Chapter 10.9, Problem 111RP

A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater is shown below. The high-pressure turbine receives 100 kg/s of steam from the steam boiler. The feedwater heater exit states for the boiler feedwater and the condensed steam are the normally assumed ideal states. The following data tables give the saturation data for the pressures and data for h and s at selected states. (a) Sketch the T-s diagram for the ideal cycle. (b) Determine the net power output of the cycle, in MW. (c) If cooling water is available at 25°C, what is the minimum flow rate of the cooling water required for the ideal cycle, in kg/s? Take c_p,water = 4.18 kJ/kg·K.

Process states and selected data
State	P, kPa	T, °C	h, kJ/kg	s, kJ/kg·K
1	20
2	1400
3	1400
4	1400
5	5000
6	5000	700	3894	7.504
7	1400		3400	7.504
8	1200		3349	7.504
9	1200	600	3692	7.938
10	245		3154	7.938
11	20		2620	7.938

To determine

Sketch the T-s diagram for the ideal cycle.

Answer to Problem 111RP

Sketch the T-s diagram for the ideal cycle is shown in Figure 1.

Explanation of Solution

Draw the $T-s$ diagram of the given ideal regenerative Rankine cycle as shown in

Figure 1.

To determine

The net power output of the cycle.

Answer to Problem 111RP

The net power output of the cycle is $\underset{\_}{144.4\text{MW}}$.

Explanation of Solution

Write the formula for work done by the pump during process 1-2.

$w_{\text{pI,in}}=v_1\left(P_2-P_1\right)$ (I)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 2.

$h_2=h_1+w_{\text{pI,in}}$ (II)

Write the formula for work done by the pump during process 3-4.

$w_{\text{pII,in}}=v_4\left(P_5-P_4\right)$ (III)

Here, the specific volume is $v$, the pressure is $P$, and the subscripts 4 and 5 indicates the process states.

Write the formula for enthalpy $\left(h\right)$ at state 5.

$h_5=h_4+w_{\text{pII,in}}$ (IV)

Write the formula for an energy balance on the open feed water heater.

$y{h}_7\left(1-y\right)h_3=1\left(h_4\right)$ (V)

Here, the fraction of steam extracted is $\left(y\right)$ from the high-pressure turbine.

Rewrite the Equation (V) to find out the value of $\left(y\right)$.

$y=\frac{h_4-h_3}{h_7-h_3}$ (VI)

Write the formula for an energy balance on the closed feed water heater.

$z{h}_{10}+\left(1-y\right)h_2=\left(1-y\right)h_3+z{h}_{12}$ (VII)

Here, the fraction of steam extracted is $\left(z\right)$ from the low-pressure turbine.

Rewrite the Equation (V) to find out the value of $\left(z\right)$.

$z=\frac{\left(1-y\right)\left(h_3-h_2\right)}{\left(h_{10}-h_{12}\right)}$ (VIII)

Write the formula for heat input in the boiler.

$q_{\text{in}}=\left(h_6-h_5\right)+\left(1-y\right)\left(h_9-h_8\right)$ (IX)

Write the formula for work output from the turbine.

$w_{\text{T,out}}=h_6-y{h}_7-\left(1-y\right)h_8+\left(1-y\right)h_9-z{h}_{10}-\left(1-y-z\right)h_{11}$ (X)

Write the formula for net work output from the cycle.

$w_{\text{net}}=w_{\text{T,out}}-\left(1-y\right)w_{\text{PI,in}}-w_{\text{PII,in}}$ (XI)

Write the net power output of the cycle.

${\dot{W}}_{\text{net}}=\dot{m}{w}_{\text{net}}$ (XII)

Conclusion:

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy $\left(h_1\right)$ and specific volume $\left(v_1\right)$ at state 1 corresponding to the pressure of $20\text{kPa}$ is $251.42\text{kJ}/\text{kg}$ and $0.001017{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $0.001017{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$, $20\text{kPa}$ for $P_1$, and $1400\text{kPa}$ for $P_2$ in Equation (I).

