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Chapter 1: Problem 13

The efficiency \(\eta\) of a pump is defined as the (dimensionless) ratio of the power developed by the flow to the power required to drive the pump: $$\eta=\frac{Q \Delta p}{\text { input power }}$$ where \(Q\) is the volume rate of flow and \(\Delta p\) is the pressure rise produced by the pump. Suppose that a certain pump develops a pressure rise of 35 lbf/in \(^{2}\) when its flow rate is \(40 \mathrm{L} / \mathrm{s} .\) If the input power is \(16 \mathrm{hp},\) what is the efficiency?

Short Answer

Expert verified
The efficiency of the pump is 81%.

Step by step solution

01

Convert Units for Pressure

The given pressure rise is 35 lbf/in虏. To facilitate calculation, convert this to Pascals (Pa). Recall that 1 lbf/in虏 = 6894.76 Pa. Therefore, convert as follows:\[\Delta p = 35 \times 6894.76 = 241316.6 \text{ Pa}\]
02

Convert Units for Input Power

The input power is given as 16 hp. To convert horsepower (hp) to watts (W), use the conversion 1 hp = 745.7 W.\[\text{Input power} = 16 \times 745.7 = 11931.2 \text{ W}\]
03

Convert Units for Flow Rate

Given that the flow rate is 40 L/s, convert this to cubic meters per second (m鲁/s), using the fact that 1 L = 0.001 m鲁.\[Q = 40 \times 0.001 = 0.04 \text{ m}^3/s\]
04

Calculate Efficiency

Use the efficiency equation \( \eta = \frac{Q \Delta p}{\text{input power}} \). Plug in the converted values:\[ \eta = \frac{0.04 \times 241316.6}{11931.2} = 0.81\]This means the efficiency \( \eta \) is 0.81 or 81%.

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Key Concepts

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

Power Conversion
In engineering, power conversion is an essential skill. It helps us switch between different units of power to get precise results. In the problem, we are given power in horsepower (hp) and need to convert it to watts (W), a standard unit of power.
To do this, remember:
  • 1 hp = 745.7 W
By knowing this, you can easily turn horsepower into watts:\[\text{Input power} = 16 \times 745.7 = 11931.2\ \text{W}\]This step is crucial because calculations for pump efficiency usually rely on metric units, making this conversion a necessity to ensure accuracy in results.
Flow Rate Conversion
Understanding flow rate conversion is key when dealing with fluid dynamics in pumps. Flow rates often need to be converted from liters per second (L/s) to cubic meters per second (m鲁/s).
This is important because many engineering and scientific calculations require the use of cubic meters.
Here's a simple conversion to keep in mind:
  • 1 L = 0.001 m鲁
Hence, to convert the given flow rate:\[Q = 40 \times 0.001 = 0.04\ \text{m}^3/s\]Whenever you're dealing with calculations of this sort, having the flow rate in cubic meters per second simplifies the process of plugging values into formulas, like the pump efficiency equation.
Pressure Unit Conversion
When working with pressure, it's often necessary to express it in a consistent unit for calculations. In this exercise, pressure is initially given in pounds-force per square inch (lbf/in虏), which is common in many industrial settings but needs conversion to Pascals (Pa) in scientific contexts.
Remember the conversion factor:
  • 1 lbf/in虏 = 6894.76 Pa
Using this, you convert the pressure rise:\[\Delta p = 35 \times 6894.76 = 241316.6\ \text{Pa}\]This step is vital as using Pascals allows us to maintain consistency across various metric units when calculating things like pump efficiency, ensuring that all measurements interplay correctly.

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Most popular questions from this chapter

Do some reading and report to the class on the life and achievements, especially vis-脙 -vis fluid mechanics, of (a) Evangelista Torricelli \((1608-1647)\) (b) Henri de Pitot \((1695-1771)\) \((c)\) Antoine Ch脙漏zy \((1718-1798)\) \((d)\) Gotthilf Heinrich Ludwig Hagen \((1797-1884)\) \((e)\) Julius Weisbach \((1806-1871)\) \((f)\) George Gabriel Stokes \((1819-1903)\) \((g)\) Moritz Weber \((1871-1951)\) (h) Theodor von K脙隆rm脙隆n ( \(1881-1963\) ) (i) Paul Richard Heinrich Blasius \((1883-1970)\) (j) Ludwig Prandtl (1875-1953) (k) Osborne Reynolds \((1842-1912)\) ( \(l\) ) John William Strutt, Lord Rayleigh \((1842-1919)\) \((m)\) Daniel Bernoulli \((1700-1782)\) \((n)\) Leonhard Euler \((1707-1783)\).

A soap bubble of diameter \(D_{1}\) coalesces with another bubble of diameter \(D_{2}\) to form a single bubble \(D_{3}\) with the same amount of air. Assuming an isothermal process, derive an expression for finding \(D_{3}\) as a function of \(D_{1}, D_{2}\) \(p_{\text {atm }},\) and \(Y.\)

Algebraic equations such as Bernoulli's relation, Eq. (1) of Ex. \(1.3,\) are dimensionally consistent, but what about differential equations? Consider, for example, the boundary-layer \(x\) -momentum equation, first derived by Ludwig Prandtl in 1904 : $$\rho u \frac{\partial u}{\partial x}+\rho v \frac{\partial u}{\partial y}=-\frac{\partial p}{\partial x}+\rho g_{x}+\frac{\partial \tau}{\partial y}$$ where \(\tau\) is the boundary-layer shear stress and \(g_{x}\) is the component of gravity in the \(x\) direction. Is this equation dimensionally consistent? Can you draw a general conclusion?

A velocity field is given by \(u=V \cos \theta, v=V \sin \theta,\) and \(w=0,\) where \(V\) and \(\theta\) are constants. Derive a formula for the streamlines of this flow.

A tank contains \(0.9 \mathrm{m}^{3}\) of helium at \(200 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) Estimate the total mass of this gas, in \(\mathrm{kg},(a)\) on earth and (b) on the moon. Also, (c) how much heat transfer, in MJ, is required to expand this gas at constant temperature to a new volume of \(1.5 \mathrm{m}^{3} ?\)
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