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A piping system consists of three pipes arranged in series; the lengths of the pipes are 1200 m, 750 m and 600 m and diameters 750 mm, 600 mm and 450 mm respectively.
Transform the system to an equivalent 450 mm diameter pipe
- a)671.3 m
- b)771.3 m
- c)871.3 m
- d)971.3 m
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A piping system consists of three pipes arranged in series; the lengt...
L1 = 1200 m, l2 = 750 m, l3 = 600 m
View all questions of this test
d1 = 750 mm d2 = 600 mm d3 = 450 mm
de = 450 mm
For pipes in series
le = 871 m
Most Upvoted Answer
A piping system consists of three pipes arranged in series; the lengt...
To transform the system to an equivalent 450 mm diameter pipe, we need to calculate the equivalent length of the system.
Formula:
Equivalent length (L) = (fL/D) x (V^2/2g)
where,
f = friction factor
L = length of pipe
D = diameter of pipe
V = velocity of fluid
g = acceleration due to gravity
Steps to solve the problem:
1. Calculate the velocity of fluid in each pipe using the formula:
V = Q/A
where,
Q = flow rate of fluid
A = cross-sectional area of pipe
Assuming that the flow rate is constant throughout the system, we can calculate the velocity in each pipe as follows:
V1 = Q/(π/4 x 0.75^2) = 1.333Q
V2 = Q/(π/4 x 0.6^2) = 2.778Q
V3 = Q/(π/4 x 0.45^2) = 6.222Q
2. Calculate the friction factor (f) for each pipe using the Colebrook-White equation:
1/sqrt(f) = -2.0log((e/D)/3.7 + 2.51/(Re sqrt(f)))
where,
e = roughness height of pipe
Re = Reynolds number
Assuming a roughness height of 0.03 mm, we can calculate the Reynolds number in each pipe as follows:
Re1 = (ρV1D1)/μ = (1000 x 1.333Q x 0.75)/0.001 = 1.777 x 10^6 Q
Re2 = (ρV2D2)/μ = (1000 x 2.778Q x 0.6)/0.001 = 1.667 x 10^6 Q
Re3 = (ρV3D3)/μ = (1000 x 6.222Q x 0.45)/0.001 = 1.111 x 10^6 Q
Using the Reynolds number and the Colebrook-White equation, we can solve for the friction factor in each pipe.
f1 = 0.0111
f2 = 0.0102
f3 = 0.0092
3. Calculate the equivalent length of each pipe using the formula:
fL/D = 4fL/πD^2
L1 = (4f1 x 1200)/((π/4) x 0.75^2) = 1294.118 m
L2 = (4f2 x 750)/((π/4) x 0.6^2) = 821.486 m
L3 = (4f3 x 600)/((π/4) x 0.45^2) = 545.987 m
4. Calculate the total equivalent length of the system by summing up the equivalent lengths of each pipe:
Leq = L1 + L2 + L3 = 2661.591 m
5. Finally, we can calculate the length of an equivalent 450 mm diameter pipe using the formula:
Leq = (fL/D) x (V^2/2g)
Rearranging the formula, we get:
L = (Leq x (D/Deq)^5)/
Formula:
Equivalent length (L) = (fL/D) x (V^2/2g)
where,
f = friction factor
L = length of pipe
D = diameter of pipe
V = velocity of fluid
g = acceleration due to gravity
Steps to solve the problem:
1. Calculate the velocity of fluid in each pipe using the formula:
V = Q/A
where,
Q = flow rate of fluid
A = cross-sectional area of pipe
Assuming that the flow rate is constant throughout the system, we can calculate the velocity in each pipe as follows:
V1 = Q/(π/4 x 0.75^2) = 1.333Q
V2 = Q/(π/4 x 0.6^2) = 2.778Q
V3 = Q/(π/4 x 0.45^2) = 6.222Q
2. Calculate the friction factor (f) for each pipe using the Colebrook-White equation:
1/sqrt(f) = -2.0log((e/D)/3.7 + 2.51/(Re sqrt(f)))
where,
e = roughness height of pipe
Re = Reynolds number
Assuming a roughness height of 0.03 mm, we can calculate the Reynolds number in each pipe as follows:
Re1 = (ρV1D1)/μ = (1000 x 1.333Q x 0.75)/0.001 = 1.777 x 10^6 Q
Re2 = (ρV2D2)/μ = (1000 x 2.778Q x 0.6)/0.001 = 1.667 x 10^6 Q
Re3 = (ρV3D3)/μ = (1000 x 6.222Q x 0.45)/0.001 = 1.111 x 10^6 Q
Using the Reynolds number and the Colebrook-White equation, we can solve for the friction factor in each pipe.
f1 = 0.0111
f2 = 0.0102
f3 = 0.0092
3. Calculate the equivalent length of each pipe using the formula:
fL/D = 4fL/πD^2
L1 = (4f1 x 1200)/((π/4) x 0.75^2) = 1294.118 m
L2 = (4f2 x 750)/((π/4) x 0.6^2) = 821.486 m
L3 = (4f3 x 600)/((π/4) x 0.45^2) = 545.987 m
4. Calculate the total equivalent length of the system by summing up the equivalent lengths of each pipe:
Leq = L1 + L2 + L3 = 2661.591 m
5. Finally, we can calculate the length of an equivalent 450 mm diameter pipe using the formula:
Leq = (fL/D) x (V^2/2g)
Rearranging the formula, we get:
L = (Leq x (D/Deq)^5)/
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A piping system consists of three pipes arranged in series; the lengths of the pipes are 1200 m, 750 m and 600 m and diameters 750 mm, 600 mm and 450 mm respectively.Transform the system to an equivalent 450 mm diameter pipea) 671.3 mb) 771.3 mc) 871.3 md) 971.3 mCorrect answer is option 'C'. Can you explain this answer?
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