Question

A tank of volume \(1 \mathrm{~m}^{3}\) initially contains steam at \(6 \mathrm{MPa}\) and \(320^{\circ} \mathrm{C}\). Steam is withdrawn slowly from the tank until the pressure drops to \(p\). Heat transfer to the tank contents maintains the temperature constant at \(320^{\circ} \mathrm{C}\). Neglecting all kinetic and potential energy effects (a) determine the heat transfer, in \(\mathrm{kJ}\), if \(p=1.5 \mathrm{MPa}\). (b) plot the heat transfer, in \(\mathrm{kJ}\), versus \(p\) ranging from \(0.5\) to \(6 \mathrm{MPa}\).

Short answer

(a) Determine initial and final states, calculate masses, then find heat transfer Q = m₁h₁ - m₂h₂. (b) Repeat for varying pressures to plot heat transfer versus pressure.

Step-by-step solution

Identify the initial state

Determine the initial properties of the steam inside the tank. The volume is given as 1 m³, initial pressure is 6 MPa, and temperature is 320°C. Using steam tables, find the specific volume \(v_1\) and enthalpy \(h_1\) at this state.

Identify the final state

Determine the properties of steam at the final pressure of 1.5 MPa and temperature remaining constant at 320°C. Using steam tables, find the specific volume \(v_2\) and enthalpy \(h_2\) at this state.

Calculate the initial mass

Use the specific volume \(v_1\) to calculate the initial mass \(m_1\) of the steam using the formula \[ m_1 = \frac{V}{v_1} \] where V is the volume of the tank.

Calculate the final mass

Use the specific volume \(v_2\) to calculate the final mass \(m_2\) of the steam using the formula \[ m_2 = \frac{V}{v_2} \]

Determine mass change

The mass withdrawn from the tank \(m_w\) can be obtained by \[ m_w = m_1 - m_2 \]

Determine the heat transfer

Using the principle of conservation of energy, the heat transfer \(Q\) can be given by \[ Q = m_1 h_1 - m_2 h_2 \] Since the enthalpy change accounts for the energy removed, this expression can be rearranged to find \(Q\).

Plot heat transfer versus pressure

To plot the heat transfer vs. pressure from 0.5 MPa to 6 MPa, repeat steps 2-6 for each pressure value. Collect the heat transfer results for each corresponding pressure to create the plot.

Key concepts

specific volume

Specific volume is a key property in thermodynamics that relates to steam as well. It is defined as the volume occupied by a unit mass of a substance. When looking at steam in a tank, the specific volume helps determine the amount of space that a given mass of steam takes up. To find it, use the formula \[ v = \frac{V}{m} \]where

• *v* is the specific volume in \( m^3/kg \)
• *V* is the total volume in \( m^3 \)
• *m* is the mass in \( kg \)

This property is crucial for understanding how much steam can be held in a tank and is often obtained from steam tables based on pressure and temperature.

enthalpy

Enthalpy is a measure of the total energy in a thermodynamic system and includes internal energy plus the energy required to displace its environment to make room for itself. It is a crucial concept when dealing with heat transfer involving steam. The enthalpy \( h \) of steam can be found using steam tables, which provide the values based on pressure and temperature.
For example, in the exercise, finding the initial and final enthalpies \( h_1 \) and \( h_2 \) involves looking these values up in the steam tables based on the given conditions. The formula to calculate the heat transfer in the tank is given by: \[ Q = m_1 h_1 - m_2 h_2 \]where

• *Q* is the heat transfer in \( kJ \)
• *\(m_1\)* and *\(m_2\)* are the initial and final masses respectively
• *h_1* and *h_2* are the enthalpies at the initial and final states respectively

This expression helps quantify the energy transferred to or from the system during the process.

heat transfer

Heat transfer is the movement of thermal energy from one object or substance to another. There are three methods of heat transfer: conduction, convection, and radiation, but in this context, we are dealing with how heat transfers to maintain the temperature of steam. As the steam is withdrawn, the pressure drops but the heat transfer ensures the temperature remains constant. The energy balance is represented as: \[ Q = m_1 h_1 - m_2 h_2 \]This equation essentially implies that the heat added to the system compensates for the energy lost by the outgoing steam. To find the exact heat transfer value \( Q \), we need the mass and enthalpy before and after the steam withdrawal.
The relationship between mass and specific volume helps determine the amount of heat needed to maintain the temperature constant as pressure changes.

steam tables

Steam tables are essential tools for engineers when dealing with steam and other properties of water at different temperatures and pressures. They contain valuable data such as temperature, pressure, specific volume, enthalpy, and entropy. These tables usually come in two forms: saturated steam tables and superheated steam tables.
In the exercise, you initially use the superheated steam tables because the problem states that the steam is at a specific temperature and pressure above the boiling point of water. The steam table entries help you find specific volume *\( v \)* and enthalpy *\( h \)* for both initial and final states. This information is then used to calculate the initial and final masses (*\( m_1 \)* and *\( m_2 \)*), which are essential for determining the heat transfer.

• To calculate \( m \): \[ m = \frac{V}{v} \]
• For energy balance: \[ Q = m_1 h_1 - m_2 h_2 \]

By following these steps, you can solve any problem involving the thermodynamic properties of steam efficiently.