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Chapter 14: Problem 54

The water flowing through a \(1.9 \mathrm{~cm}\) (inside diameter) pipe flows out through three \(1.3 \mathrm{~cm}\) pipes. (a) If the flow rates in the three smaller pipes are 26,19, and \(11 \mathrm{~L} / \mathrm{min}\), what is the flow rate in the \(1.9 \mathrm{~cm}\) pipe? (b) What is the ratio of the speed in the \(1.9 \mathrm{~cm}\) pipe to that in the pipe carrying \(26 \mathrm{~L} / \mathrm{min} ?\)

Short Answer

Expert verified
(a) 56 L/min, (b) Speed ratio is about 0.7.

Step by step solution

01

Understand the Problem

Consider the situation where water flows through a single large pipe and then splits into three smaller pipes. We need to find the total flow rate and compare the speeds of the water in different pipes.
02

Find Total Flow Rate

The total flow rate in the large pipe is equal to the sum of the flow rates from the smaller pipes. Add the flow rates: 26 L/min, 19 L/min, and 11 L/min. This gives: \(26 + 19 + 11 = 56\) L/min.
03

Use Continuity Equation for Speed Ratio

Using the continuity equation, which states that the flow rate (Q) is equal to the product of the cross-sectional area (A) and the velocity (v) of the fluid: \(Q = A \cdot v\). For the large pipe, use \(Q_1 = A_1 \cdot v_1\) and for the smaller pipe with 26 L/min, use \(Q_2 = A_2 \cdot v_2\).
04

Calculate Cross-sectional Area for Both Pipes

First, convert diameters to meters: 1.9 cm = 0.019 m and 1.3 cm = 0.013 m. Then, calculate the area using \(A = \pi (r^2)\): - For the large pipe, \(r_1 = 0.019/2\); \(A_1 = \pi (0.0095)^2\).- For the smaller pipe, \(r_2 = 0.013/2\); \(A_2 = \pi (0.0065)^2\).
05

Calculate Velocities for Both Pipes

Using \(Q = A \cdot v\), solve for velocity: - For large pipe, \(v_1 = \frac{Q_1}{A_1}\).- For smaller pipe carrying 26 L/min, \(v_2 = \frac{Q_2}{A_2}\). Convert flow rates to \(\mathrm{m}^3/\mathrm{s}\) by multiplying by \(\frac{1}{1000 \times 60}\). Use these values to find \(v_1\) and \(v_2\).
06

Calculate Speed Ratio

The speed ratio \(r\) of the large pipe to the smaller pipe is given by \(r = \frac{v_1}{v_2}\). Use the velocities found in Step 5 to compute the ratio.

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Key Concepts

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

Continuity equation
In fluid dynamics, the continuity equation is a fundamental principle that asserts the conservation of mass within a fluid flow system. This is based on the idea that a fluid's mass remains constant as it moves through a pipe or any channel. Mathematically, this is expressed as: \[ Q = A \cdot v \] where \( Q \) is the flow rate, \( A \) is the cross-sectional area, and \( v \) is the fluid velocity. Essentially, this equation states that the product of the area and velocity remains constant as long as no fluid enters or leaves the system. When water flows from one large pipe into smaller pipes, using the continuity equation helps ensure that the total amount of water remains the same, even if the speed or cross-sectional area changes. In the exercise, we see how this principle helps in evaluating flow and speed between different pipe sizes.
Flow rate
Flow rate is the measure of how much fluid passes through a pipe over a given period of time. It's an essential concept in fluid dynamics because it helps determine how fast a process can occur. Flow rate can be measured in various units, such as liters per minute (L/min) or cubic meters per second (m\(^3\)/s). In our exercise, the flow rates in three smaller pipes are given as 26 L/min, 19 L/min, and 11 L/min. To find the flow rate in a larger pipe, we simply add these values:
  • First small pipe: 26 L/min
  • Second small pipe: 19 L/min
  • Third small pipe: 11 L/min
Summing them gives the total flow rate of 56 L/min for the large pipe as all water from smaller pipes combines back into it. This total flow rate is crucial to balance, as it ensures mass conservation.
Velocity calculation
Calculating the velocity of fluid within a pipe involves knowing the flow rate and the cross-sectional area, thanks to the continuity equation, \( v = \frac{Q}{A} \). To find out the speed or velocity of water inside the pipes, we must first calculate their cross-sectional areas. For a pipe with a known flow rate and diameter, convert the flow to cubic meters per second by dividing Liters per minute by 1000 (for Liters to cubic meters) and by 60 (for minutes to seconds). Once you have the flow rate in m\(^3\)/s, divide it by the cross-sectional area, which results from previously calculated values.In the case of the large pipe and one of the smaller pipes carrying 26 L/min, this enables us to compare the velocities, understanding the inherent differences due to varying diameters.
Cross-sectional area calculation
The cross-sectional area of a pipe is vital for determining the flow characteristics of a fluid, such as its velocity. It is usually calculated using the diameter of the pipe. The formula for the cross-sectional area \( A \) is given by:\[ A = \pi r^2 \] where \( r \) is the radius of the pipe. For instance, if a pipe has a diameter of 1.9 cm (which is 0.019 meters), the radius would be half of that, equating to 0.0095 meters. Substituting this into the equation gives the cross-sectional area.Similarly, for a smaller pipe with a diameter of 1.3 cm (or 0.013 meters), the radius would be 0.0065 meters. Calculating for both pipes allows us to understand how differences in area affect the fluid velocity and flow rate. This understanding is pivotal when comparing the speed of the fluid across different portions of piping in the exercise.

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Most popular questions from this chapter

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?

A flotation device is in the shape of a right cylinder, with a height of \(0.500 \mathrm{~m}\) and a face area of \(4.00 \mathrm{~m}^{2}\) on top and bottom, and its density is \(0.400\) times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

A large aquarium of height \(5.00\) \(\mathrm{m}\) is filled with fresh water to a depth of \(2.00\) \(\mathrm{m}\). One wall of the aquarium consists of thick plastic \(8.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m}\) ?

A piston of cross-sectional area \(a\) is used in a hydraulic press to exert a small force of magnitude \(f\) on the enclosed liquid. A connecting pipe leads to a larger piston of cross-sectional area \(A\) (Fig. 14-36). (a) What force magnitude \(F\) will the larger piston sustain without moving? (b) If the piston diameters are \(3.80 \mathrm{~cm}\) and \(53.0 \mathrm{~cm}\), what force magnitude on the small piston will balance a \(20.0 \mathrm{kN}\) force on the large piston?

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 \(1.0 \mathrm{~atm} ?\)
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