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Chapter 7: Problem 15

A pump delivers water through two pipes laid in parallel. One pipe is \(100 \mathrm{~mm}\) diameter and \(45 \mathrm{~m}\) long and discharges to atmosphere at a level \(6 \mathrm{~m}\) above the pump outlet. The other pipe, \(150 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long, discharges to atmosphere at a level \(8 \mathrm{~m}\) above the pump outlet. The two pipes are connected to a junction immediately adjacent to the pump and both have \(f=0.008 .\) The inlet to the pump is \(600 \mathrm{~mm}\) below the level of the outlet. Taking the datum level as that of the pump inlet, determine the total head at the pump outlet if the flow rate through it is \(0.037 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). Losses at the pipe junction may be neglected.

Short Answer

Expert verified
Total head at the pump outlet is 7.335 m.

Step by step solution

01

- Determine the velocity of the water in each pipe

To find the velocities in each pipe, use the volumetric flow rate equation: \( Q = A \times v \), where \( Q \) is the flow rate, \( A \) is the cross-sectional area of the pipe, and \( v \) is the velocity. First, calculate the cross-sectional area of each pipe:For the 100 mm diameter pipe:\[ A_1 = \pi \times \left( \frac{0.1}{2} \right)^2 = 7.85 \times 10^{-3} \text{ m}^2 \]For the 150 mm diameter pipe:\[ A_2 = \pi \times \left( \frac{0.15}{2} \right)^2 = 1.77 \times 10^{-2} \text{ m}^2 \]Total area = \( A_1 + A_2 \).
02

- Split the total flow rate into each pipe

Using the continuity equation, divide the total flow rate \( Q = 0.037 \text{ m}^3/s \): \( Q = Q_1 + Q_2 = v_1 \times A_1 + v_2 \times A_2 \)We know the total flow rate and the cross-sectional areas, set up an equation to solve for \( v_1 \) and \( v_2 \).
03

- Calculate the flow rates in each pipe

Using the ratio of the areas to split the flow rate proportionally:\( Q_1 = \frac{A_1}{A_1 + A_2} \times Q \)\( Q_1 = \frac{7.85 \times 10^{-3}}{7.85 \times 10^{-3} + 1.77 \times 10^{-2}} \times 0.037 = 0.010 \text{ m}^3/s \)\( Q_2 = 0.037 - Q_1 = 0.027 \text{ m}^3/s \)Now use \( Q_1 = v_1 \times A_1 \) and \( Q_2 = v_2 \times A_2 \) to find velocities.
04

- Calculate the velocity in each pipe

For the 100 mm diameter pipe:\( v_1 = \frac{Q_1}{A_1} = \frac{0.010}{7.85 \times 10^{-3}} = 1.27 \text{ m/s} \)For the 150 mm diameter pipe:\( v_2 = \frac{Q_2}{A_2} = \frac{0.027}{1.77 \times 10^{-2}} = 1.53 \text{ m/s} \)
05

- Calculate the head loss due to friction in each pipe

Use the Darcy-Weisbach equation:\[ h_f = f \times \frac{L}{D} \times \frac{v^2}{2g} \]For the 100 mm diameter pipe:\[ L_1 = 45 \text{ m}, D_1 = 0.1 \text{ m}, f = 0.008 \]\[ h_{f_1} = 0.008 \times \frac{45}{0.1} \times \frac{(1.27)^2}{2 \times 9.81} = 0.296 \text{ m} \]For the 150 mm diameter pipe:\[ L_2 = 60 \text{ m}, D_2 = 0.15 \text{ m}, f = 0.008 \]\[ h_{f_2} = 0.008 \times \frac{60}{0.15} \times \frac{(1.53)^2}{2 \times 9.81} = 0.375 \text{ m} \]
06

- Calculate the elevation head for each pipe

For the 100 mm diameter pipe:\( z_1 = 6 \text{ m} \)For the 150 mm diameter pipe:\( z_2 = 8 \text{ m} \)
07

- Determine the total head at the pump outlet

The total head at the pump outlet is the sum of the velocity head, friction head, and elevation head. Take the average of both pipes:\[ Total \, Head = z_{avg} + h_{f_{avg}} = \left( \frac{z_1 + z_2}{2} \right) + \left( \frac{h_{f_1} + h_{f_2}}{2} \right) \]\[ Total \, Head = \left( \frac{6 + 8}{2} \right) + \left( \frac{0.296 + 0.375}{2} \right) = 7 + 0.335 \]\[ Total \, Head = 7.335 \text{ m} \]

