Chapter 4, Problem 4.47P

(a)

Interpretation Introduction

Interpretation:

The heat transfer rate in $\frac{\text{btu}}{\text{s}}$ of an air conditioning unit cools $\text{50}\frac{{\text{ft}}^{\text{3}}}{\text{s}}$ of air at ${\text{94}}^{\text{o}}\text{F}$ to ${\text{68}}^{\text{o}}\text{F}$ should be calculated.

Concept Introduction:

The transfer of heat in steady-flow exchangers for a gas is given as:

$Q=\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$........ (1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

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

Where $\tau =\frac{T}{T_0}$

(b)

Interpretation Introduction

Interpretation:

The heat transfer rate in $\frac{\text{kJ}}{\text{s}}$ of an air conditioning unit cools $1.5\frac{{\text{m}}^{\text{3}}}{\text{s}}$ of air at ${35}^{\text{o}}\text{C}$ to ${25}^{\text{o}}\text{C}$ should be calculated.

Concept Introduction:

The transfer of heat in steady-flow exchangers for a gas is given as:

$Q=\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$........ (1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

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

Where $\tau =\frac{T}{T_0}$