Problem 11
Question
A pump raises water from a depth of \(10 \mathrm{~m}\) and discharges it horizontally through a pipe of \(0.1 \mathrm{~m}\) diameter at a velocity of \(8 \mathrm{~ms}^{-1} .\) Calculate the work done by the pump in one second. If the water impinges directly with the same velocity on a vertical wall, find the force exerted by the water on the wall if it is assumed that none of the water bounces back (Take \(g\) as \(9.81 \mathrm{~ms}^{-2}, \pi\) as \(3.142\) and the mass of \(1 \mathrm{~m}^{3}\) of water as \(1000 \mathrm{~kg}\) ).
Step-by-Step Solution
Verified Answer
The work done by the pump is 81.74 W. The force exerted by the water on the wall is 5.03 N.
1Step 1: Find the volume flow rate
The volume flow rate (Q) can be calculated using the velocity (v) of the water and the cross-sectional area (A) of the pipe. First, compute the cross-sectional area of the pipe: \[ A = \pi \left( \frac{d}{2} \right)^2 \] Given: \[ d = 0.1 \, \mathrm{m} \] \[ A = 3.142 \left( \frac{0.1}{2} \right)^2 = 3.142 \times 0.005^2 = 3.142 \times 0.000025 = 0.00007855 \, \mathrm{m^2} \] Then, use the area to find the flow rate: \[ Q = A \times v = 0.00007855 \, \mathrm{m^2} \times 8 \, \mathrm{m/s} = 0.0006284 \, \mathrm{m^3/s} \]
2Step 2: Calculate the mass flow rate
The mass flow rate is found by multiplying the volume flow rate by the density of water: \[ \dot{m} = Q \times \rho \] Given: \[ \rho = 1000 \, \mathrm{kg/m^3} \] \[ \dot{m} = 0.0006284 \, \mathrm{m^3/s} \times 1000 \, \mathrm{kg/m^3} = 0.6284 \, \mathrm{kg/s} \]
3Step 3: Determine the height-related work done
The work done to raise the water is given by the formula \( W_h = mgh \): \[ W_h = \dot{m} \times g \times h \] Given: \[ g = 9.81 \, \mathrm{m/s^2}, \; h = 10 \, \mathrm{m} \] \[ W_h = 0.6284 \, \mathrm{kg/s} \times 9.81 \, \mathrm{m/s^2} \times 10 \, \mathrm{m} = 61.63 \, \mathrm{W} \]
4Step 4: Determine the kinetic energy work done by the pump
The kinetic energy work done per second by the pump is: \[ W_k = \frac{1}{2} \times \dot{m} \times v^2 \] \[ W_k = \frac{1}{2} \times 0.6284 \, \mathrm{kg/s} \times (8 \, \mathrm{m/s})^2 = \frac{1}{2} \times 0.6284 \times 64 = 20.11 \, \mathrm{W} \]
5Step 5: Calculate the total work done by the pump
The total work done by the pump per second is the sum of the height-related work and the kinetic energy work: \[ W_{\text{total}} = W_h + W_k = 61.63 \, \mathrm{W} + 20.11 \, \mathrm{W} = 81.74 \, \mathrm{W} \]
6Step 6: Calculate the force exerted by the water on the wall
The force exerted by the water on the wall can be found using the formula: \[ F = \dot{m} \times v \] \[ F = 0.6284 \, \mathrm{kg/s} \times 8 \, \mathrm{m/s} = 5.03 \, \mathrm{N} \]
Key Concepts
Volume Flow RateMass Flow RateWork DoneKinetic EnergyForce Exerted by Fluid
Volume Flow Rate
The volume flow rate is a measure of the volume of fluid passing through a cross-sectional area of a pipe per unit of time. It is denoted by the symbol \( Q \). In this problem, the volume flow rate can be calculated using the formula:
\[ Q = A \times v \]
where \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the fluid.
By finding the area of the pipe first, you can determine the volume flow rate. The calculation starts with the diameter of the pipe:
\[ d = 0.1 \text{ meters} \]
The cross-sectional area \( A \) can be found using the formula:
\[ A = \frac{\text{π} \times d^2}{4} \]
Substituting the diameter:
\[ A = 3.142 \times \frac{0.1^2}{4} = 0.007855 \text{ square meters} \]
Now, using the velocity \( v = 8 \text{ m/s} \), the volume flow rate becomes:
\[ Q = 0.007855 \text{ m}^2 \times 8 \text{ m/s} = 0.06284 \text{ m}^3/s \]
This result tells us the volume of water being moved per second.
\[ Q = A \times v \]
where \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the fluid.
By finding the area of the pipe first, you can determine the volume flow rate. The calculation starts with the diameter of the pipe:
\[ d = 0.1 \text{ meters} \]
The cross-sectional area \( A \) can be found using the formula:
\[ A = \frac{\text{π} \times d^2}{4} \]
Substituting the diameter:
\[ A = 3.142 \times \frac{0.1^2}{4} = 0.007855 \text{ square meters} \]
Now, using the velocity \( v = 8 \text{ m/s} \), the volume flow rate becomes:
\[ Q = 0.007855 \text{ m}^2 \times 8 \text{ m/s} = 0.06284 \text{ m}^3/s \]
This result tells us the volume of water being moved per second.
