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How is the combined pump-motor efficiency of a pump and motor system defined? Can the combined pumpmotor efficiency be greater than either the pump or the motor efficiency?

Short Answer

Expert verified
Answer: No, the combined pump-motor efficiency can never be greater than either the pump or the motor efficiency individually.

Step by step solution

01

Define Combined Pump-Motor Efficiency

The combined pump-motor efficiency is a measure of how effectively a pump and motor system converts electrical energy into mechanical energy to move fluid. It can be defined as the product of the pump efficiency and the motor efficiency. Mathematically, the combined pump-motor efficiency (η_c) can be expressed as: η_c = η_pump * η_motor where η_pump represents the efficiency of the pump and η_motor represents the efficiency of the motor.
02

Efficiency Range

The efficiency of any system ranges from 0 to 1 (or 0% to 100%), where 0 denotes no efficiency and 1 denotes perfect efficiency. This applies to both pump efficiency and motor efficiency.
03

Find if Combined Efficiency Can Be Greater

To determine if the combined pump-motor efficiency can be greater than either the pump or motor efficiency, let's examine the product of two values, each between 0 and 1: If η_pump and η_motor are both between 0 and 1, then their product (η_c) will always be less than or equal to the individual efficiency values: 0 ≤ η_pump ≤ 1 and 0 ≤ η_motor ≤ 1 Therefore, 0 ≤ η_c ≤ η_pump and 0 ≤ η_c ≤ η_motor As a result, the combined pump-motor efficiency can never be greater than either the pump or the motor efficiency individually.

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Key Concepts

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

Pump Efficiency
Pump efficiency, often denoted as \( \eta_{pump} \), is a critical metric that determines how well a pump converts the energy from its power source—typically an electric motor—into the kinetic energy of the moving fluid. This transformation involves overcoming various losses such as friction within the pump, leakage, and turbulence. High-efficiency pumps are meticulously designed to minimize these losses.

Understanding pump efficiency is essential for engineering applications because it directly impacts the energy consumption and operational costs of a pump system. Operators seeking to reduce energy costs and environmental impact strive for pumps with higher efficiency ratings. The general formula to calculate pump efficiency is given by the ratio of the hydraulic power output to the mechanical power input.
Motor Efficiency
Motor efficiency, indicated as \( \eta_{motor} \), relates to how effectively an electric motor converts the electrical energy it receives into mechanical energy to drive a machine, like a pump. Just like with pumps, motor efficiency is affected by energy losses which occur due to factors such as heat generation, friction, and electrical resistance.

Manufacturers endeavor to produce motors with high efficiency to ensure that most of the electrical energy is utilized for productive work, leading to lower operating costs and reduced greenhouse gas emissions. It is determined by the percentage of electrical energy input that is converted into mechanical energy output. Motor efficiency is a key consideration when selecting a motor for a pump system since it significantly influences the overall system efficiency.
Energy Conversion
Energy conversion is the process of changing one form of energy into another. In the context of pump and motor systems, energy conversion involves transforming electrical energy into mechanical energy. This conversion is not 100% efficient due to inherent losses in the system. Understanding these conversion processes is crucial when analyzing system performance and determining combined pump-motor efficiency.

Importantly, the first law of thermodynamics, also known as the principle of conservation of energy, states that energy cannot be created or destroyed, only transformed. Energy conversion efficiency is, therefore, a measure of how much input energy is usefully converted to the desired output form without being wasted as unwanted forms, like heat or sound.
Mechanical Energy
Mechanical energy is the energy associated with the motion and position of an object. In pump systems, mechanical energy is the end product of the electric motor's work, harnessed to move fluids against gravity, resistance, and pressure differentials. It is composed of potential and kinetic energy aspects; with pumps, this is often seen as the pressure energy and flow velocity of the fluid being moved.

The efficiency of converting mechanical energy dictates the performance of a pump to fulfill its intended use. In practical terms, higher mechanical energy at the pump output means more efficient fluid movement, which can contribute to the overall performance of the system. The study of mechanical energy helps in optimizing the design and operation of a pump for maximum output with minimal energy input.

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Most popular questions from this chapter

An average vehicle puts out nearly 20 lbm of carbon dioxide into the atmosphere for every gallon of gasoline it burns, and thus one thing we can do to reduce global warming is to buy a vehicle with higher fuel economy. A U.S. government publication states that a vehicle that gets 25 rather than 20 miles per gallon will prevent 10 tons of carbon dioxide from being released over the lifetime of the vehicle. Making reasonable assumptions, evaluate if this is a reasonable claim or a gross exaggeration.

Heat is transferred steadily through a \(0.2-\mathrm{m}\) thick \(8 \mathrm{m} \times 4 \mathrm{m}\) wall at a rate of \(2.4 \mathrm{kW}\). The inner and outer surface temperatures of the wall are measured to be \(15^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C} .\) The average thermal conductivity of the wall is \((a) 0.002 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\) \((b)0.75 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\) \((c) 1.0 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\) \((d) 1.5 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\) \((e) 3.0 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\)

The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price. Suppose a utility company is selling electric power for \(\$ 0.05 / \mathrm{kWh}\) at night and is willing to pay \(\$ 0.12 / \mathrm{kWh}\) for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir \(40 \mathrm{m}\) above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump-motor operates as a turbine- generator during reverse flow. Preliminary analysis shows that a water flow rate of \(2 \mathrm{m}^{3} / \mathrm{s}\) can be used in either direction. The combined pump-motor and turbine-generator efficiencies are expected to be 75 percent each. Disregarding the frictional losses in piping and assuming the system operates for \(10 \mathrm{h}\) each in the pump and turbine modes during a typical day, determine the potential revenue this pump-turbine system can generate per year.

A 75 -hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high efficiency motor that has an efficiency of 95.4 percent. The motor operates 4368 hours a year at a load factor of 0.75 Taking the cost of electricity to be \(\$ 0.12 / \mathrm{kWh}\), determine the amount of energy and money saved as a result of installing the high-efficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are \(\$ 5449\) and \(\$ 5520,\) respectively.

A room is cooled by circulating chilled water through a heat exchanger located in a room. The air is circulated through the heat exchanger by a 0.25 -hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25 -hp equipment is 54 percent. Determine the rate of heat supply by the fan- motor assembly to the room.

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