Chapter 11, Problem 11.33P

To determine

(a)

The value of the diameter.

Answer to Problem 11.33P

The required value of the diameter is $47cm$

Explanation of Solution

Given Information:

$\begin{array}{l}\text{BEP}\text{flow}\text{rate}\text{=}\text{2000}\text{gal/min}\\ \text{Diameter}\left(\text{d}\right)\text{=}\text{16}\text{in}\\ \text{Speed}\left(\text{n}\right)\text{=}\text{1200}\text{r/min}\end{array}$

Formula used:

${\eta }_1={\eta }_2=\frac{\gamma Q_1{H}_1}{P_1}$

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

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

Calculation:

To solve this problem, we are taking the value of specific weight of the water as $\gamma \text{=}\text{9790}N{\text{/m}}^3$.

Now we will find the efficiency;

${\eta }_1={\eta }_2=\frac{\gamma Q_1{H}_1}{P_1}$ ;

Put the values in the above equation;

$\Rightarrow \frac{\left(9790N/m^3\right)\left(7.57/60m^3/s\right)\left(24.67m\right)}{\left(48x746W\right)}$

$\Rightarrow 0.851$

$\Rightarrow 85.1\%$

Now we will use the scaling laws for BEP flow rate and head;

$\frac{Q_1}{n_1{D}_1{}^3}=\frac{7.57}{\left(1200\right){\left(0.406\right)}^3}=\frac{Q_2}{n_2{D}_2{}^3}=\frac{15.14}{n_2{D}_2{}^3}$

$\frac{g{H}_1}{n_1{}^2{D}_1{}^2}=\frac{\left(9.81\right)\left(24.67\right)}{{\left(1200\right)}^2{\left(0.406\right)}^2}=\frac{g{H}_2}{n_2{}^2{D}_2{}^2}=\frac{\left(9.81\right)\left(54.86\right)}{n_2{}^2{D}_2{}^2}$

Solve for D₂ and n₂ simultaneously:

$D_2=47cm$

$n_2=1545r/\min$.

To determine

(b)

The speed that will deliver a BEP water flow rate of $15.14m^3/\min$ and a head of $54.86m$.

Answer to Problem 11.33P

The required value of the speed is $1545r/\min$

Explanation of Solution

Given information:

$\begin{array}{l}\text{BEP}\text{flow}\text{rate}\text{=}\text{2000}\text{gal/min}\\ \text{Diameter}\left(\text{d}\right)\text{=}\text{16}\text{in}\\ \text{Speed}\left(\text{n}\right)\text{=}\text{1200}\text{r/min}\end{array}$

Formula used:

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

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

Calculation:

Explained in the above part;

Repeat:

Now we will use the scaling laws for BEP flow rate and head;

$\frac{Q_1}{n_1{D}_1{}^3}=\frac{7.57}{\left(1200\right){\left(0.406\right)}^3}=\frac{Q_2}{n_2{D}_2{}^3}=\frac{15.14}{n_2{D}_2{}^3}$

$\frac{g{H}_1}{n_1{}^2{D}_1{}^2}=\frac{\left(9.81\right)\left(24.67\right)}{{\left(1200\right)}^2{\left(0.406\right)}^2}=\frac{g{H}_2}{n_2{}^2{D}_2{}^2}=\frac{\left(9.81\right)\left(54.86\right)}{n_2{}^2{D}_2{}^2}$

Solve for D₂ and n₂ simultaneously:

$D_2=47cm$

$n_2=1545r/\min$.

To determine

(c)

The brake horsepower.

Answer to Problem 11.33P

The required value of the brake horsepower is $213hp$

Explanation of Solution

Given Information:

$\begin{array}{l}\text{BEP}\text{flow}\text{rate}\text{=}\text{2000}\text{gal/min}\\ \text{Diameter}\left(\text{d}\right)\text{=}\text{16}\text{in}\\ \text{Speed}\left(\text{n}\right)\text{=}\text{1200}\text{r/min}\end{array}$

Formula used:

$P_2=\frac{\gamma Q_2{H}_2}{{\eta }_2}$

Calculation:

We know the formula for the required horse power, i.e.: $P_2=\frac{\gamma Q_2{H}_2}{{\eta }_2}$ ;

Put the values in the above equation

$\begin{array}{l}{P}_2=\frac{\gamma Q_2{H}_2}{{\eta }_2}=\frac{\left(9790\right)\left(15.14/60\right)\left(54.86\right)}{0.853}\\ \Rightarrow 1.59kW/746\\ \Rightarrow 213hp\end{array}$.