Question

A piston-cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400 C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is (a) 1.0 MPa and 250 C and (b) 500 kPa. (c) Also determine the temperature at the final state in part (b).

Answer 1

Concepts and reason

Closed system:

A closed system is defined as a thermodynamic system in which only energy transfer takes place and there is no mass transfer to or from the system.

Examples for closed system are piston cylinder arrangement, tank etc.

Piston-cylinder or closed tank arrangements are good examples of a closed system. Mass of fluid present in the cylinder is compressed or expanded by external work or heating. Here, only energy transfer takes place by pressure or heat and no mass transfer takes place.

Piston-cylinder arrangement is a closed system if it has no valves. Piston-cylinder arrangement in automobile engines has valves from which matter transfer takes place, these comes under open system.

Internal energy:

Total energy contained in matter at specific state is called as Internal energy.

Specific enthalpy:

Total energy contained per unit mass at specific state is called as specific Internal energy.

Specific volume:

Specific volume is defined as volume per unit mass.

Saturation pressure:

The pressure at which the phase change takes place is known as saturation pressure.

Saturation temperature:

The temperature at which the phase change takes place is known as saturation temperature.

Dryness fraction:

It is the ratio of the mass of vapor to the total mass of the mixture and it is generally represented as x.

Property tables:

Properties of a substance at different states are tabulated. These are used to determine all the specific properties of a substance with any two independent properties of the substance at a specific state.

First law of thermodynamics: When there is any energy transfer takes place to or from the system then the system internal energy changes.

Fundamentals

Linear interpolation:

Linear interpolation is used to calculate the value of a property which is not available from the tables. Consider a state with a temperature ${T_2}$ at which specific enthalpy, ${h_2}$ has to be determined in which properties are at specific temperature are not given in the table.

As the properties are linearly related, use the following formulae to determine the property at specific temperature.

${u_2} = {u_1} + \left( {{u_3} - {u_1}} \right)\left( {\frac{{{T_2} - {T_1}}}{{{T_3} - {T_1}}}} \right)$

Here, temperature ${T_1}$ is the value which is slightly lower than ${T_2}$ and for which properties are available in the table and corresponding specific internal energy is ${u_1}$ , temperature ${T_2}$ is the value which is slightly higher than ${T_2}$ and for which properties are available in the table and corresponding specific internal energy is ${u_3}$ .

This can be used to determine any property in the table.

The work done in closed system can be calculated by using $\int {pdv}$ equation. Here, p is pressure and dv is change in volume.

First law to the piston cylinder arrangement:

$Q = W + dU$

Here, Q is heat transfer, W is work transfer, and dU is change in internal energy

If the work is done on the system then consider the value as negative, if the work is done by the system then consider the value as positive.

If heat transfer is positive then heat addition takes place during the process and if heat transfer is negative then heat rejection takes place during the process.

(a)

Obtain properties of steam at initial pressure ${p_1} = 1.0\,{\rm{MPa}}$ and temperature ${T_1} = 400^\circ {\rm{C}}$ from superheated steam tables.

Specific volume ${v_1} = 0.30661\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

Specific internal energy ${u_1} = 2957.9\,{\rm{kJ/kg}}$

Find the volume of the piston cylinder at stops.

$\begin{array}{c}\\{v_{stop}} = 0.4{v_1}\\\\ = 0.4 \times 0.30661\\\\ = 0.1226\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}\\\end{array}$

Obtain properties of steam at final pressure ${p_2} = 1.0\,{\rm{MPa}}$ and temperature ${T_2} = 250^\circ {\rm{C}}$ from superheated steam tables.

Specific volume ${v_2} = 0.23275\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

Specific internal energy ${u_2} = 2710.4\,{\rm{kJ/kg}}$

The process 1-2 is constant pressure process. Find the work done during the compression process.

$\begin{array}{c}\\{W_{1 - 2}} = m{p_1}\left( {{v_2} - {v_1}} \right)\\\\ = 0.6 \times 1 \times {10^3}\left( {0.23275 - 0.30661} \right)\\\\ = - 44.316\,{\rm{kJ}}\\\end{array}$

Here, negative sign indicates work is done on the system.

Obtain specific volume of steam at final pressure ${p_2} = 500\,{\rm{kPa}}$ from saturated steam tables.

${v_g} = 0.37483\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}$

Here, the process is compression process and ${v_g}$ should not be greater than ${v_{stop}}$ . Hence the process contains two processes. First process is constant pressure process from initial state to stops and the second process is from stops to final state.

Draw p-v diagram of the process.

Now find the work done during the processes.

$W = {\left( {pdV} \right)_{1 - stop}} + {\left( {pdV} \right)_{stop - 2}}$

The volume remains constant during the process from stops to final state. Hence the work done is zero during the process from stops to final state.

$\begin{array}{c}\\W = m{p_1}\left( {{v_{stop}} - {v_1}} \right) + 0\\\\ = 0.6 \times 1 \times {10^3}\left( {0.1226 - 0.30661} \right)\\\\ = - 110.406\,{\rm{kJ}}\\\end{array}$

Here, negative sign indicates work is done on the system.

c)

The final volume is equal to the volume at stops.

$\begin{array}{c}\\{v_2} = {v_{stop}}\\\\ = 0.1226\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}\\\end{array}$

Take the specific volumes at 500 kPa from saturated water tables.

$\begin{array}{l}\\{v_f} = 0.001093\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}\\\\{v_g} = 0.37483\,{{\rm{m}}^{\rm{3}}}{\rm{/kg}}\\\end{array}$

The final volume lies in between saturated liquid and saturated vapor.

${v_f} < {v_2} < {v_g}$

Hence the final state should be in the two phase region. Therefore, the temperature at the final state is equal to the saturation temperature at 500 kPa.

Obtain saturation temperature at final pressure ${p_2} = 500\,{\rm{kPa}}$ from saturated water tables.

${T_{sat}} = {T_2} = 151.83^\circ {\rm{C}}$

Ans: Part a

Therefore, the compression work is $- 44.316\,{\rm{kJ}}$