Chapter 5, Problem 140P

Water enters a pump at 350 kPa at a rate of 1 kg/s. The water leaving the pump enters a turbine in which the pressure is reduced and electricity is produced. The shaft power input to the pinup is 1 kW and the shaft power output from the turbine is 1 kW. Both the pump and turbine are 90 percent efficient. If the elevation and velocity of the water remain constant throughout the flow and the irreversible head loss is 1 m. the pressure of water at the turbine exit is

(a) 350 kPa
(b) 100 kPa
(c) 173 kPa
(d) 218 kPa
(e) 129 kPa

To determine

The pressure of water at the turbine at exit.

Answer to Problem 140P

The pressure at the outlet is $129.078\text{kPa}$ and the correct option is $\left(e\right)$.

Explanation of Solution

Given information:

The initial pressure of the pump is $350\text{kPa}$, the mass flow rate is $1\text{kg}/\text{s}$, shaft input power is $1\text{kW,}$ and shaft output power is $1\text{kW}$.

The following figure shows arrangement of the pump and the turbine.

  Figure-(1)

Write the expression for the final pressure of the water at exit using efficiency.

   $\begin{array}{l}{\eta }_{pump}=\frac{{\dot{W}}_{pump}}{{\dot{W}}_{shaft}}\\ {\dot{W}}_{pump}={\eta }_{pump}\times {\dot{W}}_{shaft}\\ \dot{m}\frac{\left(P_2-P_1\right)}{\rho }={\eta }_{pump}\times {\dot{W}}_{shaft}\\ P_2=\left(\frac{{\eta }_{pump}\times {\dot{W}}_{shaft}\times \rho }{\dot{m}}\right)+P_1\end{array}$    ...... (I)

Here, the initial pressure at inlet is $P_1$, efficiency of the pump is ${\eta }_{pump}$, work input at the shaft of the pump is ${\dot{W}}_{shaft}$, mass flow rate is $\dot{m},$ and density of water is $\rho$.

Write the expression for pressure at inlet of turbine using the energy equation for the flow with constant velocity and elevation from pump to the turbine.

   $\begin{array}{l}\frac{P_2}{\rho g}=\frac{P_3}{\rho g}+h_L\\ P_3=\left(\frac{P_2}{\rho g}-h_L\right)\rho g\end{array}$    ...... (II)

Here, the irreversible head loss is $h_L$.

Write the expression for the pressure of water at the outlet of the turbine.

   $\begin{array}{l}{\eta }_{turbine}=\frac{{\dot{W}}_{shaft}}{{\dot{W}}_{turbine}}\\ {\dot{W}}_{turbine}=\frac{{\dot{W}}_{shaft}}{{\eta }_{turbine}}\\ \dot{m}\frac{\left(P_3-P_4\right)}{\rho }=\frac{{\dot{W}}_{shaft}}{{\eta }_{turbine}}\\ P_4=P_3-\left(\frac{{\dot{W}}_{shaft}\times \rho }{\dot{m}\times {\eta }_{turbine}}\right)\end{array}$    ...... (III)

Here, the efficiency of the turbine is ${\eta }_{turbine}$ and work output at shaft is ${\dot{W}}_{shaft}$.

Calculation:

Substitute $0.9$ for ${\eta }_{pump}$, $1\text{kW}$ for ${\dot{W}}_{shaft}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1\text{kg}/\text{s}$ for $\dot{m},$ and $350\text{kPa}$ for $P_1$ in Equation (I).

   $\begin{array}{c}{P}_2=\left(\frac{0.9\times 1\text{kW}\times 1000\text{kg}/{\text{m}}^{\text{3}}}{1\text{kg}/\text{s}}\right)+350\text{kPa}\\ \text{=900}{\text{kWm}}^3/s\left(\frac{1\text{kPa}}{1{\text{kWm}}^3/s}\right)+350\text{kPa}\\ =900\text{kPa}+350\text{kPa}\\ \text{=1250}\text{kPa}\end{array}$

Substitute $1000\text{}\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1\text{m}$ for $h_L$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g,$ and $1250\text{kPa}$ for $P_2$ in Equation (II).

   $\begin{array}{c}{P}_3=\left(\frac{1250\text{kPa}}{\left(1000\text{}\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}-1\text{m}\right)\left(1000\text{}\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(\frac{1250\text{kPa}\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)}{\left(1000\text{}\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}-1\text{m}\right)\left(1000\text{}\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(127.42\text{N}\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)/\text{kg}\cdot {\text{s}}^{\text{-2}}-1\text{m}\right)\left(9810\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\right)\\ =\left(127.42\text{m}-1\text{m}\right)\left(9810\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\right)\end{array}$

   $\begin{array}{c}{P}_3=\left(127.42\text{m}-1\text{m}\right)\left(9810\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\right)\\ =1240190\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{kPa}}{1000\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =1240.190\text{kPa}\end{array}$

Substitute $0.9$ for ${\eta }_{turbine}$, $1\text{kW}$ for ${\dot{W}}_{turbine}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1\text{kg}/\text{s}$ for $\dot{m}$, $1240.190\text{}\text{kPa}$ for $P_3$ in Equation (III),

   $\begin{array}{c}{P}_4=1240.190\text{}\text{kPa}-\left(\frac{1\text{kW}\times 1000\text{kg}/{\text{m}}^{\text{3}}}{1\text{kg}/\text{s}\times 0.9}\right)\\ =1240.190\text{}\text{kPa}-\left(1111.11\text{kW}/{\text{m}}^{\text{3}}{\text{s}}^{\text{-1}}\left(\frac{1\text{}\text{kPa}}{1\text{kW}/{\text{m}}^{\text{3}}{\text{s}}^{\text{-1}}}\right)\right)\\ =1240.190\text{}\text{kPa-1111}\text{.11}\text{kPa}\\ \text{=129}\text{.078}\text{kPa}\end{array}$

Conclusion:

The pressure at the outlet is $129.078\text{kPa}$ and the correct option is $\left(e\right)$.