Chapter 5, Problem 54P

a)

To determine

The change in enthalpy $\left(\Delta h\right)$ during heating process.

Explanation of Solution

Given:

The initial temperature $\left(T_1\right)$ is 600 K.

The final temperature $\left(T_2\right)$ is 1000 K.

Calculation:

Refer the table A-2, ”Ideal gas specific heats of various common gases table”,

 Obtain the following constant values.

  $\begin{array}{l}a=28.90\\ b=-0.1571\times {10}^{-2}\\ c=0.8081\times {10}^{-5}\\ d=-2.873\times {10}^{-9}\end{array}$

Write the mean change in enthalpy $\left(\Delta \overline{h}\right)$ for an ideal gas during heating process from state 1 to state 2.

  $\Delta \overline{h}={\displaystyle \underset{1}{\overset{2}{\int }}{\overline{c}}_p\left(T\right)dT}$        (I)

Here, mean specific heat at constant volume function of temperature is ${\overline{c}}_v\left(T\right)$ and change in temperature is $dT$.

Substitute $a+bT+c{T}^2+d{T}^3$ for ${\overline{c}}_v\left(T\right)$ in Equation (I).

  $\Delta \overline{h}={\displaystyle \underset{1}{\overset{2}{\int }}\left[a+bT+c{T}^2+d{T}^3\right]dT}$        (II)

Here, constant is a, b, c and d.

Integrate the Equation (II).

  $\Delta \overline{h}=\left\{\begin{array}{l}a\left(T_2-T_1\right)+\frac{1}{2}b\left(T_2^2-T_1^2\right)+\\ \frac{1}{3}c\left(T_2^3-T_1^3\right)+\frac{1}{4}d\left(T_2^4-T_1^4\right)\end{array}\right\}$        (III)

Substitute $T_2=1000\text{K}$, $T_1=600\text{K}$, and constants (a, b, c, and d) in Equation (III).

  $\begin{array}{c}\Delta \overline{h}=\left\{\begin{array}{l}\left(28.90\right)\left(1,000\text{K}-600\text{K}\right)+\frac{1}{2}\left(-0.1571\times {10}^{-2}\right)\\ \left({\left(1,000\text{K}\right)}^2-{\left(600\text{K}\right)}^2\right)+\frac{1}{3}\left(0.8081\times {10}^{-5}\right)\left({\left(1,000\text{K}\right)}^3-{\left(600\text{K}\right)}^3\right)\\ +\frac{1}{4}\left(-2.873\times {10}^{-9}\right)\left({\left(1,000\text{K}\right)}^4-{\left(600\text{K}\right)}^4\right)\end{array}\right\}\\ =12,544\text{kJ}/\text{kmol}\end{array}$

Refer table A-1, “Molar mass properties table”,

The molar mass of nitrogen substance e $\left(M\right)$ as $28.013\text{kg}/\text{kmol}$.

Write the change in enthalpy $\left(\Delta h\right)$ during heating process.

  $\Delta h=\frac{\Delta \overline{h}}{M}$

  $\begin{array}{c}\Delta h=\frac{12,544\text{kJ}/\text{kmol}}{28.013\text{kg}/\text{kmol}}\\ =447.8\text{kJ}/\text{kg}\end{array}$

Thus, the change in enthalpy $\left(\Delta h\right)$ during heating process is $447.8\text{kJ}/\text{kg}$.

b)

To determine

The change in enthalpy $\left(\Delta h\right)$ at temperature of 500 K.

Explanation of Solution

Refer the table A-2, ”Ideal gas specific heats of various common gases table”, obtain the specific heat at constant pressure of nitrogen gas at temperature of 800 K as $1.121\text{kJ}/\text{kg}\cdot \text{K}$.

  $c_{p,avg}=1.121\text{kJ}/\text{kg}\cdot \text{K}$

Write the change in enthalpy $\left(\Delta h\right)$ at a temperature of 500 K.

  $\Delta h=c_{p,avg}\left(T_2-T_1\right)$

  $\begin{array}{c}\Delta h=\left(1.121\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1,000\text{K}-600\text{K}\right)\\ =448.4\text{kJ}/\text{kg}\end{array}$

Thus, the change in enthalpy $\left(\Delta h\right)$ at temperature of 500 K is $448.4\text{kJ}/\text{kg}$.

c)

To determine

The change in enthalpy $\left(\Delta h\right)$ at room temperature.

Explanation of Solution

Refer the table A-2, ”Ideal gas specific heats of various common gases table”, obtain the specific heat at constant volume of nitrogen gas at temperature of 300 K as $1.039\text{kJ}/\text{kg}\cdot \text{K}$.

  $c_{p,avg}=1.039\text{kJ}/\text{kg}\cdot \text{K}$

Write the change in enthalpy $\left(\Delta h\right)$ at room temperature.

  $\Delta h=c_{p,avg}\left(T_2-T_1\right)$

  $\begin{array}{c}\Delta h=\left(1.039\text{kJ}/\text{kg}\cdot \text{K}\right)\left(1,000\text{K}-600\text{K}\right)\\ =415.6\text{kJ}/\text{kg}\end{array}$

Thus, the change in enthalpy $\left(\Delta h\right)$ at room temperature is $415.6\text{kJ}/\text{kg}$.