Mass Flow Rate
The mass flow rate represents the mass of fluid passing through a cross-sectional area per unit of time. To calculate it, you use the volume flow rate and the density of the fluid. The formula for the mass flow rate \( \dot{m} \) is:
\[ \dot{m} = Q \times \rho \]
Where \( \rho \) is the density of the fluid (for water, \( \rho = 1000 \text{ kg/m}^3 \)) and \( Q \) is the volume flow rate.
Given the volume flow rate calculated earlier (\( Q = 0.06284 \text{ m}^3/s \)):
\[ \dot{m} = 0.06284 \text{ m}^3/s \times 1000 \text{ kg/m}^3 = 62.84 \text{ kg/s} \]
This means that 62.84 kilograms of water flow through the pipe each second.
\[ \dot{m} = Q \times \rho \]
Where \( \rho \) is the density of the fluid (for water, \( \rho = 1000 \text{ kg/m}^3 \)) and \( Q \) is the volume flow rate.
Given the volume flow rate calculated earlier (\( Q = 0.06284 \text{ m}^3/s \)):
\[ \dot{m} = 0.06284 \text{ m}^3/s \times 1000 \text{ kg/m}^3 = 62.84 \text{ kg/s} \]
This means that 62.84 kilograms of water flow through the pipe each second.
Work Done
Work done in a physical sense refers to the energy transferred to move an object against a force. For a pump raising water, it includes both lifting the water height-wise and increasing its speed.
To compute the work done against gravity (height-related work), use:
\[ W_h = \dot{m} \times g \times h \]
Where \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)), \( h = 10 \text{ m} \), and \( \dot{m} = 62.84 \text{ kg/s} \):
\[ W_h = 62.84 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 10 \text{ m} = 6,163 \text{ W} \]
Next, calculate the kinetic energy work done by the pump as:
\[ W_k = \frac{1}{2} \times \dot{m} \times v^2 \]
Given \( v = 8 \text{ m/s} \):
\[ W_k = \frac{1}{2} \times 62.84 \text{ kg/s} \times (8 \text{ m/s})^2 = 2,011 \text{ W} \]
Summing these, the total work done by the pump per second is:
\[ W_{\text{total}} = W_h + W_k = 6,163 \text{ W} + 2,011 \text{ W} = 8,174 \text{ W} \]
To compute the work done against gravity (height-related work), use:
\[ W_h = \dot{m} \times g \times h \]
Where \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)), \( h = 10 \text{ m} \), and \( \dot{m} = 62.84 \text{ kg/s} \):
\[ W_h = 62.84 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 10 \text{ m} = 6,163 \text{ W} \]
Next, calculate the kinetic energy work done by the pump as:
\[ W_k = \frac{1}{2} \times \dot{m} \times v^2 \]
Given \( v = 8 \text{ m/s} \):
\[ W_k = \frac{1}{2} \times 62.84 \text{ kg/s} \times (8 \text{ m/s})^2 = 2,011 \text{ W} \]
Summing these, the total work done by the pump per second is:
\[ W_{\text{total}} = W_h + W_k = 6,163 \text{ W} + 2,011 \text{ W} = 8,174 \text{ W} \]
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For a fluid in motion, it is determined per unit time using the mass flow rate. The formula to calculate the kinetic energy work is:
\[ W_k = \frac{1}{2} \times \dot{m} \times v^2 \]
Given the mass flow rate \( \dot{m} = 62.84 \text{ kg/s} \) and velocity \( v = 8 \text{ m/s} \):
\[ W_k = \frac{1}{2} \times 62.84 \text{ kg/s} \times (8 \times \text{ m/s})^2 = 2,011 \text{ W} \]
This tells us that the pump exerts 2,011 watts of energy to keep the water flowing at the required velocity.
\[ W_k = \frac{1}{2} \times \dot{m} \times v^2 \]
Given the mass flow rate \( \dot{m} = 62.84 \text{ kg/s} \) and velocity \( v = 8 \text{ m/s} \):
\[ W_k = \frac{1}{2} \times 62.84 \text{ kg/s} \times (8 \times \text{ m/s})^2 = 2,011 \text{ W} \]
This tells us that the pump exerts 2,011 watts of energy to keep the water flowing at the required velocity.
Force Exerted by Fluid
The fluid exerts force on the vertical wall when it impacts directly. This force is determined using the mass flow rate and velocity:
\[ F = \dot{m} \times v \]
Given \( \dot{m} = 62.84 \text{ kg/s} \) and \( v = 8 \text{ m/s} \):
\[ F = 62.84 \text{ kg/s} \times 8 \text{ m/s} = 503 \text{ N} \]
This means that the water exerts 503 newtons of force on the wall, demonstrating the significant impact of the fluid's momentum when it comes into contact with a solid barrier.
\[ F = \dot{m} \times v \]
Given \( \dot{m} = 62.84 \text{ kg/s} \) and \( v = 8 \text{ m/s} \):
\[ F = 62.84 \text{ kg/s} \times 8 \text{ m/s} = 503 \text{ N} \]
This means that the water exerts 503 newtons of force on the wall, demonstrating the significant impact of the fluid's momentum when it comes into contact with a solid barrier.
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