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In a 30-in diameter pipe, a circular orifice has a diameter of 20 in, and the difference in the height of the manometer levels is 2.3 ft. What is the flow rate in cubic feet per second if K is 0.97?
- a)8.24 ft3/sec
- b)16.48 ft3/sec
- c)4.12 ft3/sec
- d)1.6 ft3/sec
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a 30-in diameter pipe, a circular orifice has a diameter of 20 in, ...
The flow rate Q in a differential flow rate meter is given by
Q = K(π/4)(ds/dp)2√2gh
where K is the flow coefficient constant, dS is the diameter of the orifice, dP is the
pipe diameter and h is the difference in height between PH and PL.
Q= 0.97 (3.14/4)(20/30)2 
Q= 4.12 ft3/sec
Most Upvoted Answer
In a 30-in diameter pipe, a circular orifice has a diameter of 20 in, ...
Given data:
Diameter of the pipe (D) = 30 inches
Diameter of the orifice (d) = 20 inches
Difference in the height of the manometer levels (h) = 2.3 ft
Coefficient of discharge (K) = 0.97
To determine the flow rate, we can use the Bernoulli's equation for incompressible fluid flow, which states that the total energy of the fluid remains constant along a streamline.
1. Calculating the velocity of the fluid:
Using the equation of continuity, we can relate the velocities of the fluid at the pipe and the orifice.
- The cross-sectional area of the pipe (A1) = π(D/2)^2
- The cross-sectional area of the orifice (A2) = π(d/2)^2
- The velocity of the fluid at the pipe (v1) = ?
- The velocity of the fluid at the orifice (v2) = ?
According to the equation of continuity: A1v1 = A2v2
Substituting the values: π(D/2)^2v1 = π(d/2)^2v2
Simplifying: (D/2)^2v1 = (d/2)^2v2
Plugging in the values: (15^2)v1 = (10^2)v2
Simplifying further: v1 = 4/9v2
2. Applying Bernoulli's equation:
The Bernoulli's equation can be written as: P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
- P1 = Pressure at the pipe
- P2 = Pressure at the orifice
- ρ = Density of the fluid
- g = Acceleration due to gravity
- h1 = Height of the manometer fluid column on one side
- h2 = Height of the manometer fluid column on the other side
Since the pressure at both the pipe and the orifice is atmospheric, the pressure terms cancel out.
Simplifying further: (1/2)ρv1^2 + ρgh1 = (1/2)ρv2^2 + ρgh2
Dividing both sides by ρg: (1/2)v1^2 + h1 = (1/2)v2^2 + h2
3. Calculating the flow rate:
The flow rate (Q) can be determined using the equation: Q = A2v2
- A2 = π(d/2)^2
- v2 = (2gh2 - 2gh1)^0.5
Substituting the values and simplifying:
Q = π(d/2)^2(2gh2 - 2gh1)^0.5
4. Plugging in the values:
Q = π(20/2)^2(2(32.2)h)^0.5
Q = π(10)^2(2(32.2)(2.3))^0.5
Q ≈ 4.12 ft^3/sec
Therefore, the flow rate in cubic feet per second is approximately 4.12 ft^3/sec, which corresponds to option C.
Diameter of the pipe (D) = 30 inches
Diameter of the orifice (d) = 20 inches
Difference in the height of the manometer levels (h) = 2.3 ft
Coefficient of discharge (K) = 0.97
To determine the flow rate, we can use the Bernoulli's equation for incompressible fluid flow, which states that the total energy of the fluid remains constant along a streamline.
1. Calculating the velocity of the fluid:
Using the equation of continuity, we can relate the velocities of the fluid at the pipe and the orifice.
- The cross-sectional area of the pipe (A1) = π(D/2)^2
- The cross-sectional area of the orifice (A2) = π(d/2)^2
- The velocity of the fluid at the pipe (v1) = ?
- The velocity of the fluid at the orifice (v2) = ?
According to the equation of continuity: A1v1 = A2v2
Substituting the values: π(D/2)^2v1 = π(d/2)^2v2
Simplifying: (D/2)^2v1 = (d/2)^2v2
Plugging in the values: (15^2)v1 = (10^2)v2
Simplifying further: v1 = 4/9v2
2. Applying Bernoulli's equation:
The Bernoulli's equation can be written as: P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
- P1 = Pressure at the pipe
- P2 = Pressure at the orifice
- ρ = Density of the fluid
- g = Acceleration due to gravity
- h1 = Height of the manometer fluid column on one side
- h2 = Height of the manometer fluid column on the other side
Since the pressure at both the pipe and the orifice is atmospheric, the pressure terms cancel out.
Simplifying further: (1/2)ρv1^2 + ρgh1 = (1/2)ρv2^2 + ρgh2
Dividing both sides by ρg: (1/2)v1^2 + h1 = (1/2)v2^2 + h2
3. Calculating the flow rate:
The flow rate (Q) can be determined using the equation: Q = A2v2
- A2 = π(d/2)^2
- v2 = (2gh2 - 2gh1)^0.5
Substituting the values and simplifying:
Q = π(d/2)^2(2gh2 - 2gh1)^0.5
4. Plugging in the values:
Q = π(20/2)^2(2(32.2)h)^0.5
Q = π(10)^2(2(32.2)(2.3))^0.5
Q ≈ 4.12 ft^3/sec
Therefore, the flow rate in cubic feet per second is approximately 4.12 ft^3/sec, which corresponds to option C.
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In a 30-in diameter pipe, a circular orifice has a diameter of 20 in, and the difference in the height of the manometer levels is 2.3 ft. What is the flow rate in cubic feet per second if K is 0.97?a)8.24 ft3/secb)16.48 ft3/secc)4.12 ft3/secd)1.6 ft3/secCorrect answer is option 'C'. Can you explain this answer?
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