A pumped water storage plant has a head of 600 m and volume 5 x 10° m². How much potential energy is stored in the water? В. А. How much power is available if the storage empties in 2 hr with 85% efficiency?
A pumped water storage plant has a head of 600 m and volume 5 x 10° m². How much potential energy is stored in the water? В. А. How much power is available if the storage empties in 2 hr with 85% efficiency?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem 9.5: Hydroelectric Energy Calculation**
A pumped water storage plant has a head of 600 meters and a volume of \(5 \times 10^5\) cubic meters (m³).
**A. Calculation of Potential Energy Stored**
How much potential energy is stored in the water?
**B. Calculation of Available Power**
How much power is available if the storage empties in 2 hours with 85% efficiency?
---
**Detailed Explanation:**
This problem involves calculating the potential energy stored in a hydroelectric power plant and then determining the available power output when the plant releases the stored water over a specified period.
To solve part A:
- The potential energy (\(E_p\)) stored in the water can be calculated using the formula:
\[ E_p = mgh \]
where:
- \(m\) is the mass of the water (in kilograms),
- \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)),
- \(h\) is the height (head) in meters.
To find the mass (\(m\)), use the volume (\(V\)) and the density of water (\(\rho \approx 1000 \, \text{kg/m}^3\)):
\[ m = V \times \rho \]
For part B:
- Power (\(P\)) is the rate of energy release. Given the energy (\(E_p\)), time (\(t\)), and efficiency (\(\eta\)):
\[ P = \frac{E_p \times \eta}{t} \]
where efficiency (\(\eta\)) is a decimal (85% efficiency = 0.85). The time (\(t\)) should be converted to seconds for consistent units.
Ensure clarity and step-by-step procedures for students to follow these formulas and understand the physical principles involved in pumped storage hydroelectric systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf192c4a-3623-40a7-8b1d-18d623c8a61a%2F00a74a47-7470-4361-8c6a-21d6c13c07a1%2Fes3wq4r_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 9.5: Hydroelectric Energy Calculation**
A pumped water storage plant has a head of 600 meters and a volume of \(5 \times 10^5\) cubic meters (m³).
**A. Calculation of Potential Energy Stored**
How much potential energy is stored in the water?
**B. Calculation of Available Power**
How much power is available if the storage empties in 2 hours with 85% efficiency?
---
**Detailed Explanation:**
This problem involves calculating the potential energy stored in a hydroelectric power plant and then determining the available power output when the plant releases the stored water over a specified period.
To solve part A:
- The potential energy (\(E_p\)) stored in the water can be calculated using the formula:
\[ E_p = mgh \]
where:
- \(m\) is the mass of the water (in kilograms),
- \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)),
- \(h\) is the height (head) in meters.
To find the mass (\(m\)), use the volume (\(V\)) and the density of water (\(\rho \approx 1000 \, \text{kg/m}^3\)):
\[ m = V \times \rho \]
For part B:
- Power (\(P\)) is the rate of energy release. Given the energy (\(E_p\)), time (\(t\)), and efficiency (\(\eta\)):
\[ P = \frac{E_p \times \eta}{t} \]
where efficiency (\(\eta\)) is a decimal (85% efficiency = 0.85). The time (\(t\)) should be converted to seconds for consistent units.
Ensure clarity and step-by-step procedures for students to follow these formulas and understand the physical principles involved in pumped storage hydroelectric systems.
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