Chapter 4.8, Problem 29P

(A)

Interpretation Introduction

Interpretation:

The mass flow rate $\left({\dot{m}}_{in}\right)$ entering the turbine

Concept Introduction:

Write the expression to calculate the mass flow rate $\left({\dot{m}}_{in}\right)$ entering the turbine.

$\begin{array}{c}{\dot{m}}_{in}=\frac{V_{in}A}{\widehat{V}}\\ =\frac{V\left(\pi D^2/4\right)}{\widehat{V}}\end{array}$

Here, initial velocity of turbine is $V_{in}$, area of turbine is $A$, specific volume is $\widehat{V}$, and diameter of pipe is $D$.

Explanation of Solution

Given information:

Turbine: Pressure of $\text{14}\text{bar}$ and temperature of $\text{300°C}$.

Inlet velocity is $\text{2}\text{.5}\text{m}/\text{s}$.

Pipe diameter is 5 cm.

Refer the Appendix table A-3, “Superheated steam”, obtain the value of specific volume corresponding to pressure of $\text{14}\text{bar}$ and temperature of $\text{300°C}$.

$\widehat{V}=0.182{\text{m}}^{\text{3}}/\text{kg}$

Calculate the mass flow rate $\left({\dot{m}}_{in}\right)$ entering the turbine.

${\dot{m}}_{in}=\frac{V\left(\pi D^2/4\right)}{\widehat{V}}$        (1)

Substitute $\text{2}\text{.5}\text{m}/\text{s}$ for $V_{in}$, $0.182{\text{m}}^{\text{3}}/\text{kg}$ for $\widehat{V}$, and $0.05\text{m}$ for $D$ in Equation (1).

$\begin{array}{c}{\dot{m}}_{in}=\frac{V_{in}\left(\pi D^2/4\right)}{\widehat{V}}\\ =\frac{\left(\text{2}\text{.5}\text{m}/\text{s}\right)\left[\pi {\left(0.05\text{m}\right)}^2/4\right]}{0.182{\text{m}}^{\text{3}}/\text{kg}}\\ =0.270\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate $\left({\dot{m}}_{in}\right)$ entering the turbine is $0.270\text{kg}/\text{s}$.

(B)

Interpretation Introduction

Interpretation:

The power output $\left({\dot{W}}_S\right)$ from the turbine

Concept Introduction:

Write the steady state energy balance equation around the turbine.

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

Here, time taken 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 ${\widehat{H}}_{in}$, rate of heat is added or removed from the system is $\dot{Q}$, 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 of shaft work is added to the system is ${\dot{W}}_{\text{S}}$, and rate of work is added to the system through expansion or contraction of the system is ${\dot{W}}_{\text{EC}}$.

Explanation of Solution

Given information:

Outlet datum is 1.5 m.

Exhaust steam pressure is 0.9 bar.

Calculate the work output using steady state energy balance equation around the turbine.

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

Rewrite the steady state Equation (2) by neglecting the heat transfer and work done by expansion or contraction process.

$\begin{array}{l}0={\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)+{\dot{W}}_S\\ {\dot{W}}_S={\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)-{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)\end{array}$        (3)

Here, initial specific enthalpy is ${\widehat{H}}_{in}$, initial volume is $V_{in}$, initial enthalpy is $h_{in}$, final mass flow rate ${\dot{m}}_{out}$, final specific enthalpy ${\widehat{H}}_{out}$, final volume is $V_{out}$, and final enthalpy is $h_{out}$.

Refer the Appendix table A-3, “Superheated steam”, obtain the value of initial specific enthalpy corresponding to pressure of $\text{14}\text{bar}$ and temperature of $\text{300°C}$.

${\widehat{H}}_{in}=3,040.9\text{kJ}/\text{kg}$

Refer the Appendix table A-1, “Saturated steam-pressure increments”, obtain the value of final specific enthalpy and specific volume corresponding to pressure of $\text{0}\text{.9}\text{bar}$

${\widehat{H}}_{out\left(@{\widehat{H}}^V\right)}=2,670.3\text{kJ}/\text{kg}$

${\widehat{V}}_{out\left(@{\widehat{V}}^V\right)}=1.869{\text{m}}^3/\text{kg}$

Calculate the velocity while assuming that the mass flow rate is constant.

