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Figure P4.101 shows a pumped-hydro energy storage system delivering water at steady state from a lower reservoir to an upper reservoir using off-peak electricity (see Sec. 4.8.3). Water is delivered to the upper reservoir at a volumetric flow rate of \(150 \mathrm{~m}^{3} / \mathrm{s}\) with an increase in elevation of \(20 \mathrm{~m}\). There is no significant change in temperature, pressure, or kinetic energy from inlet to exit. Heat transfer from the pump to its surroundings occurs at a rate of \(0.6 \mathrm{MW}\) and \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the pump power required, in MW. Assuming the same volumetric flow rate when the system generates on-peak electricity using this water, will the power be greater, less, or the same as the pump power? Explain.

Step by step solution

01

- Calculate the potential energy change per second

First, determine the potential energy change per second for the water being pumped to the higher elevation. We use the formula for gravitational potential energy: \[ P.E. = \text{mass} \times g \times h \] Given that the water density \( \rho \) is constant at approximately \( 1000 \text{ kg/m}^3 \), the mass flow rate \( \dot{m} \) can be calculated as: \[ \dot{m} = \rho \times \dot{V} \] where \( \dot{V} = 150 \text{ m}^3/\text{s} \) and \( \rho = 1000 \text{ kg/m}^3 \), thus: \[ \dot{m} = 1000 \times 150 = 150000 \text{ kg/s} \] The potential energy change per second is then: \[ \text{Potential Energy per second} = 150000 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 20 \text{ m} = 29430000 \text{ W} = 29.43 \text{ MW} \]

02

- Account for heat transfer

The problem states that the heat transfer from the pump to its surroundings is 0.6 MW. Since heat loss reduces the pump's efficiency, we must add this heat transfer to the potential energy change to find the total power required by the pump: \[ \text{Total Pump Power} = 29.43 \text{ MW} + 0.6 \text{ MW} = 30.03 \text{ MW} \]

03

- Determine power generation

When the system generates on-peak electricity, the power generated will be affected by system efficiencies and losses similar to the pump. Assuming the same conditions and neglecting other losses, the power generated will likely be less than the power required by the pump due to conversion losses. However, if no losses are considered for this hypothetical scenario, the generated power would be equivalent to the change in potential energy, 29.43 MW.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. For the pumped-hydro system, it's the energy gained by water when it's pumped to a higher elevation.
We calculate it using the formula: \( P.E. = \text{mass} \times g \times h \).
In our example, water with a mass flow rate of 150,000 kg/s is lifted 20 meters. Given the gravitational acceleration, g, as 9.81 m/s^2, the potential energy change per second is: \[ 150,000 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 20 \text{ m} = 29.43 \text{ MW} \]. This shows how the energy required to lift the water is directly proportional to the height and mass.

Mass Flow Rate

Mass flow rate is a measure of the amount of mass moving through a system per unit of time. In pumped-hydro systems, it's crucial to know how much water is being moved.
We obtain the mass flow rate \( \dot{m} \) by multiplying the water's density \( \rho\) by the volumetric flow rate \( \dot{V}\): \[ \dot{m} = \rho \times \dot{V} \].
Given that water’s density is roughly 1000 kg/m^3 and the volumetric flow rate is 150 m^3/s, the mass flow rate calculates as: \[ 1000 \times 150 = 150,000 \text{ kg/s} \]. This mass flow rate is critical for determining the energy requirements and efficiency of the system.

System Efficiency

System efficiency in a pumped-hydro system measures how well the system converts input energy into useful output energy. Efficiency is reduced by factors like heat loss.
In our scenario, we know heat transfer from the pump to its surroundings is 0.6 MW. The total power requirement, including heat loss, adds to the potential energy required to lift the water: \[ 29.43 \text{ MW} + 0.6 \text{ MW} = 30.03 \text{ MW} \].
Hence, any inefficiencies, like heat loss, increase the total energy required to pump the water.

Heat Transfer

Heat transfer in the pumped-hydro system is an energy loss mechanism. Here, the pump loses 0.6 MW of energy as heat.
This loss means not all the electrical energy used by the pump contributes to raising the water’s elevation. Instead, some energy dissipates as thermal energy. Thus, the pump needs more power than just the calculated potential energy.
Considering our calculated potential energy change (29.43 MW), and adding the heat loss (0.6 MW), we get the total power required for operation: \[ 30.03 \text{ MW} \]. This shows the importance of accounting for heat transfer to understand real energy requirements.

Power Generation

Power generation in a pumped-hydro system converts the stored gravitational potential energy back into electrical energy during peak demand.
If the system's efficiency remains identical to when pumping, losses will still occur, implying that some potential energy will convert to heat or other losses.
Assuming ideal conditions, neglecting other losses, the power generated aligns with the potential energy change (29.43 MW).
Practically, conversion inefficiencies mean the actual power generated will be slightly less than the potential energy calculated due to these inherent losses. Therefore, managing efficiency and minimizing losses is key to maximizing power output.