out} - h_{in} \,)\) In this case, as work interaction does not happen during the process, \(\dot{W} = 0\): \(\dot{Q} = \dot{m}(\, h_{out} - h_{in} \,)\) Now, we can find the amount of heat transfer: \(\dot{Q} = \dot{m}(h_{out} - h_{1})\)

Calculate the reversible work

Now, let's calculate the reversible work by applying an exergy balance for an open system: \(\dot{W}_{rev} = \dot{Q} \left( 1 - \frac{T_0}{T_s} \right)\) Where \(T_0\) is the surrounding temperature, and \(T_s\) is the temperature of the heat source. The reversible work can be calculated as: \(\dot{W}_{rev} = \dot{Q} \left( 1 - \frac{T_0}{T_s} \right)\)

Calculate exergy destruction

To calculate the exergy destruction, we can use the second-law efficiency equation: \(\eta_{II} = \frac{\dot{W}_{rev}}{\dot{Q} - \dot{W}_{rev}}\) Rearrange the equation to make \(E_x\) the subject: \(E_x = \dot{Q} - \dot{W}_{rev}\) Now, we can find exergy destruction: \(E_x = \dot{Q} - \dot{W}_{rev}\) These calculations will give us the required values for \((a)\) the heat transfer \((\dot{Q})\), \((b)\) the reversible work (\(\dot{W}_{rev}\)), and exergy destruction (\(E_x\)) for the given process.