$\begin{array}{l}{\dot{m}}_{in}={\dot{m}}_{out}\\ \frac{V_{in}}{{\widehat{V}}_{in}}=\frac{V_{out}}{{\widehat{V}}_{out}}\\ \frac{V_{in}}{{\widehat{V}}_{in}}\left(A_{inlet}\right)=\frac{V_{out}}{{\widehat{V}}_{out}}\left(A_{outlet}\right)\end{array}$

$\begin{array}{l}\frac{V_{in}}{{\widehat{V}}_{in}}\left(\pi D_{inlet}^2/4\right)=\frac{V_{out}}{{\widehat{V}}_{out}}\left(\pi D_{outet}^2/4\right)\\ V_{out}=\frac{\left(V_{in}\right)\left(\pi D_{inlet}^2/4\right)\left({\widehat{V}}_{out}\right)}{\left({\widehat{V}}_{in}\right)\left(\pi D_{outet}^2/4\right)}\\ V_{out}=\frac{\left(V_{in}\right)\left(D_{inlet}^2\right)\left({\widehat{V}}_{out}\right)}{\left({\widehat{V}}_{in}\right)\left(D_{outet}^2\right)}\end{array}$        (4)

Here, initial mass is $M_{in}$, final mass is $M_{out}$, initial specific volume is ${\widehat{V}}_{in}$, final velocity is $V_{out}$, final specific volume is ${\widehat{V}}_{out}$, area of inlet is $A_{inlet}$, area of outlet is $A_{outlet}$, diameter of inlet is $D_{inlet}$, and diameter of outlet is $D_{outlet}$.

Substitute $\text{2}\text{.5}\text{m}/\text{s}$ for $v_{in}$, $\text{5}\text{cm}$ for $D_{inlet}$, $1.869{\text{m}}^3/\text{kg}$ for ${\widehat{V}}_{out}$, $0.182{\text{m}}^3/\text{kg}$ for ${\widehat{V}}_{in}$, and $\text{20}\text{cm}$ for $D_{outlet}$ in Equation (4).

$\begin{array}{c}{V}_{out}=\frac{\left(v_{in}\right)\left(D_{inlet}^2\right)\left({\widehat{V}}_{out}\right)}{\left({\widehat{V}}_{in}\right)\left(D_{outet}^2\right)}\\ =\frac{\left(\text{2}\text{.5}\text{m}/\text{s}\right){\left(\text{5}\text{cm}\right)}^2\left(1.869{\text{m}}^3/\text{kg}\right)}{\left(0.182{\text{m}}^3/\text{kg}\right){\left(20\text{cm}\right)}^2}\\ =\frac{\left(\text{2}\text{.5}\text{m}/\text{s}\right){\left[\left(\left(\text{5}\text{cm}\right)0.01\text{m}/1\text{cm}\right)\right]}^2\left(1.869{\text{m}}^3/\text{kg}\right)}{\left(0.182{\text{m}}^3/\text{kg}\right){\left[\left(20\text{cm}\right)\left(0.01\text{m}/1\text{cm}\right)\right]}^2}\end{array}$

$\begin{array}{c}=\frac{\left(\text{2}\text{.5}\text{m}/\text{s}\right){\left(\text{0}\text{.05}\text{m}\right)}^2\left(1.869{\text{m}}^3/\text{kg}\right)}{\left(0.182{\text{m}}^3/\text{kg}\right){\left(0.2\text{m}\right)}^2}\\ V_{out}=1.605\text{m}/\text{s}\end{array}$

Substitute $\text{0}\text{.270}\text{kg}/\text{s}$ for ${\dot{m}}_{out}$, $2,670.3\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}$, $1.605\text{m}/\text{s}$ for $V_{out}$, $\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.5\text{m}$ for $h_{out}$. $0$ for $g{h}_{in}$, $\text{0}\text{.270}\text{kg}/\text{s}$ for ${\dot{m}}_{in}$, $3,040.9\text{kJ}/\text{kg}$ for ${\widehat{H}}_{in}$, and $\text{2}\text{.5}\text{m}/\text{s}$ for $V_{in}$ in Equation (3).

