Chapter 11, Problem 11.31P

To determine

(a)

The best efficiency point and the maximum efficiency for given parameters.

Answer to Problem 11.31P

The required maximum efficiency of $85\%$ occurs at Q $2000gal/\min$

Explanation of Solution

Given Information:

$Temperature\left(T\right)={20}^{o}C$

Formula used:

$\eta =\frac{\gamma QH}{P}$

Calculation:

To solve this problem, first we will convert the given data into efficiency, like at $Q=400gal/\min$.

We know that, $\eta =\frac{\gamma QH}{P}$ ;

Put the values in the above equation;

$\Rightarrow \frac{\left(62.4lbf/f{t}^3\right)\left(400/448.8f{t}^3/s\right)\left(115ft\right)}{\left(36\times 550ft\cdot lbf/s\right)}$

$\Rightarrow 0.32$

$\Rightarrow 32\%$

After conversion the table look like:

Q gal/min	0	400	800	1200	1600	2000	$2400$
$\eta$ %	0	$32\%$	55%	70%	80%	85%	82%

Therefore the maximum efficiency of $85\%$ occurs at Q $2000gal/\min$.

To determine

(b)

The most effective flow rate, resulting head and brake horsepower.

Answer to Problem 11.31P

The required values are:

Flow rate = $24000gal/\min$

Head = $729ft$

BHP = $5180hp$

Explanation of Solution

Given information:

$\begin{array}{l}\text{Temperature}\left(\text{T}\right)\text{=}{\text{20}}^{\text{o}}\text{C}\\ {\text{N}}_{\text{r}}\text{=}\text{50\%}\text{increased}\end{array}$

Formula used:

$C_Q{}^{*}=\frac{Q_1}{n_1{D}_1{}^3}$ = $\frac{Q_2}{n_2{D}_2{}^3}$

$C_H*=\frac{g{H}_1}{n_1{}^2{D}_1{}^2}=\frac{g{H}_2}{n_2{}^2{D}_2{}^2}$

$C_P{}^{*}=\frac{P_1}{\rho n_1{}^3{D}_1{}^5}=\frac{P_2}{\rho n_2{}^3{D}_2{}^5}$

Calculation:

Since we don’t know the values of $C_Q{}^{*}$, $C_H{}^{*}$ and $C_P{}^{*}$, yet we can set them equal for conditions 1 and 2;

$C_Q{}^{*}=\frac{Q_1}{n_1{D}_1{}^3}$ = $\frac{Q_2}{n_2{D}_2{}^3}$

Put the values in the above equation;

$\Rightarrow \frac{Q_2}{\left(1.5n_1\right){\left(2D_1\right)}^3}$

And, $Q_2=12Q_1$ ;

$\Rightarrow 12\left(2000gpm\right)$

$\Rightarrow 24000gal/\min$

$C_H*=\frac{g{H}_1}{n_1{}^2{D}_1{}^2}=\frac{g{H}_2}{n_2{}^2{D}_2{}^2}=\frac{g{H}_2}{{\left(1.5n_1\right)}^2{\left(2D_1\right)}^2}$

$or{H}_2=9H_1=9\left(81ft\right)=729ft$

$\begin{array}{l}{C}_P{}^{*}=\frac{P_1}{\rho n_1{}^3{D}_1{}^5}=\frac{P_2}{\rho n_2{}^3{D}_2{}^5}\\ \Rightarrow \frac{P_2}{\rho {\left(1.5n_1\right)}^3{(2D_1)}^5}\end{array}$

$\Rightarrow P_2=108P_1=108\left(48hp\right)$

$\Rightarrow 5180hp$.