Chapter 6, Problem 162RQ

To determine

Plot the following graphs for the pressure varies from $10\text{kPa}$ to $50\text{kPa}$ and the exit area varies from $1000{\text{cm}}^{\text{2}}$ to $3000{\text{cm}}^{\text{2}}$.

• Exit velocity versus exit pressure power output of turbine
• Turbine power output versus exit pressure power output of turbine

Explanation of Solution

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for inlet mass flow rate.

  $\dot{m}=\frac{A_1{V}_1}{v_1}$        (I)

Here, the cross-sectional area is $A$, the velocity is $V$ and the specific volume is $v$; the subscript 1 indicates the inlet condition.

At inlet:

The steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

Obtain the inlet enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ corresponding to the pressure of $7\text{MPa}$ and the temperature of $600°\text{C}$.

  $\begin{array}{l}{h}_1=3650.6\text{kJ/kg}\\ v_1=0.055665{\text{m}}^3\text{/kg}\end{array}$

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for exit mass flow rate.

  $\dot{m}=\frac{A_2{V}_2}{v_2}$        (II)

Here, the cross-sectional area is $A$, the velocity is $V$ and the specific volume is $v$; the subscript 2 indicates the exit condition.

Rearrange the Equation (II) to obtain exit velocity $\left(V_2\right)$.

  $V_2=\frac{\dot{m}{v}_2}{A_2}$        (III)

At exit:

Consider the exit pressure $\left(P_2\right)$ is $25\text{kPa}$.

The steam is with the quality of $95\%$.

Write the formula for exit enthalpy $\left(h_2\right)$ of the steam with the consideration of its quality.

  $h_2=h_f+x{h}_{fg}$        (IV)

Write the formula for exit specific volume $\left(v_2\right)$ of the steam with the consideration of its quality.

  $v_2=v_f+x\left(v_g-v_f\right)$        (V)

Here, the enthalpy is $h$, the quality of steam is $x$, and subscript $f$ indicates that fluid state, $g$ indicates the gaseous state, $fg$ indicates the mixed state (vaporization).

Refer Table A-5, “Saturated water—Pressure table”.

Obtain the following corresponding to the pressure of $25\text{kPa}$.

  $\begin{array}{l}{v}_f=0.001020{\text{m}}^3\text{/kg}\\ v_g=6.2034{\text{m}}^3\text{/kg}\\ h_f=271.96\text{kJ/kg}\\ h_{fg}=2345.5\text{kJ/kg}\end{array}$

Consider the steam flows at steady state. Hence, the inlet and exit mass flow rates are equal.

  ${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the energy rate balance equation for one inlet and one outlet system.

  $\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$        (VI)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The steam flows at steady state through the turbine. Hence, the rate of change in net energy of the system becomes zero.

  $\Delta {\dot{E}}_{\text{system}}=0$

Heat loss occurs at the rate of $20\text{kJ/kg}$. Neglect the potential energy changes. Here, the work done is by the system (turbine) and the work done on the system is zero i.e. ${\dot{W}}_1=0$.

The Equations (VI) reduced as follows to obtain the work output $\left({\dot{W}}_2\right)$.

  $\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)={\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ {\dot{W}}_2=\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)-{\dot{Q}}_2\end{array}$

  $\begin{array}{l}{\dot{W}}_2=\dot{m}\left(h_1-h_2+\frac{V_1{}^2-V_2{}^2}{2}\right)-{\dot{Q}}_2\\ {\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-{\dot{Q}}_2\end{array}$        (VII)

Here, ${\dot{Q}}_2=\dot{m}{Q}_2$.

Rewrite the Equation (VII) as follows.

  ${\dot{W}}_2=-\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)-\dot{m}{Q}_2$        (VIII)

Calculation:

Substitute $271.96\text{kJ/kg}$ for $h_f$, $95\%$ for $x$ and $2345.5\text{kJ/kg}$ for $h_{fg}$ in Equation (IV).

  $\begin{array}{c}{h}_2=271.96\text{kJ/kg}+95\%\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+\frac{95}{100}\left(2345.5\text{kJ/kg}\right)\\ =271.96\text{kJ/kg}+2228.225\text{kJ/kg}\\ =2500.185\text{kJ/kg}\end{array}$

Substitute $0.001020{\text{m}}^3\text{/kg}$ for $v_f$, $95\%$ for $x$ and $6.2034{\text{m}}^3\text{/kg}$ for $v_g$ in

Equation (V).

  $\begin{array}{c}{v}_2=0.001020{\text{m}}^3\text{/kg}+95\%\left(6.2034{\text{m}}^3\text{/kg}-0.001020{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+\frac{95}{100}\left(6.20238{\text{m}}^3\text{/kg}\right)\\ =0.001020{\text{m}}^3\text{/kg}+5.892261{\text{m}}^3\text{/kg}\\ =5.893281{\text{m}}^3\text{/kg}\end{array}$

Substitute $150{\text{cm}}^2$ for $A_1$, $60\text{m/s}$ for $V_1$ ,and $0.055665{\text{m}}^3\text{/kg}$ for $v_1$ in Equation (I).

  $\begin{array}{c}\dot{m}=\frac{\left(150{\text{cm}}^2\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =\frac{\left(150{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\left(60\text{m/s}\right)}{0.055665{\text{m}}^3\text{/kg}}\\ =16.1681\text{kg/s}\\ \simeq 16.17\text{kg/s}\end{array}$

Consider the exit area $\left(A_2\right)$ of the turbine is $1400{\text{cm}}^3$.

