Question

A pump is positioned at 2 m above the water of the reservoir. The inlet of the pipe connected to the pump is positioned at 6m beneath the water of the reservoir. When a pump draws 220 m3/hour of water at 20 °C from a reservoir, the total friction head loss is 5 m. The diameter of the pipe connected to the inlet and exit nozzle of the pump is 12 cm and 5 cm, respectively. The flow discharges through the exit nozzle to the atmosphere. Calculate the pump power in kW delivered to the water.

Answer 1

The pump delivers 32.737 kilowatts to the water.

Step-by-step explanation:

We can describe the system by applying the Principle of Energy Conservation and the Work-Energy Theorem, the pump system, which works at steady state and changes due to temperature are neglected, is represented by the following model:

`\dot W_(in) + \dot m \cdot g \cdot (z_(1)-z_(2)) + (1)/(2)\cdot \dot m \cdot (v_(1)^(2)-v_(2)^(2))+\dot m \cdot [(u_(1)+P_(1)\cdot \\u_(1))-(u_(2)+P_(2)\cdot \\u_(2))]-\dot E_(losses) = 0` (Eq. 2)

Where:

`\dot m` - Mass flow, measured in kilograms per second.

`g` - Gravitational acceleration, measured in meters per square second.

`z_(1)`,
`z_(2)` - Initial and final heights, measured in meters.

`v_(1)`,
`v_(2)` - Initial and final flow speeds at pump nozzles, measured in meters per second.

`u_(1)`,
`u_(2)` - Initial and final internal energies, measured in joules per kilogram.

`P_(1)`,
`P_(2)` - Initial and final pressures, measured in pascals.

`\\u_(1)`,
`\\u_(2)` - Initial and final specific volumes, measured in cubic meters per kilogram.

Then, we get this expression:

`\dot W_(in) + \dot m \cdot g \cdot (z_(1)-z_(2)) +(1)/(2)\cdot \dot m \cdot (v_(1)^(2)-v_(2)^(2)) +\dot m\cdot \\u \cdot (P_(1)-P_(2))-\dot E_(losses) = 0` (Ec. 3)

We note that specific volume is the reciprocal of density:

`\\u = (1)/(\rho)` (Ec. 4)

Where
`\rho` is the density of water, measured in kilograms per cubic meter.

The initial pressure of water (
`P_(1)`), measured in pascals, can be found by Hydrostatics:

`P_(1) = P_(atm) + \rho\cdot g \cdot \Delta z` (Ec. 5)

Where:

`P_(atm)` - Atmospheric pressure, measured in pascals.

`\Delta z` - Depth of the entrance of the inlet pipe with respect to the limit of the water reservoir.

If we know that
`p_(atm) = 101325\,Pa`,
`\rho = 1000\,(kg)/(m^(3))`,
`g = 9.807\,(m)/(s^(2))` and
`\Delta z = 6\,m`, then:

`P_(1) = 101325\,Pa+\left(1000\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(m)/(s^(2)))\cdot (6\,m)`

`P_(1) = 160167\,Pa`

And the specific volume of water (
`\\u`), measured in cubic meters per kilogram, is: (
`\rho = 1000\,(kg)/(m^(3))`)

`\\u = (1)/(1000\,(kg)/(m^(3)) )`

`\\u = 1* 10^(-3)\,(m^(3))/(kg)`

The power losses due to friction is found by this expression:

`\dot E_(losses) = \dot m \cdot g\cdot h_(losses)`

Where
`h_(losses)` is the total friction head loss, measured in meters.

The mass flow is obtained by this:

`\dot m = \rho \cdot \dot V` (Ec. 6)

Where
`\dot V` is the volumetric flow, measured in cubic meters per second.

If we know that
`\rho = 1000\,(kg)/(m^(3))` and
`\dot V = 0.061\,(m^(3))/(s)`, then:

`\dot m = \left(1000\,(kg)/(m^(3))\right)\cdot \left(0.061\,(m^(3))/(s) \right)`

`\dot m = 61\,(kg)/(s)`

Then, the power loss due to friction is: (
`h_(losses) = 5\,m`)

`\dot E_(losses) = \left(61\,(kg)/(s)\right)\cdot \left(9.807\,(m)/(s^(2)) \right) \cdot (5\,m)`

`\dot E_(losses) = 2991.135\,W`

Now, we calculate the inlet and outlet speed by this formula:

`v = (\dot V)/((\pi)/(4)\cdot D^(2) )` (Ec. 7)

Inlet nozzle (
`\dot V = 0.061\,(m^(3))/(s)`,
`D = 0.12\,m`)

`v_(1) = (0.061\,(m^(3))/(s) )/((\pi)/(4)\cdot (0.12\,m)^(2) )`

`v_(1) \approx 5.394\,(m)/(s)`

Oulet nozzle (
`\dot V = 0.061\,(m^(3))/(s)`,
`D = 0.05\,m`)

`v_(2) = (0.061\,(m^(3))/(s) )/((\pi)/(4)\cdot (0.05\,m)^(2) )`

`v_(2) \approx 31.067\,(m)/(s)`

(
`\dot m = 61\,(kg)/(s)`,
`g = 9.807\,(m)/(s^(2))`,
`z_(2) = 2\,m`,
`z_(1) = -6\,m`,
`v_(2) \approx 31.067\,(m)/(s)`,
`v_(1) \approx 5.394\,(m)/(s)`,
`P_(2) = 101325\,Pa`,
`P_(1) = 160167\,Pa`,
`\dot E_(losses) = 2991.135\,W`)

`\dot W_(in) = \left(61\,(kg)/(s)\right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot [2\,m-(-6\,m)]+(1)/(2)\cdot \left(61\,(kg)/(s)\right) \cdot \left[\left(31.067\,(m)/(s) \right)^(2)-\left(5.394\,(m)/(s) \right)^(2)\right] +\left(61\,(kg)/(s)\right)\cdot \left(1* 10^(-3)\,(m^(3))/(kg) \right)\cdot (101325\,Pa-160167\,Pa)+2991.135\,W`

`\dot W_(in) = 32737.518\,W`

The pump delivers 32.737 kilowatts to the water.