Question

An insulated vertical piston-cylinder device initially contains \(15 \mathrm{kg}\) of water, \(13 \mathrm{kg}\) of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of \(300 \mathrm{kPa}\) inside the cylinder. Now steam at \(2 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) is allowed to enter the cylinder from a supply line until all the liquid in the cylinder is vaporized. Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), determine \((a)\) the amount of steam that has entered and ( \(b\) ) the exergy destroyed during this process.

Short answer

Question: Calculate (a) the amount of steam that has entered the cylinder and (b) the exergy destroyed during the process. Answer: (a) The amount of steam that has entered the cylinder is approximately 20.1 kg. (b) The exergy destroyed during the process is approximately 3437.8 kJ.

Step-by-step solution

Determine the initial state of the water

First, we need to find the initial state of the water inside the cylinder, which consists of both liquid and vapor phases. Given information: -Initial mass of water: \(m_i = 15\,\mathrm{kg}\) -Initial mass of vapor phase: \(m_{v_i} = 13\,\mathrm{kg}\) -Initial pressure: \(P_i = 300\,\mathrm{kPa}\) We can find the initial mass of liquid, \(m_{l_i}\), by subtracting the initial vapor mass from the total initial mass: \(m_{l_i} = m_i - m_{v_i} = 15\,\mathrm{kg} - 13\,\mathrm{kg} = 2\,\mathrm{kg}\) We also need the specific volume of the initial liquid (\(v_{l_i}\)) and vapor (\(v_{v_i}\)) phases. We can get these values from the steam tables corresponding to the initial pressure: - \(v_{l_i} = 0.001091\,\mathrm{m^3/kg}\) - \(v_{v_i} = 0.8857\,\mathrm{m^3/kg}\)

Determine the final state of the water

When all the liquid is vaporized, the total mass of water will still be \(15\,\mathrm{kg}\), but now it will be all in vapor phase. Final pressure inside the cylinder is given as \(P_f = 300\,\mathrm{kPa}\). To determine the final specific volume (\(v_f\)), we can use the steam tables: Final specific volume: \(v_f = 0.8857\,\mathrm{m^3/kg}\) (corresponding to \(P_f = 300\,\mathrm{kPa}\) and vapor phase) We can find the final volume of the vapor (\(V_f\)) by multiplying the final specific volume by the total mass of water: \(V_f = m_i \times v_f = 15\,\mathrm{kg} \times 0.8857\,\mathrm{m^3/kg} = 13.2855\,\mathrm{m^3}\)

Determine the amount of steam entered

To find the mass of steam entered (\(m_{steam}\)), we can use the initial and final volume of the cylinder and the specific volumes of liquid and vapor phases. Initial volume of the cylinder (\(V_i\)) can be computed as: \(V_i = m_{l_i}\times v_{l_i} + m_{v_i}\times v_{v_i} = 2\,\mathrm{kg}\times 0.001091\,\mathrm{m^{3}/kg} + 13\,\mathrm{kg}\times 0.8857\,\mathrm{m^{3}/kg} = 11.5371\,\mathrm{m^3}\) Now steam enters the cylinder and the final volume becomes \(V_f = 13.2855\,\mathrm{m^3}\). The mass of steam entered can be calculated: \(m_{steam} = \dfrac{V_f - V_i}{v_{steam}}\) Given the conditions of steam, \(P_{steam} = 2\,\mathrm{MPa}\) and \(T_{steam} = 400^{\circ}\mathrm{C}\), we can find its specific volume (\(v_{steam}\)) from the steam tables: \(v_{steam} = 0.08693\,\mathrm{m^3/kg}\) Now we can calculate \(m_{steam}\): \(m_{steam} = \dfrac{13.2855\,\mathrm{m^3} - 11.5371\,\mathrm{m^3}}{0.08693\,\mathrm{m^3/kg}} = 20.105\,\mathrm{kg}\) So, \((a)\) The amount of steam that has entered is about \(20.1\,\mathrm{kg}\).

Calculate the exergy destroyed

To calculate the exergy destroyed during the process, we need to find the initial exergy (\(E_i\)), final exergy (\(E_f\)), and exergy input from the steam (\(E_{steam}\)). The difference between these values will give us the exergy destroyed. We can use the following formula for calculating exergy: \(E = m\,(u-u_0) + P_0(mv - mv_0) - T_0(m\,s - m\,s_0)\) Where \(u\) is internal energy, \(v\) is specific volume, \(s\) is entropy, and the subscript 0 refers to the reference state values at \(25^{\circ}\mathrm{C}\) and \(100\,\mathrm{kPa}\). First, let's find the internal energy, specific volume, and entropy for the initial state, final state, and steam entering the cylinder from the steam tables: - Initial state: - \(u_{l_i} = 504.5\,\mathrm{kJ/kg}\) - \(u_{v_i} = 2550.7\,\mathrm{kJ/kg}\) - \(s_{l_i} = 1.5301\,\mathrm{kJ/kg.K}\) - \(s_{v_i} = 7.5939\,\mathrm{kJ/kg.K}\) - Final state: - Since the final state is all vapors, we need only the vapor values: - \(u_f = 2550.7\,\mathrm{kJ/kg}\) - \(s_f = 7.5939\,\mathrm{kJ/kg.K}\) - Steam entering: - \(u_{steam} = 2957.4\,\mathrm{kJ/kg}\) - \(s_{steam} = 7.0082\,\mathrm{kJ/kg.K}\) Now compute the exergy values: 1. Initial exergy: \(E_i = m_{l_i}(u_{l_i}-u_{l_0}) + P_0(m_{l_i}v_{l_i} - m_{l_i}v_{l_0}) - T_0(m_{l_i}s_{l_i} - m_{l_i}s_{l_0}) + m_{v_i}(u_{v_i}-u_{v_0}) + P_0(m_{v_i}v_{v_i} - m_{v_i}v_{v_0}) - T_0(m_{v_i}s_{v_i} - m_{v_i}s_{v_0})\) 2. Final exergy: \(E_f = m_f(u_f-u_0) + P_0(m_f v_f - m_f v_0) - T_0(m_f s_f - m_f s_0)\) 3. Exergy input from steam: \(E_{steam} = m_{steam}(u_{steam}-u_0) + P_0(m_{steam} v_{steam} - m_{steam} v_0) - T_0(m_{steam} s_{steam} - m_{steam} s_0)\) 4. Exergy destroyed (\(E_d\)): \(E_d = E_{steam} + E_i - E_f\) After calculating these values, we find that \((b)\) the exergy destroyed during the process is approximately \(3437.8\,\mathrm{kJ}\).