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Key Concepts

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

Darcy-Weisbach equation
The Darcy-Weisbach equation helps us determine the frictional head loss in a pipe due to the flow of fluid. This is essential in fluid mechanics to understand energy losses in systems like pipelines. The equation is given by: \[ h_f = f \times \frac{L}{D} \times \frac{v^2}{2g} \] Where:
  • h_f is the head loss due to friction (m)
  • f is the Darcy friction factor (dimensionless)
  • L is the length of the pipe (m)
  • D is the diameter of the pipe (m)
  • v is the velocity of the fluid (m/s)
  • g is the acceleration due to gravity (\(9.81 \text{ m/s}^2\))
In this exercise, we used the Darcy-Weisbach equation to calculate the head loss due to friction for each of the two pipes in the system. This is a critical step in determining the total head required at the pump outlet.
Volumetric flow rate
Volumetric flow rate is a measure of the volume of fluid passing through a cross-section of a pipe per unit time. It is denoted by \(Q\) and expressed in cubic meters per second (\(m^3/s\)). The relationship between volumetric flow rate, cross-sectional area \(A\), and velocity \(v\) is given by: \[ Q = A \times v \] In this exercise, we have a total flow rate of 0.037 \( \text{m}^3/\text{s} \). We used this together with the cross-sectional areas of the two pipes to find the individual flow rates in each pipe. The cross-sectional area \(A\) for a circular pipe is calculated as: \[ A = \frac{\text{蟺} D^2}{4} \] Where \(D\) is the diameter of the pipe. By splitting the total flow rate into each pipe using the areas, we determined how much water flows through each.
Continuity equation
The continuity equation is a fundamental principle in fluid mechanics, stating that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another, assuming incompressible flow. Mathematically, it expresses: \[ Q = Q_1 + Q_2 \] Where:
  • Q is the total flow rate
  • Q_1 and Q_2 are the flow rates in each individual pipe
In this problem, we used the continuity equation to ensure that the total flow rate of 0.037 \( \text{m}^3/\text{s} \) was split correctly between the two pipes based on their respective areas. This allows for the calculation of the individual velocities and subsequent head losses due to friction.
Head loss due to friction
Head loss due to friction is the loss of energy (or pressure) in a pipe due to the friction between the moving fluid and the pipe walls. This is an essential factor to consider in fluid mechanics systems because it affects the required pump energy. The Darcy-Weisbach equation helps calculate this as previously explained. In our exercise, the calculated head loss due to friction for the 100 mm diameter pipe was 0.296 m, and for the 150 mm diameter pipe, it was 0.375 m. Summarizing, understanding head loss due to friction enables us to determine the total head requirements at the pump outlet by combining these with elevation heads and other losses, if any, to ensure efficient fluid distribution.

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Most popular questions from this chapter

Calculate the power required to pump sulphuric acid (dynamic viscosity \(0.04 \mathrm{~Pa} \cdot \mathrm{s}\), relative density \(1.83\) ) at \(45 \mathrm{~L} \cdot \mathrm{s}^{-1}\) from a supply tank through a glass-lined \(150 \mathrm{~mm}\) diameter pipe, \(18 \mathrm{~m}\) long, into a storage tank. The liquid level in the storage tank is \(6 \mathrm{~m}\) above that in the supply tank. For laminar flow \(f=16 / \mathrm{Re}\); for turbulent flow \(f=0.0014\left(1+100 \mathrm{Re}^{-1 / 3}\right)\) if \(\mathrm{Re}<10^{7}\). Take all losses into account.

A pipe \(600 \mathrm{~mm}\) diameter and \(1 \mathrm{~km}\) long with \(f=0.008\) connects two reservoirs having a difference in water surface level of \(30 \mathrm{~m}\). Calculate the rate of flow between the reservoirs and the shear stress at the wall of the pipe. If the upstream half of the pipe is tapped by several side pipes so that one-third of the quantity of water now entering the main pipe is withdrawn uniformly over this length, calculate the new rate of discharge to the lower reservoir. Neglect all losses other than those due to pipe friction.

A water main with a constant gauge pressure of \(300 \mathrm{kPa}\) is to supply water through a pipe \(35 \mathrm{~m}\) long to a tank of uniform plan area \(6 \mathrm{~m}^{2}\), open to atmosphere at the top. The pipe is to enter the base of the tank at a level \(2.9 \mathrm{~m}\) above that of the main. The depth of water in the tank is to be increased from \(0.1 \mathrm{~m}\) to \(2.7 \mathrm{~m}\) in not more than 15 minutes. Assuming that \(f\) has the constant value \(0.007\), and neglecting energy losses other than pipe friction, determine the diameter of pipe required.

A hose pipe of \(75 \mathrm{~mm}\) bore and length \(450 \mathrm{~m}\) is supplied with water at \(1.4 \mathrm{MPa}\). A nozzle at the outlet end of the pipe is \(3 \mathrm{~m}\) above the level of the inlet end. If the jet from the nozzle is to reach a height of \(35 \mathrm{~m}\) calculate the maximum diameter of the nozzle assuming that \(f=0.01\) and that losses at inlet and in the nozzle are negligible. If the efficiency of the supply pump is \(70 \%\) determine the power required to drive it.

In a heat exchanger there are 200 tubes each \(3.65 \mathrm{~m}\) long and \(30 \mathrm{~mm}\) outside diameter and \(25 \mathrm{~mm}\) bore. They are arranged axially in a cylinder of \(750 \mathrm{~mm}\) diameter and are equally spaced from one another. A liquid of relative density \(0.9\) flows at a mean velocity of \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) through the tubes and water flows at \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) between the tubes in the opposite direction. For all surfaces \(f\) may be taken as \(0.01\). Neglecting entry and exit losses, calculate (a) the total power required to overcome fluid friction in the exchanger and \((b)\) the saving in power if the two liquids exchanged places but the system remained otherwise unaltered.
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