18.8
Application of the Energy Equation
The application of the energy equation to centrifugal pumps is a fundamental principle in fluid dynamics and engineering. In this scenario, the energy…
If the flow rate of a centrifugal pump that applies a pressure of 7500 joules per second to lift water from a reservoir at a lower level to a reservoir at an upper level has to be calculated, the energy equation can be used.
The level difference between the reservoirs is 10 meters, and the head loss is 5 meters.
As the surfaces of both reservoirs are open to the atmosphere, the pressure and velocity in both the inlet and the outlet are zero, further simplifying the energy equation.
Here, the head of the pump is obtained by the quotient of the net shaft power input and the product of the unit weight of the water and the flow rate.
If the head of the pump is substituted for the simplified energy equation, the pump's flow rate can be obtained.
Further, substituting the values provided in the expression for the flow rate gives the final value indicating the rate at which the water is pumped.
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Q1: How does the energy equation apply to centrifugal pump flow rate calculations?
The energy equation relates the pump's shaft power input to the flow rate by accounting for gravitational head and head loss. The pump head is calculated by dividing net shaft power by the product of flow rate and water's specific weight. Substituting this into the simplified energy equation allows engineers to solve for the pump's flow rate directly using conservation of energy in control volume principles.
Q2: Why do pressure and velocity terms cancel out in this pump system?
Both reservoirs are open to the atmosphere, so inlet and outlet pressures equal atmospheric pressure and cancel in the energy equation. Similarly, water surfaces in both reservoirs have negligible velocity. These simplifications reduce the energy equation to depend only on gravitational head and pump head, making calculations more straightforward.
Q3: What role does head loss play in determining pump flow rate?
Head loss, caused by friction and other resistances, represents energy dissipated in the system. In this example, the 5-meter head loss reduces the effective energy available for lifting water. The energy equation accounts for this loss, ensuring the calculated flow rate reflects the actual energy balance between pump input, gravitational lift, and system resistance.
Q4: How is pump head calculated from shaft power input?
Pump head equals the net shaft power input divided by the product of flow rate and water's specific weight. In this case, with 7500 joules per second of power input, the head represents the height equivalent of energy the pump can deliver. This relationship is central to solving for flow rate using the energy equation.
Q5: What does the 10-meter level difference represent in the energy equation?
The 10-meter elevation difference between reservoirs represents the gravitational head the pump must overcome. This gravitational potential energy requirement is a key term in the energy equation. Combined with the 5-meter head loss, the total head needed is 15 meters, which the pump must supply through its shaft power input.
Q6: Why is specific weight of water important in pump flow rate calculations?
Specific weight, the weight per unit volume of water, directly relates shaft power to pump head through the energy equation. Dividing power by the product of specific weight and flow rate yields the pump head. This parameter ensures dimensional consistency and accounts for water's physical properties in converting mechanical power into hydraulic head.
Q7: How do you solve for flow rate once the energy equation is simplified?
After substituting pump head into the simplified energy equation and rearranging algebraically, flow rate becomes the unknown variable. Substituting known values—7500 joules per second power, 10-meter elevation, 5-meter head loss, and water's specific weight—yields the numerical flow rate. This final calculation determines the rate at which water is pumped between reservoirs.