A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 1.5 ft3/s of water. Does the water temperature at the inlet have any significant effect on the required flow power?

To determine the power input required, you would use the formula for hydraulic power, P, given by:

\[ P = \text{flow rate (Q)} \times \text{pressure increase}(\Delta P) \times \text{pump efficiency} (\eta) \]

However, since we don't have the pump efficiency, we'll have to assume the pump operates at 100% efficiency for this calculation (this is an idealization and in reality, pump efficiency is less than 100%). Also, the pressure increase should be in units consistent with the flow rate, so the following unit conversions will be performed:

\[ 1 \text{ hp} = 550 \text{ ft}\cdot\text{lbf/s} \] \[ 1 \text{ psia} = 144 \text{ lbf/ft}^2 \ (because 1 \text{ psi} = 144 \text{ lbf/ft}^2) \]

Given: - Initial pressure \( P1 = 15 \text{ psia} \) - Final pressure \( P2 = 70 \text{ psia} \) - Flow rate \( Q = 1.5 \text{ ft}^3/\text{s} \)

First, calculate the pressure increase in pounds force per square foot (lbf/ft^2):

\[ \Delta P = P2 - P1 = (70 - 15) \text{ psia} \times 144 \text{ lbf/ft}^2/\text{psia} = 55 \text{ psia} \times 144 \text{ lbf/ft}^2/\text{psia} = 7920 \text{ lbf/ft}^2 \]

Now, we find the hydraulic power in foot-pounds force per second (ft*lbf/s):

\[ P = Q \times \Delta P = 1.5 \text{ ft}^3/\text{s} \times 7920 \text{ lbf/ft}^2 = 11880 \text{ ft}\cdot\text{lbf/s} \]

Finally, convert this power to horsepower (hp):

\[ P_{\text{hp}} = \frac{11880 \text{ ft}\cdot\text{lbf/s}}{550 \text{ ft}\cdot\text{lbf/s/hp}} \approx 21.6 \text{ hp} \]

So, the power input required to pump 1.5 ft^3/s of water from 15 psia to 70 psia, assuming 100% efficiency, is approximately 21.6 hp.

Extra: The water temperature can affect the required power in several ways. First, the density of water changes with temperature, which can influence the mass flow rate for a given volumetric flow rate. A higher density means more mass is being moved with the same volume, potentially requiring more power. However, for the pressures and temperature ranges typically found in water pumping applications, this effect is typically minor.

Water's viscosity also changes with temperature. Higher temperatures typically decrease water's viscosity, allowing it to flow more easily through pipes, which might slightly reduce energy losses due to friction.

In reality, efficiency plays a significant role in the power required by the pump. Pump efficiency depends on several factors, including pump design, operating point relative to the pump's best efficiency point (BEP), and the condition of the pump and the system it's operating in (such as pipe roughness, bends, and other fittings). However, this was not factored into the calculation since we assumed 100% efficiency.

In most practical situations, for reasonable temperature variations that one might encounter in typical water pumping scenarios, the effects on required power due to temperature changes of the water are usually considered negligible. The primary drivers of power requirement are the pressure increase and the volumetric flow rate, as these dictate the amount of work the pump must perform on the water.