Substitute $16.17\text{kg/s}$ for $\dot{m}$, $5.893281{\text{m}}^3\text{/kg}$ for $v_2$ , and $1400{\text{cm}}^2$ for $A_2$ in

Equation (III).

  $\begin{array}{c}{V}_2=\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2}\\ =\frac{\left(16.17\text{kg/s}\right)\left(5.893281{\text{m}}^3\text{/kg}\right)}{1400{\text{cm}}^2\times \frac{1\text{}{\text{m}}^2}{{10}^4{\text{cm}}^2}}\\ =680.6739\text{m/s}\end{array}$

Substitute $16.17\text{kg/s}$ for $\dot{m}$, $2500.185\text{kJ/kg}$ for $h_2$, $3650.6\text{kJ/kg}$ for $h_1$, $680.6739\text{m/s}$ for $V_2$, $60\text{m/s}$ for $V_1$, and $20\text{kJ/kg}$ for $Q_2$ in Equation (VIII).

  $\begin{array}{c}{\dot{W}}_2=-16.17\text{kg/s}\left[\begin{array}{l}2500.185\text{kJ/kg}-3650.6\text{kJ/kg}\\ +\frac{{\left(680.6739\text{m/s}\right)}^2-{\left(60\text{m/s}\right)}^2}{2}\end{array}\right]-16.17\text{kg/s}\left(20\text{kJ/kg}\right)\\ =\left[\begin{array}{c}-16.17\text{kg/s}\left(-1150.415\text{kJ/kg}+229858.4791{\text{m}}^2{\text{/s}}^2\times \frac{1\text{}\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}\right)\\ -323.4\text{kJ/s}\end{array}\right]\\ =-16.17\text{kg/s}\left(-920.5565\text{kJ/kg}\right)-323.4\text{kJ/s}\\ =\text{14885}\text{.3989}\text{kJ/s}-323.4\text{kJ/s}\end{array}$

  $\begin{array}{c}=14561.9989\text{kJ/s}\times \frac{1\text{kW}}{1\text{kJ/s}}\\ \simeq 14562\text{kW}\end{array}$

The exit velocity $\left(V_2\right)$ and the power output $\left({\dot{W}}_2\right)$ for exit pressure $\left(P_2\right)$ of $25\text{kPa}$ and the exit area $\left(A_2\right)$ of $1400{\text{cm}}^2$ is as follows.

  $\begin{array}{l}{V}_2=680.6739\text{m/s}\\ {\dot{W}}_2=14562\text{kW}\end{array}$

Using excel spread sheet, the exit velocity $\left(V_2\right)$ and the power output $\left({\dot{W}}_2\right)$ is calculated for the exit pressure varies from $10\text{kPa}$ to $50\text{kPa}$, for the exit area of $1000{\text{cm}}^{\text{2}}$ and shown in Table 1.

S.No.	$P_2\left(\text{kPa}\right)$	$V_2\left(\text{m/s}\right)$	${\dot{W}}_2\left(\text{kW}\right)$
1	10	2253.540216	–22171.1196
2	15	1539.230498	–514.857057
3	20	1174.871104	7295.806083
4	25	952.9435377	10965.91684
5	30	803.2150134	12968.62817
6	40	613.4390747	14943.44488
7	50	497.7670121	15822.49054

Table 1

Similarly, the exit velocity $\left(V_2\right)$ and the power output $\left({\dot{W}}_2\right)$ is calculated for the exit pressure varies from $10\text{kPa}$ to $50\text{kPa}$, for the exit area of $2000{\text{cm}}^{\text{2}}$ and shown in Table 2.

S.No.	$P_2\left(\text{kPa}\right)$	$V_2\left(\text{m/s}\right)$	${\dot{W}}_2\left(\text{kW}\right)$
1	10	1126.770108	8623.292217
2	15	769.6152491	13851.56455
3	20	587.435552	15665.73428
4	25	476.4717689	16472.41662
5	30	401.6075067	16880.6829
6	40	306.7195374	17225.27947
7	50	248.883506	17324.918

Table 2

Similarly, the exit velocity $\left(V_2\right)$ and the power output $\left({\dot{W}}_2\right)$ is calculated for the exit pressure varies from $10\text{kPa}$ to $50\text{kPa}$, for the exit area of $3000{\text{cm}}^{\text{2}}$ and shown in Table 3.

S.No.	$P_2\left(\text{kPa}\right)$	$V_2\left(\text{m/s}\right)$	${\dot{W}}_2\left(\text{kW}\right)$
1	10	751.180072	14325.96107
2	15	513.0768327	16512.01299
3	20	391.6237013	17215.72099
4	25	317.6478459	17492.1388
5	30	267.7383378	17605.13749
6	40	204.4796916	17647.84143
7	50	165.9223374	17603.1453

Table 3

Refer Table 1, 2, and 3.

Plot the graph for the exit pressure $\left(P_2\right)$ against power output of turbine $\left({\dot{W}}_2\right)$ as shown in Figure 1.

Refer Table 1, 2, and 3.

Plot the graph for the exit pressure $\left(P_2\right)$ against exit velocity $\left(V_2\right)$ as shown in Figure 2.