Chapter 5.7, Problem 19P

(A)

Interpretation Introduction

Interpretation:

The flow rate of refrigerant. Is it the same or different in the two cycles?

Concept Introduction:

The equation of energy balance on the condenser is,

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{\text{S}}+W_{\text{EC}}+\dot{Q}\end{array}\right]$

Here, time is $t$, total mass is $M$, specific internal energy is $\widehat{U}$, velocity is $V$, acceleration due to gravity is $g$, height is $h$, initial mass flow rate is ${\dot{m}}_{in}$, initial specific enthalpy is ${\widehat{H}}_{in}$, initial velocity is $V_{in}$, initial height of the gas is $h_{in}$, final mass flow rate is ${\dot{m}}_{out}$, final height of the gas is $h_{out}$, rate at which shaft work is added to the system is ${\dot{W}}_{\text{S}}$, rate at which work is added to the system through expansion or contraction of the system is ${\dot{W}}_{\text{EC}}$, and rate at which heat is added to the system is $\dot{Q}$.

(B)

Interpretation Introduction

Interpretation:

The compressor work for each of the two possible compressors.

Concept Introduction:

The equation of generalized entropy balance is,

$\frac{d\left(M\widehat{S}\right)}{dt}={\displaystyle \sum _{j=1}^{j=J}{\dot{m}}_{j,in}{\widehat{S}}_j}-{\displaystyle \sum _{k=1}^{k=K}{\dot{m}}_{k,out}{\widehat{S}}_k}+{\displaystyle \sum _{n=1}^{n=N}\frac{{\dot{Q}}_n}{T_n}}+{\dot{S}}_{gen}$

Here, time is t, mass of the system is M, specific entropy of the system is $\widehat{S}$, mass flow rates of individual streams entering and leaving the system is ${\dot{m}}_{j,in}$, ${\dot{m}}_{k,out}$, specific entropies of streams entering and leaving the system is ${\widehat{S}}_j,{\widehat{S}}_k$, actual rate at which heat is added to or removed from the system at one particular location is ${\dot{Q}}_n$, the temperature of the system at the boundary where the heat transfer labelled n occurs is $T_n$, and the rate at which entropy is generated within the boundaries of the system is ${\dot{S}}_{gen}$.

The expression to obtain the compressor work rate with 80% efficiency is,

$\frac{{\dot{W}}_{S,actual}}{\dot{m}}=\frac{\frac{{\dot{W}}_{S,reversible}}{\dot{m}}}{{\eta }_{compressor}}$

The expression to obtain the compressor work with 80% efficiency is,

${\dot{W}}_{S,actual,80\%}=\frac{{\dot{W}}_{S,actual}}{\dot{m}}\left(\dot{m}\right)$

(C)

Interpretation Introduction

Interpretation:

The coefficient of performance for the cycle.

Concept Introduction:

The expression to obtain the coefficient of performance $\left(\text{C}\text{.O}\text{.P}\right)$ of compressor work with 80% efficiency is,

$\text{C}\text{.O}\text{.P}=\frac{{\dot{Q}}_C}{{\dot{W}}_{S,actual,80\%}}$

The expression to obtain the coefficient of performance $\left(\text{C}\text{.O}\text{.P}\right)$ of compressor work with 70% efficiency is,

$\text{C}\text{.O}\text{.P}=\frac{{\dot{Q}}_C}{{\dot{W}}_{S,actual,70\%}}$

(D)

Interpretation Introduction

Interpretation:

How long the system would the system have to run in order for the higher-efficiency compressor to be cost effective?

Concept Introduction:

The expression to obtain the time $\left(t\right)$ that the system has to run in order for the higher efficiency compressor to be cost effective is,

$t=\frac{\$5,000/\text{cost}}{{\dot{W}}_{S,actual,70\%}-{\dot{W}}_{S,actual,80\%}}$