$\begin{array}{c}{w}_{\text{pI,in}}=\left(0.001017{\text{m}}^{\text{3}}/\text{kg}\right)\left(1400\text{kPa}-20\text{kPa}\right)\\ =\left(0.001017{\text{m}}^{\text{3}}/\text{kg}\right)\left(1380\text{kPa}\right)\\ =1.40346\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =1.40346\text{kJ}/\text{kg}\end{array}$

        $\cong 1.4035\text{kJ}/\text{kg}$

Substitute $251.42\text{kJ}/\text{kg}$ for $h_1$, and $1.4035\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$ in Equation (II).

$\begin{array}{c}{h}_2=251.42\text{kJ/kg}+1.4035\text{kJ}/\text{kg}\\ =252.8\text{kJ}/\text{kg}\end{array}$

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy $\left(h_4\right)$ and specific volume $\left(v_4\right)$ at state 4 corresponding to the pressure of $1400\text{kPa}$ is $829.96\text{kJ}/\text{kg}$ and $0.001149{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $0.001149{\text{m}}^{\text{3}}/\text{kg}$ for $v_4$, $1400\text{kPa}$ for $P_4$, and $5000\text{kPa}$ for $P_5$ in Equation (III).

$\begin{array}{c}{w}_{\text{pII,in}}=\left(0.001149{\text{m}}^{\text{3}}/\text{kg}\right)\left(5000\text{kPa}-1400\text{kPa}\right)\\ =\left(0.001149{\text{m}}^{\text{3}}/\text{kg}\right)\left(3600\text{kPa}\right)\\ =4.1364\text{kPa}\cdot {\text{m}}^3/\text{kg}\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =4.1364\text{kJ}/\text{kg}\end{array}$

         $\cong 4.14\text{kJ}/\text{kg}$

Substitute $829.96\text{kJ}/\text{kg}$ for $h_3$, and $4.14\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$ in Equation (IV).

$\begin{array}{c}{h}_5=829.96\text{kJ}/\text{kg}+4.14\text{kJ}/\text{kg}\\ =834.1\text{kJ}/\text{kg}\end{array}$

Refer Table A-5, “Saturated water-temperature Table”, and write the enthalpy at state 12 at pressure of $\text{245}\text{kPa}$ using an interpolation method.

$h_3=h_{12}=h_{f@\text{245}\text{kPa}}$ (XIII)

Here, enthalpy of saturation liquid at pressure of $\text{245}\text{kPa}$ is $h_{f@\text{245}\text{kPa}}$.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (XIV)

Here, the variables denote by x and y is pressure and specific enthalpy at state 12 respectively.

Show the specific enthalpy at state 12 corresponding to temperature as in Table (1).

Pressure at state 12 $\text{kPa}$	Specific enthalpy at state 12 $h_{12}\left(\text{kJ}/\text{kg}\right)$
225 $\left(x_1\right)$	520.71 $\left(y_1\right)$
245 $\left(x_2\right)$	$\left(y_2=?\right)$
250 $\left(x_3\right)$	535.35 $\left(y_3\right)$

Substitute $225\text{kPa},245\text{kPa},\text{and}\text{}250\text{kPa}$ for $x_1,x_2\text{and}{x}_3$ respectively, $520.71\text{kJ}/\text{kg}$ for $y_1$ and $535.35\text{kJ}/\text{kg}$ for $y_3$ in Equation (XIV).

$\begin{array}{c}{y}_2=\left[\begin{array}{l}\frac{\left(245-225\right)\text{kPa}\times \left(535.35-520.71\right)\text{kJ}/\text{kg}}{\left(250-225\right)\text{kPa}}\\ +520.71\text{kJ}/\text{kg}\end{array}\right]\\ =532.422\text{kJ}/\text{kg}\\ \cong 532\text{kJ}/\text{kg}\end{array}$

Substitute $532\text{kJ}/\text{kg}$ for $h_{f@\text{245}\text{kPa}}$ in Equation (XIII).

$h_{12}=\text{532}\text{kJ}/\text{kg}$

Here, the throttle valve operation of specific enthalpy at the state 13 is equal to specific enthalpy at the state 12.

Substitute $830\text{kJ}/\text{kg}$ for $h_4$, $532\text{kJ}/\text{kg}$ for $h_3$, and $3400\text{kJ}/\text{kg}$ for $h_7$ in Equation (VI).

