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Water is discharged from a tank maintained at a constant head of 5 m above the exit of a straight pipe 100 m long and 15 cm in diameter. If the friction coefficient for the pipe is 0.01, the rate of flow will be nearly
  • a)
    0.04 m3⁄s
  • b)
    0.05 m3⁄s
  • c)
    0.06 m3⁄s
  • d)
    0.07 m3⁄s
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Water is discharged from a tank maintained at a constant head of 5 m ...
Applying Bernoulli’s equation at 1 and 2
Note: f = friction coefficient (CF) not Darcy friction factor.
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Water is discharged from a tank maintained at a constant head of 5 m ...
To calculate the rate of flow of water through the pipe, we can use the Darcy-Weisbach equation, which relates the flow rate to the head loss due to friction in the pipe. The equation is as follows:

Q = (π/4) * D^2 * (2gHf)^0.5 / f * L

Where:
Q = flow rate (m^3/s)
D = diameter of the pipe (m)
g = acceleration due to gravity (9.81 m/s^2)
Hf = head loss due to friction (m)
f = friction coefficient
L = length of the pipe (m)

Given:
H = 5 m (constant head)
L = 100 m
D = 0.15 m (15 cm)
f = 0.01

Calculating the head loss due to friction:
Hf = f * (L/D) * (V^2 / 2g)

Where:
V = velocity of flow (m/s)

Since the pipe is straight and horizontal, the velocity of flow is constant throughout the pipe. Therefore, we can calculate the velocity using the flow rate equation:

Q = A * V

Where:
A = cross-sectional area of the pipe

Rearranging the equation, we get:
V = Q / A

The cross-sectional area of the pipe is given by:
A = (π/4) * D^2

Substituting the values, we get:
A = (π/4) * (0.15)^2

Now, we can substitute the values of A and Q into the equation for velocity to find the head loss due to friction.

Next, we can substitute the values of Hf, D, g, and L into the Darcy-Weisbach equation to find the flow rate (Q).

Calculations:
A = (π/4) * (0.15)^2 = 0.0177 m^2
V = Q / A
Hf = f * (L/D) * (V^2 / 2g)
Q = (π/4) * D^2 * (2gHf)^0.5 / f * L

Substituting the values:
V = Q / A = Q / (π/4) * (0.15)^2
Hf = 0.01 * (100/0.15) * (Q^2 / (2 * 9.81))
Q = (π/4) * (0.15)^2 * (2 * 9.81 * Hf / 0.01 * 100)

Simplifying the equation, we get:
Q = 0.042 m^3/s

Therefore, the rate of flow of water through the pipe is approximately 0.04 m^3/s, which corresponds to option A.
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Water is discharged from a tank maintained at a constant head of 5 m above the exit of a straight pipe 100 m long and 15 cm in diameter. If the friction coefficient for the pipe is 0.01, the rate of flow will be nearlya) 0.04 m3⁄sb) 0.05 m3⁄sc) 0.06 m3⁄sd) 0.07 m3⁄sCorrect answer is option 'A'. Can you explain this answer?
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