Chapter 5, Problem 5.47P

(a)

Interpretation Introduction

Interpretation:

What is the minimum power requirement in hp of air.

Concept Introduction:

The minimum power requirement is calculated by following formula if air is assumed as an ideal gas:

  $W_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S$

    ....(1)

Answer to Problem 5.47P

  $W_{\text{ideal}}=1.848\text{hp}$

Explanation of Solution

Given information:

It is given that air at${70}^{\circ }\text{F}\text{and 1 atm}$ is cooled at the rate of$100000\frac{{\text{ft}}^3}{\text{hr}}$ to${20}^{\circ }\text{F}$ by refrigeration. The surrounding temperature is given as${70}^{\circ }\text{F}$ .

Hence

  $T_0={70}^{\circ }\text{F}+459.67=529.67\text{Rankine}$

  $T={20}^{\circ }\text{F}+459.67=479.67\text{Rankine}$

  $T_{\sigma }={70}^{\circ }\text{F}+459.67=529.67\text{Rankine}$

  $\stackrel{•}{V}=100000\frac{{\text{ft}}^3}{\text{hr}}$

From equation (1),

  $W_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S$

For an ideal gas$\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}$ where,

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

    ....(2)

And

  $\tau =\frac{T}{T_0}$

Values of above constants for air in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{479.67\text{Rankine}}{529.67\text{Rankine}}=0.906\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=3.355\times 529.67\times (0.906-1)+\frac{0.575\times {10}^{-3}}{2}\times {529.67}^2\times ({0.906}^2-1)\\ +\frac{0}{3}\times {529.67}^3({0.906}^3-1)+\frac{-0.016\times {10}^5}{529.67}(\frac{0.906-1}{0.906})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=-181.18\text{Rankine}\end{array}$

Therefore,

  $\begin{array}{l}\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}\\ \Delta H=1.987\frac{\text{btu}}{\text{lb-mol Rankine}}\times -181.18\text{Rankine}\\ \Delta H=-360\frac{\text{btu}}{\text{lb-mol }}\end{array}$

And

For an ideal isobaric gas$\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$ where,

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$

    ....(3)

And

  $\tau =\frac{T}{T_0}$

Values of above constants for air in equation (3) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{479.67\text{Rankine}}{529.67\text{Rankine}}=0.906\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.355+\left[0.575\times {10}^{-3}\times 529.67+\left(\frac{-0.016\times {10}^5}{{0.906}^2\times {529.67}^2}\right)\left(\frac{0.906+1}{2}\right)\right]\left(\frac{0.906-1}{\ln 0.906}\right)\right]\ln 0.906\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=-0.3592\end{array}$

Therefore,

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=1.987\frac{\text{btu}}{\text{lb-mol Rankine}}\times -0.3592\\ \Delta S=-0.714\frac{\text{btu}}{\text{lb-mol Rankine}}\end{array}$

Hence

  $\begin{array}{l}{W}_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S\\ W_{\text{ideal}}=-360\frac{\text{btu}}{\text{lb-mol }}+529.67\text{Rankine}\times 0.714\frac{\text{btu}}{\text{lb-mol Rankine}}\\ W_{\text{ideal}}=18.184\frac{\text{btu}}{\text{lb-mol }}\end{array}$

And we know that for an ideal gas, ideal gas law is applicable and hence

  $\begin{array}{l}P\stackrel{•}{V}=\stackrel{•}{n}RT\\ \stackrel{•}{n}=\frac{P\stackrel{•}{V}}{RT}=\frac{1\cancel{\text{atm}}\times 100000\frac{{\text{ft}}^3}{\text{hr}}}{0.730\frac{{\text{ft}}^3.\cancel{\text{atm}}}{\text{lb-mol Rankine}}\times 529.67\text{Rankine}}=258.626\frac{\text{lb-mol}}{\text{hr}}\end{array}$

Hence

  $\begin{array}{l}{W}_{\text{ideal}}=18.184\frac{\text{btu}}{\text{lb-mol }}\times 258.626\frac{\text{lb-mol}}{\text{hr}}=4702.953\frac{\text{btu}}{\text{hr}}\\ W_{\text{ideal}}=4702.953\frac{\text{btu}}{\text{hr}}\times \frac{0.000393\text{hp}}{1\frac{\text{btu}}{\text{hr}}}=1.848\text{hp}\\ W_{\text{ideal}}=1.848\text{hp}\end{array}$

(b)

Interpretation Introduction

Interpretation:

What is the minimum power requirement in$\text{kW}$ of air.

