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Chapter 14: Problem 55

How much work is done by pressure in forcing \(1.4 \mathrm{~m}^{3}\) of water through a pipe having an internal diameter of \(13 \mathrm{~mm}\) if the difference in pressure at the two ends of the pipe is \(1.0 \mathrm{~atm} ?\)

Short Answer

Expert verified
The work done is 141,855 J.

Step by step solution

01

Understand the Problem

We are asked to find the work done by pressure in forcing 1.4 m鲁 of water through a pipe. We are given the pressure difference as 1.0 atm and need to convert it to Pascals (Pa) for standard units.
02

Convert Pressure to Standard Units

Convert the given pressure difference from atmospheres to Pascals: 1 atm = 101,325 Pa, so 1.0 atm = 101,325 Pa.
03

Use the Work Done by Pressure Formula

The formula for work done by pressure (W) is given by:\[ W = P \cdot V \]where \( P \) is the pressure difference and \( V \) is the volume of water. Here, \( P = 101,325 \) Pa and \( V = 1.4 \) m鲁.
04

Substitute Values and Calculate

Substitute the values into the formula:\[ W = 101,325 \times 1.4 \]Now, calculate the work done:
05

Calculate Work Done

Calculate \( 101,325 \times 1.4 = 141,855 \). Therefore, the work done by the pressure is 141,855 Joules (J).

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Key Concepts

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

Pressure Difference
In fluid dynamics, understanding the pressure difference is crucial when calculating work done by pressure. Pressure difference refers to the variation in pressure between two points in a system. In this exercise, it's about the pressure differences at either end of a pipe.
It's expressed in units like atmospheres (atm) or Pascals (Pa). For standardized calculations, these need to be consistent.
For instance, 1 atm is equivalent to 101,325 Pa, which is the standard conversion factor used to switch between these units. Recognizing these variations is key in determining the correct force exerted along the piping system.
Work Formula
The formula to compute the work done by pressure is a simple one. It revolves around the relationship between pressure, volume, and work. This can be expressed as:
  • \[ W = P \times V \]
  • Where \( W \) stands for the work done, \( P \) is the pressure difference, and \( V \) is the volume of fluid.
Plugging these variables into the equation allows us to calculate how much energy is expended when a fluid is moved due to pressure differences. This formula is highly useful in various fields where fluid dynamics play a critical role, like engineering and physics.
Unit Conversion
Unit conversion is an essential step in solving problems involving physical quantities. In this context, converting pressure units from atmospheres to Pascals is pivotal. This ensures that all units in your equations are compatible and in standard form for accurate calculations.
To convert atmospheres to Pascals, you use the conversion factor of 1 atm = 101,325 Pa. Applying this, a pressure of 1.0 atm equals 101,325 Pa. Unit conversion thus ensures you accurately compute the work done in the process, eliminating errors from mismatched units.
Work Calculation
Once you have all components in compatible units, it's time to calculate the work done by pressure. Using the work formula \( W = P \times V \), you substitute the converted values. Here, the pressure difference \( P \) is 101,325 Pa, and the volume \( V \) is 1.4 m鲁.
Substituting, we have:
  • \[ W = 101,325 \times 1.4 \]
  • This results in a work done (\( W \)) of 141,855 Joules (J).
This calculation demonstrates how energy is converted into work in moving water through a pipe due to pressure differences. Understanding this helps bridge the gap between theoretical physics and practical engineering scenarios.

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Most popular questions from this chapter

(a) If this longnecked, gigantic sauropod had a head height of \(21 \mathrm{~m}\) and a heart height of \(9.0 \mathrm{~m}\), what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 80 torr (just enough to perfuse the brain with blood)? Assume the blood had a density of \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (b) What was the blood pressure (in torr or \(\mathrm{mm} \mathrm{Hg}\) ) at the feet?

Water is moving with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) through a pipe with a cross-sectional area of \(4.0 \mathrm{~cm}^{2}\). The water graduall descends \(10 \mathrm{~m}\) as the pipe cross-sectional area increases to \(8.0 \mathrm{~cm}^{2} .\) (a) What is the speed at the lower level? (b) If the pressure at the upper level is \(1.5 \times 10^{5} \mathrm{~Pa}\), what is the pressure at the lower level?

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is \(60.0 \mathrm{~cm}\), and the density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\) Find the inner diameter.

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density \(1.30 \times 10^{3}\) \(\mathrm{kg} / \mathrm{m}^{3}\). The area of each base is \(4.00 \mathrm{~cm}^{2}\), but in one vessel the liquid height is \(0.854\) \(\mathrm{m}\) and in the other it is \(1.560 \mathrm{~m}\). Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. \(14-50\) ); the cross-sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of cross- sectional area \(a\) with speed \(v .\) A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta p\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. \(14-50\), show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(64 \mathrm{~cm}^{2}\) in the pipe and 32 \(\mathrm{cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and 41 \(\mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?
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