Chapter 11, Problem 11.74P

To determine

(a)

The discharge and brake horsepower required for one pump.

Answer to Problem 11.74P

The discharge required for one pump is $Q=14.7154kgal/\min$ and the brake horsepower is $P=0.009076hp$.

Explanation of Solution

Given Information:

Pump diameter, d = 32in

Fig 11.7a

From the figure, the best curve for 32-inch diameter:

$H_P=a+b{Q}^2$ ………………………. (Relation 1)

Here, a & b = constants

From the curve find constants by taking 2 points

1^st point:

Let, $H_P=500$ and $Q=0$ in relation 1,

$500=a+b{\left(0\right)}^2$

$a=500$

2^nd point,

Let, $a=500$, $H_P=450$, $Q=12$ in relation 1,

$450=500+b{\left(12\right)}^2$

$450=500+144b$

$b=\frac{450-500}{144}$

$b=-0.3472$

Now, substitute, $a=500$ and $b=-0.3472$ in relation 1,

$H_P=500-0.3472Q^2$

Suction head = Pump head for single pump. So,

$H_P=H_S$

We know, $H_S=100+1.5Q^2$

Substituting these values, we get,

$500-0.3472Q^2=100+1.5Q^2$

$\left(1.5+0.3472\right)Q^2=500-100$

$Q^2=216.5439$

$Q=14.7154kgal/\min$

Brake horsepower is given by:

$P=\frac{\rho Qg\left(100+1.5Q^2\right)}{0.746\times 3.6\times {10}^6}$

Substituting, $\begin{array}{l}H=100+1.5Q^2\\ Q=14.7154kgal/\min \\ \rho =1.94slug/f{t}^3\end{array}$

We get,

$P=\frac{\left\{\begin{array}{l}1.94\times \left(14.7154\times 0.227125\frac{m^3/h}{kgal/\min }\right)\times 32.2\\ \times \left(100+1.5{\left(14.7154\frac{kgal}{\min }\times 0.227125\frac{m^3/h}{kgal/\min }\right)}^2\right)\end{array}\right\}}{0.746\times 3.6\times {10}^6}$

$P=0.009076hp$

Conclusion:

The discharge required for one pump is $Q=14.7154kgal/\min$ and the brake horsepower is = $P=0.009076hp$.

To determine

(b)

The discharge and brake horsepower required for two pumps in parallel.

Answer to Problem 11.74P

The discharge required for two pumps in parallel is $Q=15.877kgal/\min$ and the brake horse power is $P=0.01023hp$.

Explanation of Solution

Given Information:

Pump diameter, d = 32 inch

Fig 11.7a

For two pumps in parallel, flow rate is half. Hence:

$H_P=H_S$

Substitute,

$H_P=500-0.3472{\left(\frac{Q}{2}\right)}^2$

$H_S=100+1.5Q^2$

$500-0.3472{\left(\frac{Q}{2}\right)}^2=100+1.5Q^2$

$500-0.0868Q^2=100+1.5Q^2$

$\left(1.5+0.0868\right)Q^2=500-100$

$Q^2=216.5439$

$Q=15.877kgal/\min$

Brake horsepower is given by:

$P=\frac{\rho Qg\left(100+1.5Q^2\right)}{0.746\times 3.6\times {10}^6}$

Substitute,

$\begin{array}{l}H=100+1.5Q^2\\ Q=15.877kgal/\min \\ \rho =1.94slug/f{t}^3\end{array}$

$P=\frac{\left\{\begin{array}{l}1.94\times \left(15.877\times 0.227125\frac{m^3/h}{kgal/\min }\right)\times 32.2\\ \times \left(100+1.5{\left(15.877\frac{kgal}{\min }\times 0.227125\frac{m^3/h}{kgal/\min }\right)}^2\right)\end{array}\right\}}{0.746\times 3.6\times {10}^6}$

$P=0.01023hp$

Conclusion:

The discharge required for two pumps in parallel is $Q=15.877kgal/\min$ and the brake horse power is $P=0.01023hp$.

To determine

(c)

The discharge and brake horsepower required for two pumps in series.

Answer to Problem 11.74P

The discharge required for two pumps in series is $Q=20.2518kgal/\min$ and the brake horse power is $P=0.01409hp$.

Explanation of Solution

Given Information:

Pump diameter, d = 32in

Fig 11.7a

For two pumps in series, flow rate is two times the actual. Hence:

$2H_P=H_S$

Substitute,

$H_P=500-.03472Q^2$

$H_S=100+1.5Q^2$

$2\left(500-0.3472Q^2\right)=100+1.5Q^2$

$1000-0.6944Q^2=1000-100$

$Q^2=410.1348$

$Q=20.2518kgal/\min$

Brake horse power is given by:

$P=\frac{\rho Qg\left(100+1.5Q^2\right)}{0.746\times 3.6\times {10}^6}$

Substitute,

$H=100+1.5Q^2$, $Q=20.2518kgal/\min$ and $\rho =1.94slug/f{t}^3$

$P=\frac{\left\{\begin{array}{l}1.94\times \left(20.2518\times 0.227125\frac{m^3/h}{kgal/\min }\right)\times 32.2\\ \times \left(100+1.5{\left(20.2518\frac{kgal}{\min }\times 0.227125\frac{m^3/h}{kgal/\min }\right)}^2\right)\end{array}\right\}}{0.746\times 3.6\times {10}^6}$

$P=0.01409hp$

The configuration with 2 pumps in series is the best since, it has the best efficiency.

Conclusion:

The discharge required for two pumps in series is $Q=20.2518kgal/\min$ and the brake horse power is $P=0.01409hp.$.