Homework Assignment: The World's Largest Water Lift
Home Work Assignment 3the Worlds Largest Water Lift The Edmonston P
Home Work Assignment 3the Worlds Largest Water Lift The Edmonston P
Home Work Assignment #3 The world’s largest water lift, the Edmonston Pumping Plant located at the northern edge of the Tehachapi Mountains, delivers as much as 2.5 million acre-feet (per year) of water to 13 Southern California water contractors / districts. Through this process, the Edmonston plant consumes 40 percent of all electricity used by the State Water Project, making it the project’s biggest user of electric power. To deliver the water, fourteen pumps lift the water 2,000 feet at a maximum flow rate of 10,000 acre-feet per day to a gravity transfer station reservoir at the top of the mountain, where it is split into two aqueducts that serve Southern California. Water from the West Branch Aqueduct is stored in Pyramid Lake and Castaic Lake for distribution to Los Angeles and surrounding cities.
The East Branch Aqueduct passes through Palmdale and Lancaster, and stores water in Silverwood Lake and Lake Perris for distribution to Inland Empire cities such as San Bernardino and Riverside. Assuming that each motor / pump combination operates at 65% total efficiency at full load, and there is an assumed 2% fluid flow total energy loss to friction (hL) of the water flow through the piping from suction reservoir to mountain top discharge reservoir. Determine the maximum power demand in kW of the plant while at full flow capacity. Use the energy equation below and show your work.
P1/Ƴ + v12/2g + Z1 + hp = P2/Ƴ + v22/2g + Z2 + hT + hL
Paper For Above instruction
The Edmonston Pumping Plant stands as the largest water lift facility in the world, playing a pivotal role in California’s water infrastructure. Given its capacity to lift and deliver significant volumes of water across challenging terrains, analyzing its power demands provides critical insights into the energy efficiency and sustainability of such large-scale water transportation systems. This paper computes the maximum power demand of the Edmonston plant during full flow operation by applying fundamental fluid mechanics principles and energy equations, considering system efficiencies and energy losses.
The primary data includes a maximum flow rate of 10,000 acre-feet per day and a vertical lift of 2,000 feet. The plant’s pumps operate at 65% efficiency, with a 2% energy loss attributable to friction within the piping system. The goal is to determine the power required to achieve this flow under these conditions, expressed in kilowatts (kW).
The analysis begins with the fundamental energy conservation equation for open channel and pressurized flow, incorporating elevation differences, pressure head, velocity head, and head losses:
P1/Ƴ + v12/2g + Z1 + hp = P2/Ƴ + v22/2g + Z2 + hT + hL
Where, P1 and P2 are the pressures at the inlet and outlet respectively; v1 and v2 are the velocities of water at these points; Z1 and Z2 are the elevation heads; hp is the pressure head; hT is the total dynamic head; and hL represents the head loss due to friction.
Given that the flow is from a reservoir at atmospheric pressure and the discharge at the top is similarly open to atmospheric conditions, the pressure heads at both points are effectively zero (P1 = P2 = 0). The elevation head difference (Z2 - Z1) equals 2000 feet, which needs to be converted into meters for SI consistency.
The flow rate of 10,000 acre-feet per day is converted into cubic meters per second (m3/s) for use in fluid velocity calculations. Using known conversion factors:
- 1 acre-foot ≈ 1233.5 m3
- 1 day = 86400 seconds
The total annual volume of water is 2.5 million acre-feet, but for the power calculation, the maximum flow rate is considered. After the conversion, the volumetric flow rate (Q) is computed as:
Q = (10,000 acre-feet/day) × (1233.5 m3/acre-foot) / 86400 seconds ≈ 142.8 m3/s
The velocity (v) of water in the pipe is obtained from Q and the cross-sectional area (A). Without specific pipe diameter, standard assumptions or typical pipe dimensions can be used; however, in this context, the velocity can be approximated if necessary or assumed as constant. For this calculation, considering the energy head and flow, we can focus on the total dynamic head (Htotal), represented as:
Htotal = Z2 + hp + hT + hL
The total head comprises the elevation head (2000 feet) and the energy losses. Head loss due to friction (hL) is 2% of the total energy head, thus:
hL = 0.02 × Htotal
Conversely, the total head loss can be estimated iteratively or simplified by recognizing the dominant head component as the elevation head. Converting 2000 feet to meters (1 foot ≈ 0.3048 meters):
2000 feet × 0.3048 ≈ 609.6 meters
The total dynamic head (Htotal) is approximately 609.6 meters, accounting for the pressure and velocity heads which are comparatively smaller at high flow rates and long pipe conditions.
The power required (Pmech) to pump water at the given flow rate can be calculated with the formula:
Pmech = ρ × g × Q × Htotal
where ρ is the density of water (approximately 1000 kg/m3) and g is gravitational acceleration (9.81 m/s2). Incorporating head loss:
Effective power input, considering efficiency (η), is:
Pinput = Pmech / η
and accounting for the 65% efficiency:
Pactual = Pinput / 0.65
Calculations:
Pmech = 1000 kg/m3 × 9.81 m/s2 × 142.8 m3/s × 609.6 m ≈ 8.52 × 108 W
Pmech ≈ 852 MW
Effective power input considering efficiency:
Pinput = 852 MW / 0.65 ≈ 1,310.77 MW
Finally, adjusting for efficiency:
Pactual = 1,310.77 MW / 0.65 ≈ 2,017.35 MW
Converting to kilowatts gives approximately 2,017,350 kW. This figure represents the maximum power demand of the Edmonston plant during full flow operation, considering the energy losses and operational efficiencies.
This estimation underscores the immense energy requirement of such large-scale water pumping systems and highlights the importance of improving efficiency and exploring renewable energy integration to reduce the environmental impact.
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