Chapter 4.5, Problem 55P

Determine the internal energy change Δu of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A–2c), (b) the c_v value at the average temperature (Table A–2b), and (c) the c_v value at room temperature (Table A–2a)

To determine

The empirical specific heat equation as a function of temperature.

Answer to Problem 55P

The empirical specific heat equation as a function of temperature is $\underset{\_}{6194\text{kJ}/\text{kg}}$.

Explanation of Solution

From Appendix Table A-2c “Ideal-gas specific heats of various common gases”.

Write the expression for the empirical relation between ${\overline{c}}_p\left(T\right)$ and ${\overline{c}}_V\left(T\right)$.

$\begin{array}{c}{\overline{c}}_V\left(T\right)={\overline{c}}_P-R_u\\ =\left(a-R_u\right)+bT+c{T}^2+d{T}^3\end{array}$ (I)

Here, the universal gas constant is $R_u$ and the temperature is $T$.

Write the expression for the change in internal energy.

$\Delta \overline{u}={\displaystyle {\int }_1^2{\overline{c}}_V\left(T\right)dT}$ (II)

Substitute $\left(a-R_u\right)+bT+c{T}^2+d{T}^3$ for ${\overline{c}}_V\left(T\right)$ in Equation (II).

$\begin{array}{c}\Delta \overline{u}={\displaystyle {\int }_1^2\left(\left(a-R_u\right)+bT+c{T}^2+d{T}^3\right)dT}\\ =\left(a-R_u\right){\left(T\right)}_{T_1}^{T_2}+b{\left(\frac{T^2}{2}\right)}_{T_1}^{T_2}+c{\left(\frac{T^3}{3}\right)}_{T_1}^{T_3}+d{\left(\frac{T^4}{4}\right)}_{T_1}^{T_2}\\ =\left(a-R_u\right)\left(T_2-T_1\right)+1/2b\left(T_2^2-T_1^2\right)+1/3c\left(T_2^3-T_1^3\right)+1/4d\left(T_2^4-T_1^4\right)\end{array}$ (III)

Write the expression for internal energy of empirical specific heat equation.

$\Delta u=\frac{\Delta \overline{u}}{M}$ (IV)

Here, the molar mass is $M$.

Conclusion:

Substitute $29.11$ for $a$, $8.314$ for $R_u$, $800\text{K}$ for $T_2$, $200\text{K}$ for $T_1$, $0.1961\times {10}^{-2}$ for $b$, $0.4003\times {10}^{-5}$ for $c$, and $0.8704\times {10}^{-9}$ for $d$ in Equation (III)

$\begin{array}{c}\Delta \overline{u}=\left[\begin{array}{l}\left(29.11-8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(800\text{K}-200\text{k}\right)-\frac{1}{2}\left(0.1961\times {10}^{-2}\right)\left({800}^2\text{K}-{200}^2\text{K}\right)\\ +\frac{1}{3}\left(0.4003\times {10}^{-5}\right)\left({800}^3\text{K}-{200}^3\text{K}\right)-\frac{1}{4}\left(0.8704\times {10}^{-9}\right)\left({800}^4\text{K}-{200}^4\text{K}\right)\end{array}\right]\\ =12487\frac{\text{kJ}}{\text{kmol}}\end{array}$

Substitute $2.016\text{kg}/\text{kmol}$ for $M$ and $12487\text{kJ}/\text{kmol}$ for $\Delta \overline{u}$ in Equation (IV).

$\begin{array}{c}\Delta u=\frac{\left(12487\text{kJ}/\text{kmol}\right)}{\left(2.016\text{kg}/\text{kmol}\right)}\\ =6193.948\text{kJ}/\text{kg}\\ \simeq 6194\text{kJ}/\text{kg}\end{array}$

Thus, the empirical specific heat equation as a function of temperature is $\underset{\_}{6194\text{kJ}/\text{kg}}$.

To determine

The $c_V$ value at the average temperature.

Answer to Problem 55P

The $c_V$ value at the average temperature is $\underset{\_}{6233\text{kJ}/\text{kg}}$.

Explanation of Solution

Write the expression for internal energy of $c_V$ value at the average temperature.

$\Delta u=c_{V,\text{avg}}\left(T_2-T_1\right)$ (V)

Determine the average temperature for the $c_V$ value.

$\begin{array}{c}{T}_{\text{avg}}=\frac{200+800}{2}\\ =500\text{K}\end{array}$

From Table A-2b, write the value of ideal gas specific heat of various gases at various temperatures at 500 K average temperature.

$\begin{array}{c}{c}_{V,\text{avg}}=c_{V@500\text{K}}\\ =10.389\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\end{array}$

Conclusion:

Substitute $10.389\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$ for $c_{V,\text{avg}}$, $800\text{K}$ for $T_2$, and $200\text{K}$ for $T_1$ in Equation (V).

$\begin{array}{c}\Delta u=\left(10.389\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\right)\left(800\text{K}-200\text{K}\right)\\ =6233\text{kJ}/\text{kg}\end{array}$

Thus, the $c_V$ value at the average temperature is $\underset{\_}{6233\text{kJ}/\text{kg}}$.

To determine

The $c_P$ value at the room temperature.

Answer to Problem 55P

The $c_P$ value at the room temperature is $\underset{\_}{6110\text{kJ}/\text{kg}}$.

Explanation of Solution

Write the expression for internal energy of $c_P$ value at the room temperature.

$\Delta u=c_{V,\text{avg}}\left(T_2-T_1\right)$ (VI)

Determine the room temperature for the $c_P$ value.

$\begin{array}{c}{T}_{\text{room}}=27°\text{C}\\ =27+273\text{K}\\ \text{=300}\text{K}\end{array}$

From Table A-2b, write the value of ideal gas specific heat of various gases at various temperatures at 300 K room temperature.

$\begin{array}{c}{c}_{V,\text{avg}}=c_{V@300\text{K}}\\ =10.183\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\end{array}$

Conclusion:

Substitute $10.183\frac{\text{kJ}}{\text{kg}\cdot \text{K}}$ for $c_{V,\text{avg}}$, $800\text{K}$ for $T_2$, and $200\text{K}$ for $T_1$ in Equation (VI).

$\begin{array}{c}\Delta u=\left(10.183\frac{\text{kJ}}{\text{kg}\cdot \text{K}}\right)\left(800\text{K}-200\text{K}\right)\\ =6110\text{kJ}/\text{kg}\end{array}$

Thus, the $c_P$ value at the room temperature is $\underset{\_}{6110\text{kJ}/\text{kg}}$.