Question

The pump of a water distribution system is powered by a 15 -kW electric motor whose efficiency is 90 percent. The water flow rate through the pump is \(50 \mathrm{L} / \mathrm{s}\). The diameters of the inlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the pressures at the inlet and outlet of the pump are measured to be \(100 \mathrm{kPa}\) and \(300 \mathrm{kPa}\) (absolute), respectively, determine the mechanical efficiency of the pump.

Short answer

Question: What is the mechanical efficiency of the pump given the following information: 1. Power of the electric motor: 15 kW 2. Efficiency of the electric motor: 90% 3. Water flow rate through the pump: 50 L/s 4. Pressures at the inlet and outlet: 100 kPa and 300 kPa (absolute) 5. The elevation difference across the pump is negligible. Answer: The mechanical efficiency of the pump is approximately 7.6%.

Step-by-step solution

Calculate input shaft power

The input shaft power can be determined using the electric motor power and efficiency. Since efficiency is the ratio of useful output power to input power, we can set up the following equation: Input shaft power = (Electric motor power) × (Electric motor efficiency) Let's plug in the known values and calculate the input shaft power. Input shaft power = (15 kW) × (0.9) Input shaft power = 13.5 kW

Calculate work done on the fluid

The work done on the fluid can be calculated using the following formula, which is derived using the Bernoulli's equation for fluid flow: Work done on the fluid = Flow rate × (outlet pressure - inlet pressure) / \(\rho g\) Where \(\rho\) represents the density of water (approximately \(1000 \mathrm{kg/m^3}\)), and \(g\) represents the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). First, we need to convert the flow rate from L/s to m³/s: Flow rate = \(50 \frac{\mathrm{L}}{\mathrm{s}} \times \frac{1 \mathrm{m^3}}{1000 \mathrm{L}} = 0.05 \frac{\mathrm{m^3}}{\mathrm{s}}\) Now, let's plug in the known values and calculate the work done on the fluid: Work done on the fluid = (0.05 \(\frac{\mathrm{m^3}}{\mathrm{s}}\)) × (300 kPa - 100 kPa) / (1000 \(\frac{\mathrm{kg}}{\mathrm{m^3}}\) × 9.81 \(\frac{\mathrm{m}}{\mathrm{s^2}}\)) Work done on the fluid = 0.05 × (200 × 10³ N/m²) / (1000 × 9.81 N/kg) Work done on the fluid = 1.02 kW

Determine the mechanical efficiency of the pump

Finally, we can calculate the mechanical efficiency of the pump using the following formula: Mechanical efficiency = (Work done on the fluid) / (Input shaft power) Plugging in the values, we get: Mechanical efficiency = (1.02 kW) / (13.5 kW) Mechanical efficiency ≈ 0.076 or 7.6% Therefore, the mechanical efficiency of the pump is approximately 7.6%.

Key concepts

Pump Performance

Understanding the performance of a pump is critical to ensuring efficient operation in fluid systems. Pump performance encapsulates several parameters including flow rate, pressure increase, power consumption, and efficiency. The flow rate indicates how much fluid the pump can move over a period of time and is often measured in liters per second (L/s) or cubic meters per hour (m³/h).

Pressure increase, on the other hand, is the ability of the pump to raise the pressure of the fluid from the inlet to the outlet. It is typically measured in pascals (Pa) or kilopascals (kPa). Power consumption relates to the energy required to operate the pump, generally represented in kilowatts (kW). Efficiency, a critical aspect of pump performance, represents the ratio of useful fluid work to the input power and is expressed as a percentage. The higher the efficiency, the more effectively a pump converts the input power into fluid movement and pressure.

Bernoulli's Equation

Bernoulli's equation is a fundamental principle in fluid mechanics that describes the conservation of energy in a flowing fluid. It is derived from the conservation of mechanical energy and assumes steady, incompressible flow with no friction losses.

The equation states that the sum of static pressure, dynamic pressure, and gravitational potential energy per unit volume is constant along a streamline. Mathematically, it can be represented as: \[ P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \] where \( P \) is the fluid pressure, \( \rho \) is the fluid density, \( v \) is the fluid velocity, and \( g \) is the acceleration due to gravity. The term \( h \) represents the height above a reference point. Bernoulli's equation helps us understand how pressure varies with fluid speed and elevation, and is crucial in calculating the work done by pumps on the fluid as demonstrated in the exercise provided.

Fluid Mechanics

Fluid mechanics is the study of fluids (liquids and gases) and the forces acting upon them. It has two main branches - fluid statics, which deals with fluids at rest, and fluid dynamics, which involves fluids in motion.

In the context of pumps, fluid dynamics is particularly important as it involves understanding the behavior of the fluid as it moves through the system. Important variables to consider include flow velocity, pressure changes, fluid density, and viscosity. Fluid mechanics also involves the study of various principles and equations like Bernoulli's principle, the Continuity equation, and the Navier-Stokes equations, all relevant in the design and analysis of pumping systems. The concepts from fluid mechanics provide the basis for understanding pump operation and for predicting the movement and attributes of the fluid within a pump system.

Energy Conversion Efficiency

Energy conversion efficiency in the context of pumps refers to the effectiveness with which the pump converts electrical (or other forms of) energy into kinetic and potential energy within the fluid.

The efficiency of a pump is described by two key values: the motor efficiency and the mechanical efficiency. Motor efficiency accounts for the conversion from electrical energy to mechanical energy and is a measure of how much input electrical power is converted to output shaft power. Mechanical efficiency, on the other hand, compares the power added to the fluid (as increased pressure and flow) to the input shaft power. This latter value, which was calculated in our exercise, highlights the losses within the pump itself, due to factors such as friction and turbulence. When designed and operated optimally, pumps can achieve high levels of energy conversion efficiency, leading to reduced operational costs and energy consumption.