Chapter 11, Problem 11.75P

To determine

(a)

The flow rate and power required when both the pumps are running.

Answer to Problem 11.75P

When both the pumps are running the flow-rate is $Q=74{\text{m}}^{\text{3}}\text{/min}\left({\text{37 m}}^{\text{3}}\text{/min for each pump}\right)$ and power is $P=1547.65\text{bhp}\text{.}$

Explanation of Solution

Given information:

Water temperature = 20° C

The pump size = 35 inch = 88.9 cm

The density of water = 998 kg/m³

All other system losses can be neglected.

Calculation:

Let us assume following values.

The density of water = 998 kg/m³.

The value of dynamic viscosity = $\mu =1.003\times {10}^{-3}{\text{Ns/m}}^2$

For the cast iron, $\epsilon =2.59\times {10}^{-4}m,\text{so }\frac{\epsilon }{d}=0.000425$

The 35-inch pump has the curve fit head relation

$H_p\left(\text{m}\right)=71.63-3.372\times {10}^{-5}{Q}^3,{\text{ Q is in m}}^3\text{/min}$.

The two pumps are connected in parallel, so, each pipe takes half of the total volume flow rate.

$H_p=fcn\left(\frac{Q}{2}\right)=H_{sys}=\Delta z+\frac{fL{V}^2}{d\times 2g}$ ....(1)

The velocity of flow is given by $V=\frac{Q}{\frac{\pi }{4}\times d^2}=\frac{Q}{\frac{\pi }{4}\times {\left(0.61\right)}^2}=33.771Q$ m/s

And, the change in elevation is, $\Delta z=30.5m$

Putting values in equation (1):

$H_p=fcn\left(\frac{Q}{2}\right)=H_{sys}=30.5+\frac{0.0163\times 1609.3\times {\left(33.771Q\right)}^2}{0.6096\times 2\times 9.81}$

The value of Q is to be iterated till both pump head and system head become equal.

By iterative calculations, at $Q=74{\text{m}}^{\text{3}}\text{/min}\left({\text{37 m}}^{\text{3}}\text{/min for each pump}\right)$

Now, the total power required is

$P=\frac{\rho gQH}{\eta }$

Under BEP operation conditions,

$P=\frac{998\times 9.81\times \left(\frac{74}{60}\right)\times 69.8}{0.73}=1154550W$

$P=\frac{1154550}{746}=1547.65\text{bhp}$

Conclusion:

Thus, when both the pumps are running the flow-rate is:

$Q=74{\text{m}}^{\text{3}}\text{/min}\left({\text{37 m}}^{\text{3}}\text{/min for each pump}\right)$ and power is $P=\frac{1154550}{746}=1547.65\text{bhp}$.

To determine

(b)

The flow-rate and power, when one of the pumps is shut off and isolated.

Answer to Problem 11.75P

When one of the pumps is shut off and isolated, the flow-rate is $Q=66{\text{m}}^{\text{3}}\text{/min}$ and power is $P=1025.47\text{bhp}\text{.}$

Explanation of Solution

Given information:

RPM of the pump = 3500 rpm.

Water temperature = 20° C

Diameter of the pipe = 5 inch = 12.7 cm.

Length of the pipe = 2195 m.

All other system losses can be neglected.

Calculation

Let us assume following values

The density of water = 998 kg/m³.

The value of dynamic viscosity = $\mu =1.003\times {10}^{-3}{\text{Ns/m}}^2$

For the cast iron, $\epsilon =2.59\times {10}^{-4}m,\text{so }\frac{\epsilon }{d}=0.000425$

The 35-inch pump has the curve fit head relation

$H_p\left(\text{m}\right)=71.63-3.372\times {10}^{-5}{Q}^3,{\text{ Q is in m}}^3\text{/min}$.

The two pumps are connected in parallel, so, each pipe takes half of the total volume flow rate.

$H_p=fcn\left(Q\right)=H_{sys}=\Delta z+\frac{fL{V}^2}{d\times 2g}$ ....(1)

The velocity of flow is given by $V=\frac{Q}{\frac{\pi }{4}\times d^2}=\frac{Q}{\frac{\pi }{4}\times {\left(0.61\right)}^2}=33.771Q$ m/s

And, the change in elevation is, $\Delta z=30.5m$

Putting values in equation (1):

$H_p=fcn\left(Q\right)=H_{sys}=30.5+\frac{0.0163\times 1609.3\times {\left(33.771Q\right)}^2}{0.6096\times 2\times 9.81}$

The value of Q is to be iterated till both pump head and system head become equal.

By iterative calculations, at $Q=66{\text{m}}^{\text{3}}\text{/min}.$

Now, the total power required is:

$P=\frac{\rho gQH}{\eta }$

Under BEP operation conditions for single pump,

$P=\frac{998\times 9.81\times \left(\frac{66}{60}\right)\times 61.8}{0.87}=765000W$

$P=\frac{765000}{746}=1025.47\text{bhp}$

Conclusion:

Thus, when one of the pumps is shut off and isolated, the flow-rate is $Q=66{\text{m}}^{\text{3}}\text{/min}$ and power is $P=1025.47\text{bhp}\text{.}$.