Chapter 3, Problem 3.174P

To determine

(a)

The power extracted by the turbine.

Answer to Problem 3.174P

The power extracted by the turbine is $305.04\text{kW}$.

Explanation of Solution

Given information:

The flow rate is $15000\text{gal}/\text{min}$ . The friction head loss is $17\text{ft}$ . The magnitude of $z_1$ is $150\text{ft}$ and the magnitude of $z_2$ is $25\text{ft}$.

Write the expression for the Bernoulli’s equation between the inlet and outlet of the turbine.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_T+h_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f+h_T\end{array}$ ...(I)

Here, the pressure at Section (1) is $p_1$, the velocity at Section (1) is $V_1$, the height of the Section (1) from the datum is $z_1$, the pressure at outlet is $p_2$, the velocity at Section (2) is $V_2$, the height of the Section (2) from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$, the head loss due to turbine is $h_T$ and the acceleration due to gravity is $g$.

Since the pressure at the inlet and outlet are atmospheric pressure so the pressure at section (1) and section (2) are same.

$p_1=p_2=p_{\text{atm}}$

Since the diameter of the pipe is constant, so the velocity at section (1) and section (2) is same.

$V_1=V_2$

Substitute $p_{\text{atm}}$ for $p_1$, $p_{\text{atm}}$ for $p_2$ and $V_2$ for $V_1$ in Equation (I).

$\begin{array}{l}\left(\frac{p_{\text{atm}}-p_{\text{atm}}}{\rho g}\right)+\left(\frac{V_2^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f+h_T\\ \left(z_1-z_2\right)=h_f+h_T\end{array}$ ...(II)

Write the expression for the turbine power.

$P=\rho Qg{h}_T$ ...(III)

Here, the power extracted by the turbine is $P$.

Calculation:

Substitute $150\text{ft}$ for $z_1$, $25\text{ft}$ for $z_2$ and $17\text{ft}$ for $h_f$ in Equation (II).

$\begin{array}{l}\left\{\left(150\text{ft}\right)-\left(25\text{ft}\right)\right\}=\left(17\text{ft}\right)+h_T\\ h_T=\left(125\text{ft}\right)-\left(17\text{ft}\right)\\ h_T=108\text{ft}\end{array}$

Substitute $108\text{ft}$ for $h_T$, $15000\text{gal}/\text{min}$ for $Q$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $1.94\text{slug}/{\text{ft}}^3$ for $\rho$ in Equation (III).

$\begin{array}{c}P=\left(1.94\text{slug}/{\text{ft}}^3\right)\left(15000\text{gal}/\text{min}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(108\text{ft}\right)\\ =\left(1.94\text{slug}/{\text{ft}}^3\right)\left(15000\text{gal}/\text{min}\right)\left(\frac{1{\text{ft}}^3/\text{s}}{450\text{gal}/\text{min}}\right)\left(3477.6{\text{ft}}^2/{\text{s}}^2\right)\\ =\left(64.67\text{slug}/\text{s}\right)\left(3477.6{\text{ft}}^2/{\text{s}}^2\right)\\ =224896.392\text{lbf}\cdot \text{ft}/\text{s}\end{array}$

Convert the turbine power from $\text{lbf}\cdot \text{ft}/\text{s}$ into $\text{kW}$.

$\begin{array}{c}P=\left(224896.392\text{lbf}\cdot \text{ft}/\text{s}\right)\left(\frac{1\text{hp}}{550\text{lbf}\cdot \text{ft}/\text{s}}\right)\left(\frac{0.746\text{kW}}{1\text{hp}}\right)\\ =\left(408.90\text{hp}\right)\left(\frac{0.746\text{kW}}{1\text{hp}}\right)\\ \approx 305.04\text{kW}\end{array}$

Conclusion:

The power extracted by the turbine is $305.04\text{kW}$.

To determine

(b)

The power delivered by the pump.

Answer to Problem 3.174P

The power delivered by the pump is $401.07\text{kW}$.

Explanation of Solution

Given information:

The flow rate is $15000\text{gal}/\text{min}$ . The friction head loss is $17\text{ft}$ . The magnitude of $z_1$ is $150\text{ft}$ and the magnitude of $z_2$ is $25\text{ft}$.

Write the expression for the Bernoulli’s equation between the inlet and outlet of the turbine.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h_P=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f-h_P\end{array}$ ...(IV)

Here, the pressure at Section (1) is $p_1$, the velocity at Section (1) is $V_1$, the height of the Section (1) from the datum is $z_1$, the pressure at outlet is $p_2$, the velocity at Section (2) is $V_2$, the height of the Section (2) from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$, the pump head is $h_P$ and the acceleration due to gravity is $g$.

Since the pressure at the inlet and outlet are atmospheric pressure so the pressure at section (1) and section (2) are same.

$p_1=p_2=p_{\text{atm}}$

Since the diameter of the pipe is constant, so the velocity at section (1) and section (2) is same.

$V_1=V_2$

Substitute $p_{\text{atm}}$ for $p_1$, $p_{\text{atm}}$ for $p_2$ and $V_2$ for $V_1$ in Equation (I).

$\begin{array}{l}\left(\frac{p_{\text{atm}}-p_{\text{atm}}}{\rho g}\right)+\left(\frac{V_2^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f-h_P\\ \left(z_1-z_2\right)=h_f-h_P\end{array}$ ...(V)

Write the expression for the pump power.

$P_P=\rho Qg{h}_P$ ...(VI)

Here, the power extracted by the pump is $P_P$.

Calculation:

Substitute $150\text{ft}$ for $z_1$, $25\text{ft}$ for $z_2$ and $17\text{ft}$ for $h_f$ in Equation (V).

$\begin{array}{l}\left\{\left(25\text{ft}\right)-\left(150\text{ft}\right)\right\}=\left(17\text{ft}\right)-h_P\\ h_P=\left(17\text{ft}\right)+\left(125\text{ft}\right)\\ h_P=142\text{ft}\end{array}$

Substitute $142\text{ft}$ for $h_P$, $15000\text{gal}/\text{min}$ for $Q$, $32.2\text{ft}/{\text{s}}^2$ for $g$ and $1.94\text{slug}/{\text{ft}}^3$ for $\rho$ in Equation (VI).

$\begin{array}{c}{P}_P=\left(1.94\text{slug}/{\text{ft}}^3\right)\left(15000\text{gal}/\text{min}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(142\text{ft}\right)\\ =\left(1.94\text{slug}/{\text{ft}}^3\right)\left(15000\text{gal}/\text{min}\right)\left(\frac{1{\text{ft}}^3/\text{s}}{450\text{gal}/\text{min}}\right)\left(4572.4{\text{ft}}^2/{\text{s}}^2\right)\\ =\left(64.67\text{slug}/\text{s}\right)\left(4572.4{\text{ft}}^2/{\text{s}}^2\right)\\ =295697.108\text{lbf}\cdot \text{ft}/\text{s}\end{array}$

Convert the pump power from $\text{lbf}\cdot \text{ft}/\text{s}$ into $\text{kW}$.

$\begin{array}{c}{P}_P=\left(295697.108\text{lbf}\cdot \text{ft}/\text{s}\right)\left(\frac{1\text{hp}}{550\text{lbf}\cdot \text{ft}/\text{s}}\right)\left(\frac{0.746\text{kW}}{1\text{hp}}\right)\\ =\left(537.63\text{hp}\right)\left(\frac{0.746\text{kW}}{1\text{hp}}\right)\\ =401.07\text{kW}\end{array}$

Conclusion:

The power extracted by the pump is $401.07\text{kW}$.