$\begin{array}{c}y=\frac{\left(830\text{kJ}/\text{kg}-532\text{kJ}/\text{kg}\right)}{\left(3400\text{kJ}/\text{kg}-532\text{kJ}/\text{kg}\right)}\\ =\frac{\left(298\text{kJ}/\text{kg}\right)}{\left(2868\text{kJ}/\text{kg}\right)}\\ =0.1039\end{array}$

Substitute $252.8\text{kJ}/\text{kg}$ for $h_2$, 0.1039 for $y$, $532\text{kJ}/\text{kg}$ for $h_3$, and $3154\text{kJ}/\text{kg}$ for $h_{10}$ in Equation (VIII).

$\begin{array}{c}z=\frac{\left(1-0.1039\right)\left(532\text{kJ}/\text{kg}-252.8\text{kJ}/\text{kg}\right)}{\left(3154\text{kJ}/\text{kg}-532\text{kJ}/\text{kg}\right)}\\ =\frac{\left(250.1911\text{kJ}/\text{kg}\right)}{\left(2622\text{kJ}/\text{kg}\right)}\\ =0.09542\end{array}$

Substitute $3894\text{kJ}/\text{kg}$ for $h_6$, $834.1\text{kJ}/\text{kg}$ for $h_5$, 0.1039 for $y$, $3692\text{kJ}/\text{kg}$ for $h_9$, and $3349\text{kJ}/\text{kg}$ for $h_8$ in Equation (IX).

$\begin{array}{c}{q}_{\text{in}}=\left(3894-834.1\right)\text{kJ}/\text{kg}+\left(1-0.1039\right)\left(3692-3349\right)\text{kJ}/\text{kg}\\ =3059.9\text{kJ}/\text{kg}+\left(0.8961\right)\times \left(343\text{kJ}/\text{kg}\right)\\ =3059.9\text{kJ}/\text{kg}+307.36\text{kJ}/\text{kg}\\ =3367\text{kJ}/\text{kg}\end{array}$

Substitute $3894\text{kJ}/\text{kg}$ for $h{6}_{}$, $3400\text{kJ}/\text{kg}$ for $h_7$, 0.09542 for $z$, 0.1039 for $y$, $3692\text{kJ}/\text{kg}$ for $h_9$, $3349\text{kJ}/\text{kg}$ for $h_8$, $3154\text{kJ}/\text{kg}$ for $h_{10}$, and $2620\text{kJ}/\text{kg}$ for $h_{11}$ in Equation (X).

$\begin{array}{c}{w}_{\text{T,out}}=\left[\begin{array}{l}\left(3894\text{kJ}/\text{kg}\right)-\left(0.1039\right)\left(3400\text{kJ}/\text{kg}\right)\\ -\left(1-0.1039\right)\left(3349\text{kJ}/\text{kg}\right)+\left(1-0.1039\right)\left(3692\text{kJ}/\text{kg}\right)\\ -\left(0.09542\right)\left(3154\text{kJ}/\text{kg}\right)-\left(1-0.1039-0.09542\right)\left(2620\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(3894\text{kJ}/\text{kg}\right)-\left(353.26\text{kJ}/\text{kg}\right)\\ -\left(0.8961\right)\left(3349\text{kJ}/\text{kg}\right)+\left(0.8961\right)\left(3692\text{kJ}/\text{kg}\right)\\ -\left(0.09542\right)\left(3154\text{kJ}/\text{kg}\right)-\left(0.80068\right)\left(2620\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(3894\text{kJ}/\text{kg}\right)-\left(353.26\text{kJ}/\text{kg}\right)\\ -\left(3001.039\text{kJ}/\text{kg}\right)+\left(3308.401\text{kJ}/\text{kg}\right)\\ -\left(300.9547\text{kJ}/\text{kg}\right)-\left(2097.782\text{kJ}/\text{kg}\right)\end{array}\right]\\ =1449\text{kJ}/\text{kg}\end{array}$

Substitute $1449\text{kJ}/\text{kg}$ for $w_{\text{T,out}}$, $4.14\text{kJ}/\text{kg}$ for $w_{\text{pII,in}}$, $1.41\text{kJ}/\text{kg}$ for $w_{\text{pI,in}}$, and 0.1039 for $y$ in Equation (XI).

