Chapter 4.5, Problem 123RP

A piston–cylinder device contains helium gas initially at 100 kPa, 10°C, and 0.2 m³. The helium is now compressed in a polytropic process (PVⁿ = constant) to 700 kPa and 290°C. Determine the heat loss or gain during this process.

FIGURE P4–123

To determine

The heat transfer during the process of piston-cylinder device.

Answer to Problem 123RP

The heat transfer during the process of piston-cylinder device is $\underset{\_}{-6.50\text{kJ}}$.

Explanation of Solution

Write the expression for mass of helium.

$m=\frac{P_1{V}_1}{R{T}_1}$ (I)

Here, the initial pressure of helium is $P_1$, the initial volume of helium is $V_1$, the universal gas constant is $R$, and the initial temperature is $T_1$.

Write the expression of ideal gas relation for volume after compression.

$\begin{array}{l}\frac{P_1{V}_1}{R{T}_1}=\frac{P_2{V}_2}{R{T}_2}\\ V_2=\frac{T_2{P}_1}{T_1{P}_2}{V}_1\end{array}$ (II)

Here, the final pressure of helium is $P_1$, the final volume of helium is $V_1$, and the final temperature is $T_1$.

Write the expression of polytropic index by using ideal gas relation.

$\begin{array}{l}{P}_2{V}_2^n=P_1{V}_1^n\\ \left(\frac{P_2}{P_1}\right)={\left(\frac{V_1}{V_2}\right)}^n\end{array}$ (III)

Here, the polytropic index exponent.

Write the expression of boundary work for the polytropic process.

$\begin{array}{c}{W}_{\text{b,in}}=-{\displaystyle {\int }_1^{2}PdV}\\ =-\frac{P_2{V}_2-P_1{V}_1}{1-n}\\ =-\frac{mR\left(T_2-T_1\right)}{1-n}\end{array}$ (IV)

Here, the final temperature of helium is $T_2$.

Write the expression for the energy balance equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (V)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (V) and write energy balance relation of piston-cylinder device.

$\begin{array}{l}{Q}_{\text{in}}+W_{\text{b,in}}=\Delta U\\ Q_{\text{in}}=m\left(u_1-u_2\right)-W_{\text{b,in}}\\ Q_{\text{in}}=m{c}_V\left(T_1-T_2\right)-W_{\text{b,in}}\end{array}$ (VI)

Here, the heat to be transfer into the system is $Q_{\text{in}}$, the constant-volume specific heat of helium is $c_V$, the work to be done by the system is $W_{\text{out}}$, and the change in the internal energy of a system is $\Delta U$.

Conclusion:

From the Table A-1 “Molar mass, gas constant, and critical-point properties”, obtain the value of gas constant of helium for piston-cylinder device as $2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

From the Table A-2 “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of helium for piston-cylinder device as $3.1156\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $100\text{kPa}$ for $P_1$, $0.2{\text{m}}^{\text{3}}$ for $V_1$, $2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, and $283\text{K}$ for $T_1$ in Equation (I).

$\begin{array}{c}m=\frac{\left(100\text{kPa}\right)\left(0.2{\text{m}}^{\text{3}}\right)}{\left(2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(283\text{K}\right)}\\ =\frac{20\text{kPa}\cdot {\text{m}}^{\text{3}}}{587.7627\text{kPa}\cdot {\text{m}}^3/\text{kg}}\\ =0.03403\text{kg}\end{array}$

Substitute $563\text{K}$ for $T_2$, $283\text{K}$ for $T_1$, $100\text{kPa}$ for $P_1$, $700\text{kPa}$ for $P_2$, and $0.2{\text{m}}^{\text{3}}$ for $V_1$ in Equation (II).

$\begin{array}{c}{V}_2=\frac{\left(563\text{K}\right)\left(100\text{kPa}\right)}{\left(283\text{K}\right)\left(700\text{kPa}\right)}\times \left(0.2{\text{m}}^{\text{3}}\right)\\ =\frac{56300\text{kPa}\cdot \text{K}}{\text{198100}\text{kPa}\cdot \text{K}}\times \left(0.2{\text{m}}^{\text{3}}\right)\\ =\left(0.2842\right)\times \left(0.2{\text{m}}^{\text{3}}\right)\\ =0.05684{\text{m}}^{\text{3}}\end{array}$

Substitute $100\text{kPa}$ for $P_1$, $700\text{kPa}$ for $P_2$, $0.2{\text{m}}^{\text{3}}$ for $V_1$, and $0.05684{\text{m}}^{\text{3}}$ for $V_2$ in Equation (III).

$\begin{array}{l}\frac{700\text{kPa}}{100\text{kPa}}={\left(\frac{0.2{\text{m}}^{\text{3}}}{0.05684{\text{m}}^{\text{3}}}\right)}^n\\ 7={\left(3.518649\right)}^n\\ n=1.547\end{array}$

Substitute $0.03403\text{kg}$ for $m$, $2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, $1.547$ for $n$, $563\text{K}$ for $T_2$, and $283\text{K}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{W}_{\text{b,in}}=-\frac{\left(0.03403\text{kg}\right)\left(2.0769\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\left(563\text{K}-283\text{K}\right)}{1-1.547}\\ =-\frac{\left(0.03403\text{kg}\right)\left(2.0769\text{kJ}/\text{kg}\cdot \text{K}\right)\left(280\text{K}\right)}{\left(-0.547\right)}\\ =\frac{-19.7895\text{kJ}}{-0.547}\\ =36.178\text{kJ}\end{array}$

……$\cong 36.18\text{kJ}$

Substitute $0.03403\text{kg}$ for $m$, $3.1156\text{kJ}/\text{kg}\cdot \text{K}$ for $c_V$, $563\text{K}$ for $T_2$, $283\text{K}$ for $T_1$, and $36.18\text{kJ}$ for $W_{\text{b,in}}$ in Equation (VI).

$\begin{array}{c}{Q}_{\text{in}}=\left(0.03403\text{kg}\right)\left(3.1156\text{kJ}/\text{kg}\cdot \text{K}\right)\left(563\text{K}-283\text{K}\right)-\left(36.18\text{kJ}\right)\\ =\left(0.03403\text{kg}\right)\left(3.1156\text{kJ}/\text{kg}\cdot \text{K}\right)\left(280\text{K}\right)-\left(36.18\text{kJ}\right)\\ =-6.493\text{kJ}\\ \cong -6.50\text{kJ}\end{array}$

Thus, the heat transfer during the process of piston-cylinder device is $\underset{\_}{-6.50\text{kJ}}$.