Concept Introduction:

The minimum power requirement is calculated by following formula if air is assumed as an ideal gas:

  $W_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S$

    ....(1)

Answer to Problem 5.47P

  $W_{\text{ideal}}=2.019\text{kW}$

Explanation of Solution

Given information:

It is given that air at${25}^{\circ }\text{C}\text{and 1 atm}$ is cooled at the rate of$3000\frac{{\text{m}}^3}{\text{hr}}$ to$-8^{\circ }\text{C}$ by refrigeration. The surrounding temperature is given as${25}^{\circ }\text{C}$ .

Hence

  $T_0={25}^{\circ }\text{C}+273.15=298.15\text{K}$

  $T=-8^{\circ }\text{C}+273.15=265.15\text{K}$

  $T_{\sigma }={25}^{\circ }\text{C}+273.15=298.15\text{K}$

  $\stackrel{•}{V}=3000\frac{{\text{m}}^3}{\text{hr}}$

From equation (1),

  $W_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S$

For an ideal gas$\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}$ where,

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

    ....(2)

And

  $\tau =\frac{T}{T_0}$

Values of above constants for air in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{265.15\text{K}}{298.15\text{K}}=0.8893\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=3.355\times 298.15\times (0.8893-1)+\frac{0.575\times {10}^{-3}}{2}\times {298.15}^2\times ({0.8893}^2-1)\\ +\frac{0}{3}\times {298.15}^3({0.8893}^3-1)+\frac{-0.016\times {10}^5}{298.15}(\frac{0.8893-1}{0.8893})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}=-115.41\text{K}\end{array}$

Therefore,

  $\begin{array}{l}\Delta H=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}dT}\\ \Delta H=8.314\frac{\text{J}}{\text{mol K}}\times -115.41\text{K}\\ \Delta H=-959.52\frac{\text{J}}{\text{mol}}\end{array}$

And

For an ideal isobaric gas$\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$ where,

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$

    ....(3)

And

  $\tau =\frac{T}{T_0}$

Values of above constants for air in equation (3) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{265.15\text{K}}{298.15\text{K}}=0.8893\end{array}$

And

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.355+\left[0.575\times {10}^{-3}\times 298.15+\left(\frac{-0.016\times {10}^5}{{0.889}^2\times {298.15}^2}\right)\left(\frac{0.889+1}{2}\right)\right]\left(\frac{0.889-1}{\ln 0.889}\right)\right]\ln 0.889\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=-0.411\end{array}$

Therefore,

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times -0.411\\ \Delta S=-3.417\frac{\text{J}}{\text{mol K}}\end{array}$

Hence

  $\begin{array}{l}{W}_{\text{ideal}}=\Delta H-T_{\sigma }\Delta S\\ W_{\text{ideal}}=-959.52\frac{\text{J}}{\text{mol}}-298.15\text{K}\times -3.417\frac{\text{J}}{\text{mol K}}\\ W_{\text{ideal}}=59.275\frac{\text{J}}{\text{mol}}\end{array}$

And we know that for an ideal gas, ideal gas law is applicable and hence

  $\begin{array}{l}P\stackrel{•}{V}=\stackrel{•}{n}RT\\ \stackrel{•}{n}=\frac{P\stackrel{•}{V}}{RT}=\frac{1\cancel{\text{atm}}\times 3000\frac{{\text{m}}^3}{\text{hr}}}{8.205\times {10}^{-5}\frac{{\text{m}}^3\cancel{\text{atm}}}{\text{mol K}}\times 298.15\text{K}}=122633.142\frac{\text{mol}}{\text{hr}}\end{array}$

Hence

  $\begin{array}{l}{W}_{\text{ideal}}=59.275\frac{\text{J}}{\text{mol}}\times 122633.142\frac{\text{mol}}{\text{hr}}=7269079.494\frac{\text{J}}{\text{hr}}\\ W_{\text{ideal}}=7269079.494\frac{\text{J}}{\text{hr}}\times \frac{1\text{hr}}{3600\text{s}}\times \frac{1\frac{\text{kJ}}{\text{s}}}{1000\frac{\text{J}}{\text{s}}}\times \frac{1\text{kW}}{1\frac{\text{kJ}}{\text{s}}}=2.019\text{kW}\\ W_{\text{ideal}}=2.019\text{kW}\end{array}$