$\begin{array}{c}{\dot{W}}_S=\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\left[\left(2,670.3\frac{\text{kJ}}{\text{kg}}\right)+\frac{{\left(1.605\frac{\text{m}}{\text{s}}\right)}^2}{2}+\left(9.8\frac{\text{m}}{{\text{s}}^2}\times 1.4\text{ m}\right)\right]-\left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+\frac{{\left(\text{2}\text{.5}\frac{\text{m}}{\text{s}}\right)}^2}{2}+0\right]\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\left[\left(2,670.3\frac{\text{kJ}}{\text{kg}}\right)+1.30\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}+\left(14.7\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right)\right]-\left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+3.125\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right]\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\begin{array}{l}\left\{\begin{array}{l}\left(2,670.3\frac{\text{kJ}}{\text{kg}}\right)+\left[1.30\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right]+\\ \left[\left(14.7\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right)\right]\end{array}\right\}\\ -\left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+\left[\left(3.125\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right)\right]\right]\end{array}\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\left[\begin{array}{l}\left(2,670.3\frac{\text{kJ}}{\text{kg}}\right)+0.0013\frac{\text{kJ}}{\text{kg}}+\\ 0.0147\frac{\text{kJ}}{\text{kg}}\end{array}\right]-\left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+0.00313\frac{\text{kJ}}{\text{kg}}\right]\right\}\end{array}$

${\dot{W}}_S=-100.05\text{kJ}/\text{s}$

Thus, the power output $\left({\dot{W}}_S\right)$ from the turbine is $-100.05\text{kJ}/\text{s}$.

(C)

Interpretation Introduction

Interpretation:

The efficiency of the turbine

Concept Introduction:

Write the expression to calculate the efficiency of the turbine $\left(\eta \right)$.

$\eta =\frac{{\dot{W}}_S}{{\dot{W}}_{S,rev}}$

Here, rate of actual shaft work is ${\dot{W}}_S$, and rate of reversible shaft work is ${\dot{W}}_{S,rev}$.

Write the entropy balance equation around the boiler.

$\frac{d\left(M\widehat{S}\right)}{dt}={\dot{m}}_{in}{\widehat{S}}_{in}-{\dot{m}}_{out}{\widehat{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$

Here, total mass is $M$, specific entropy is $\widehat{S}$, initial specific entropy is ${\widehat{S}}_{in}$, final specific entropy is ${\widehat{S}}_{out}$, and rate at which entropy is generated within the boundaries of the system is ${\dot{S}}_{gen}$.

Explanation of Solution

Calculate the reversible entropy using entropy balance equation around the boiler.

$\frac{d\left(M\widehat{S}\right)}{dt}={\dot{m}}_{in}{\widehat{S}}_{in}-{\dot{m}}_{out}{\widehat{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$        (5)

Rewrite the Equation (5) by neglecting the surrounding, entropy generation, and reversible entropy term.

$0={\dot{m}}_{in}{\widehat{S}}_{in}-{\dot{m}}_{out}{\widehat{S}}_{out}$

Here, initial specific entropy is ${\widehat{S}}_{in}$, and final specific entropy is ${\widehat{S}}_{out}$.

Refer the Appendix table A-3, “superheated steam”, write the value of initial specific entropy corresponding to pressure of $\text{14}\text{bar}$ and temperature of $\text{300°C}$.