$\begin{array}{c}{w}_{\text{net}}=1449\text{kJ}/\text{kg}-\left(1-0.1039\right)\left(1.41\text{kJ}/\text{kg}\right)-\left(4.14\text{kJ}/\text{kg}\right)\\ =\left(1449\text{kJ}/\text{kg}\right)-\left(1.2635\text{kJ}/\text{kg}\right)-\left(4.14\text{kJ}/\text{kg}\right)\\ =1443.596\text{kJ}/\text{kg}\\ \cong 1444\text{kJ}/\text{kg}\end{array}$

Substitute $100\text{kg}/\text{s}$ for $\dot{m}$ and $1444\text{kJ}/\text{kg}$ for $w_{\text{net}}$ in Equation (XII).

$\begin{array}{c}{\dot{W}}_{\text{net}}=\left(100\text{kg}/\text{s}\right)\times \left(1444\text{kJ}/\text{kg}\right)\\ =144400\text{kW}\\ =144400\text{kW}\times \left(\frac{\text{1}\text{MW}}{1000\text{kW}}\right)\\ =144.4\text{MW}\end{array}$

Thus, the net power output of the cycle is $\underset{\_}{144.4\text{MW}}$.

To determine

The minimum flow rate of the cooling water.

Answer to Problem 111RP

The minimum flow rate of the cooling water is $\underset{\_}{1311\text{kg}/\text{s}}$.

Explanation of Solution

Write the formula for heat rejected from the condenser.

$q_{\text{out}}=\left(1-y-z\right)h{11}_{}+z{h}_{13}-\left(1-y\right)h_1$ (XV).

The mass flow rate cooling water will be minimum when the cooling water exit temperature is a maximum as

$T_{w,2}=T_1=T_{sat@20\text{kPa}}=60.1°\text{C}$.

Write the formula for an energy balance on the condenser.

$\begin{array}{l}\dot{m}{q}_{\text{out}}={\dot{m}}_w{c}_{\text{p,w}}\left(T_{w,2}-T_{w,1}\right)\\ {\dot{m}}_w=\frac{\dot{m}{q}_{\text{out}}}{c_{\text{p,w}}\left(T_{w,2}-T_{w,1}\right)}\end{array}$ (XVI)

Conclusion:

Substitute 0.1039 for $y$, 0.09542 for $z$, $2620\text{kJ}/\text{kg}$ for $h_{11}$, $532\text{kJ}/\text{kg}$ for $h_{13}$, and $251.4\text{kJ}/\text{kg}$ for $h_1$ in Equation (XV).

$\begin{array}{c}{q}_{\text{out}}=\left[\begin{array}{l}\left(1-0.1039-0.09542\right)\left(2620\text{kJ}/\text{kg}\right)\\ +\left(0.09542\right)\left(532\text{kJ}/\text{kg}\right)\\ -\left(1-0.1039\right)\left(251.4\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left[\left(2097.782\text{kJ}/\text{kg}\right)+\left(50.76344\text{kJ}/\text{kg}\right)-\left(225.2795\text{kJ}/\text{kg}\right)\right]\\ =1923\text{kJ}/\text{kg}\end{array}$

Substitute $100\text{kg}/\text{s}$ for $\dot{m}$, $1923\text{kJ}/\text{kg}$ for $q_{\text{out}}$, $4.18\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{\text{p,w}}$, $60.1\text{K}$ for $T_{\text{w,2}}$, and $25\text{K}$ for $T_{\text{w,1}}$ in Equation (XVI).

$\begin{array}{c}{\dot{m}}_w=\frac{\left(100\text{kg}/\text{s}\right)\left(1923\text{kJ}/\text{kg}\right)}{\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left(60.1-25\right)\text{K}}\\ =\frac{\left(192300\text{kJ}/\text{s}\right)}{\left(4.18\text{kJ}/\text{kg}\cdot \text{K}\right)\left(35.1\right)\text{K}}\\ =1310.67\text{kg}/\text{s}\\ \cong 1311\text{kg}/\text{s}\end{array}$

Thus, the minimum flow rate of the cooling water is $\underset{\_}{1311\text{kg}/\text{s}}$.