Chapter 5.5, Problem 179RP

Reconsider Prob. 5–178. Using appropriate software, investigate the effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine. Let the exit pressure vary from 10 to 50 kPa (with the same quality), and let the exit area vary from 1000 to 3000 cm². Plot the exit velocity and the power outlet against the exit pressure for the exit areas of 1000, 2000, and 3000 cm², and discuss the

results.

5–178 Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent. A heat loss of 20 kJ/kg occurs during the process. The inlet area of the turbine is 150 cm2, and the exit area is 1400 cm². Determine (a) the mass flow rate of the steam, (b) the exit velocity, and (c) the power output.

To determine

Plot the exit pressure against power output of turbine and exit pressure against exit velocity for varying pressure from $10\text{kPa}$ to $50\text{kPa}$, for the varying exit area of $1000{\text{cm}}^{\text{2}}$ to $3000{\text{cm}}^{\text{2}}$.

Answer to Problem 179RP

The plot for the exit pressure against power output of turbine and exit pressure against exit velocity for varying pressure from $10\text{kPa}$ to $50\text{kPa}$, for the varying exit area of $1000{\text{cm}}^{\text{2}}$ to $3000{\text{cm}}^{\text{2}}$ is plotted and shown in Figure 1 and Figure 2 respectively.

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 refrigerant flows at steady state through the compressor. 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)

Conclusion:

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.