${\widehat{S}}_{in}=6.955\text{kJ}/\text{kg}\cdot \text{K}$

As the initial and final specific entropy is same,

$\begin{array}{c}{\widehat{S}}_{in}={\widehat{S}}_{out}\\ =6.955\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Calculate the work output using steady state energy balance equation around the boiler. $\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_{S,rev}+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$        (6)

Rewrite the steady state Equation (6) by neglecting the heat transfer and work done by expansion or contraction process.

${\dot{W}}_{S,rev}={\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)-{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)$        (7)

Refer the Appendix table A-3, “Superheated steam”, obtain the value of final specific enthalpy corresponding to pressure of $\text{1}\text{bar}$ and an initial specific entropy of $6.955\text{kJ}/\text{kg}\cdot \text{K}$.

${\widehat{H}}_{out}=2,524.4\text{kJ}/\text{kg}$

Substitute $\text{0}\text{.270}\text{kg}/\text{s}$ for ${\dot{m}}_{out}$, $2,524.4\text{kJ}/\text{kg}$ for ${\widehat{H}}_{out}$, $1.605\text{m}/\text{s}$ for $V_{out}$, $14.7{\text{m}}^{\text{2}}/{\text{s}}^2$ for $g{h}_{out}$, $0$ for $g{h}_{in}$, $\text{0}\text{.270}\text{kg}/\text{s}$ for ${\dot{m}}_{in}$, $3,040.9\text{kJ}/\text{kg}$ for ${\widehat{H}}_{in}$, and $\text{2}\text{.5}\text{m}/\text{s}$ for $V_{in}$ in Equation (7).

$\begin{array}{c}{\dot{W}}_{S,rev}=\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\begin{array}{l}\left[\left(2,524.4\text{kJ}/\text{kg}\right)+\frac{{\left(1.605\frac{\text{m}}{\text{s}}\right)}^2}{2}+\left(14.7\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right)\right]-\\ \left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+\frac{{\left(\text{2}\text{.5}\frac{\text{m}}{\text{s}}\right)}^2}{2}+0\right]\end{array}\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\begin{array}{l}\left[\left(2,524.4\text{kJ}/\text{kg}\right)+1.30\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}+\left(14.7\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right)\right]-\\ \left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+3.125\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right]\end{array}\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\begin{array}{c}\left\{\begin{array}{l}\left(2,524.4\text{kJ}/\text{kg}\right)+\left[1.30\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right]+\\ \left[\left(14.7\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right)\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right]\end{array}\right\}-\\ \left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+\left[\left(3.125\frac{{\text{m}}^{\text{2}}}{{\text{s}}^2}\right)\left(\frac{\text{J}}{{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kJ}}{1,000J}\right)\right]\right]\end{array}\right\}\\ =\left(\text{0}\text{.270}\frac{\text{kg}}{\text{s}}\right)\left\{\begin{array}{l}\left[\begin{array}{l}\left(2,524.4\text{kJ}/\text{kg}\right)+0.0013\frac{\text{kJ}}{\text{kg}}+\\ 0.0147\frac{\text{kJ}}{\text{kg}}\end{array}\right]-\\ \left[\left(3,040.9\frac{\text{kJ}}{\text{kg}}\right)+0.00313\frac{\text{kJ}}{\text{kg}}\right]\end{array}\right\}\end{array}$

${\dot{W}}_{S,rev}=-139.45\text{kJ}/\text{s}$

Calculate the efficiency of the turbine $\left(\eta \right)$.

$\eta =\frac{{\dot{W}}_S}{{\dot{W}}_{S,rev}}$        (8)

Substitute $-100\text{kJ}/\text{s}$ for ${\dot{W}}_S$ and $-139.4\text{kJ}/\text{s}$ for ${\dot{W}}_{S,rev}$ in Equation (8).

$\begin{array}{c}\eta =\frac{{\dot{W}}_S}{{\dot{W}}_{S,rev}}\\ =\frac{-100\text{kJ}/\text{s}}{-139.4\text{kJ}/\text{s}}\\ =0.717\\ =71.7\%\end{array}$

Thus, the efficiency of the turbine $\left(\eta \right)$ is $71.7\%$.