Chapter 5.7, Problem 8P

(A)

Interpretation Introduction

Interpretation:

The operating temperature of the condenser and the efficiency of the Carnot cycle.

Concept Introduction:

The expression to obtain the Carnot efficiency is,

$\eta =1-\frac{T_C}{T_H}$

Here, temperature of a low-temperature and high temperature heat reservoir is $T_C$ and $T_H$ respectively.

(B)

Interpretation Introduction

Interpretation:

The actual efficiency of this Rankine cycle and compare it to the Carnot efficiency.

Concept Introduction:

The expression to obtain the general expression for an entropy balance equation 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 specific entropy of an outlet for a reversible process is,

${\widehat{S}}_{out,rev}=\left(1-q_{rev}\right){\widehat{S}}_L+q_{rev}{\widehat{S}}_V$

Here, mole fraction of the system for reversible process is $q_{rev}$.

The energy balance for an adiabatic steady state turbine is,

$\frac{{\dot{W}}_{S,turbine}}{\dot{m}}={\widehat{H}}_{out}-{\widehat{H}}_{in}$

Here, specific enthalpy at inlet and outlet is ${\widehat{H}}_{in}$ and ${\widehat{H}}_{out}$, mass flow rate is $\dot{m}$, and rate at which shaft work is added to the system is ${\dot{W}}_S$.

The expression to obtain the pump work is,

$\frac{{\dot{W}}_{S,pump}}{\dot{m}}=\widehat{V}\left(P_{out}-P_{in}\right)$

Here, specific volume is $\widehat{V}$, inlet, and outlet pressure is $P_{in}$ and $P_{out}$ respectively.

The simplification equation of energy balance around whole engine is,

$0=\frac{{\dot{Q}}_C}{\dot{m}}+\frac{{\dot{Q}}_H}{\dot{m}}+\frac{{\dot{W}}_{S,pump}}{\dot{m}}+\frac{{\dot{W}}_{S,turbine}}{\dot{m}}$

Here, rate of heat exchange with low temperature reservoir is ${\dot{Q}}_H$.

The expression of efficiency of heat engine is,

$\begin{array}{c}{\eta }_{\text{H}\text{.E}}=\frac{\left|W_{\text{net}}\right|}{\left|Q_{\text{added}}\right|}\\ =\frac{\left|\frac{{\dot{W}}_{S,turbine}}{\dot{m}}+\frac{{\dot{W}}_{S,pump}}{\dot{m}}\right|}{\frac{{\dot{Q}}_H}{\dot{m}}}\end{array}$

Here, net work done on turbine and pump is $W_{net}$, mass flow rate is $\dot{m}$, heat added to the heat engine system is $Q_{added}$, rate of shaft work for turbine and pump is ${\dot{W}}_{S,turbine}$ and ${\dot{W}}_{S,pump}$, and rate of heat transfer to the heat engine is ${\dot{Q}}_H$.

(C)

Interpretation Introduction

Interpretation:

What effect do you expect this action had?