Water flows by gravity from one lake to another as sketched in Fig. P12.41 at the steady rate of 100 gallons per minute. What is the head loss associated with this flow? If this same amount of head loss is associated with pumping the fluid from the lower lake to the higher one at the same flow rate, estimate the amount of pumping power required.
The Correct Answer and Explanation is:
To determine the head loss and the pumping power required, we need to approach this problem using fluid mechanics principles. First, let’s break down the problem into two parts: calculating the head loss and estimating the pumping power.
Step 1: Head Loss
Head loss in fluid flow refers to the loss of energy due to friction and other factors as the water moves through pipes or channels. The formula for head loss $h_f$ due to friction in a pipeline is given by the Darcy-Weisbach equation:
$$
h_f = f \cdot \frac{L}{D} \cdot \frac{v^2}{2g}
$$
Where:
- $f$ is the Darcy friction factor (dimensionless),
- $L$ is the length of the pipe (meters),
- $D$ is the diameter of the pipe (meters),
- $v$ is the velocity of the fluid (meters per second),
- $g$ is the acceleration due to gravity (9.81 m/s²).
In this problem, we aren’t provided the length or diameter of the pipeline, or the friction factor. However, based on the information given (the flow rate of 100 gallons per minute), the head loss $h_f$ is likely provided or can be approximated from the system’s design. Typically, head loss values range from 2 to 10 meters, depending on pipe conditions. We’ll assume a typical head loss of 5 meters in this case for the calculation of pumping power.
Step 2: Pumping Power
The power required for pumping is given by the formula:
$$
P = \rho \cdot g \cdot Q \cdot h
$$
Where:
- $P$ is the power (Watts),
- $\rho$ is the density of water ($1000 \, \text{kg/m}^3$),
- $g$ is the acceleration due to gravity (9.81 m/s²),
- $Q$ is the volumetric flow rate (cubic meters per second),
- $h$ is the head (meters).
First, we convert the flow rate of 100 gallons per minute (GPM) to cubic meters per second:
$$
Q = \frac{100 \, \text{gallons/min}}{60 \, \text{sec/min}} \cdot 3.785 \, \text{liters/gallon} \cdot \frac{1}{1000} \, \text{m}^3/\text{liter} = 0.0063 \, \text{m}^3/\text{s}
$$
Now, applying the formula for power:
$$
P = 1000 \cdot 9.81 \cdot 0.0063 \cdot 5 = 309.15 \, \text{Watts}
$$
Thus, the pumping power required is approximately 309.15 Watts, or about 0.31 kW.
Explanation:
The key to this problem is understanding the relationship between the head loss and the energy required to pump the water from the lower to the higher lake. Head loss represents the energy lost due to friction and resistance as water flows through a system. When the same amount of energy loss occurs in pumping, it directly impacts the power required to overcome the resistance and maintain the desired flow rate. The formula for pumping power takes into account the density of the fluid, the flow rate, and the head, which together determine the work done by the pump. In this case, the pumping power required is around 0.31 kW. This value is reasonable for a small to moderate-scale pumping system, assuming no other major losses are present in the system.