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In a 0.18 m, 250 m long horizontal pipe a lubricating oil of specific gravity 0.9 and dynamic viscosity 0.15 Pa-s is pumped at the rate of 0.04 m3/s. The power required to maintain the flow (in KW) is
  • a)
    2.25
  • b)
    2.4
Correct answer is between '2.25,2.4'. Can you explain this answer?
Verified Answer
In a 0.18 m, 250 m long horizontal pipe a lubricating oil of specific...
Given,
Length of pipe, L = 250 m
Diameter of pipe, D = 0.18 m
Dynamic viscosity, μ = 0.15 Pa-s
Flow rate, Q = 0.04 m3/s
The pressure drop in pipe is given by
Average velocity,
The power required to maintain the flow,
P = Q× ΔP = 0.04×58148.15 = 2325.926 W = 2.326 KW
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Most Upvoted Answer
In a 0.18 m, 250 m long horizontal pipe a lubricating oil of specific...
Given Data:
- Length of pipe (L): 250 m
- Diameter of pipe (d): 0.18 m
- Specific gravity of lubricating oil (ρ): 0.9
- Dynamic viscosity of oil (μ): 0.15 Pa-s
- Flow rate (Q): 0.04 m3/s

Calculations:
1. Calculate the cross-sectional area of the pipe:
- Area (A) = π(d/2)2
= π(0.18/2)2
= 0.02545 m2
2. Calculate the velocity of oil in the pipe:
- Velocity (V) = Q/A
= 0.04 / 0.02545
= 1.571 m/s
3. Calculate the Reynolds number:
- Reynolds number (Re) = (ρ * V * d) / μ
= (0.9 * 1.571 * 0.18) / 0.15
= 1.899
4. Calculate the friction factor (f) using the Colebrook equation or Moody chart corresponding to Re and pipe roughness.
5. Calculate the head loss (hL) using the Darcy-Weisbach equation:
- Head loss (hL) = f * (L/d) * (V^2 / 2g)
6. Calculate the power required to maintain the flow:
- Power = ρ * Q * g * hL / 1000 (to convert W to KW)

Result:
The power required to maintain the flow is between 2.25 KW and 2.4 KW.
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In a 0.18 m, 250 m long horizontal pipe a lubricating oil of specific gravity 0.9 and dynamic viscosity 0.15 Pa-s is pumped at the rate of 0.04 m3/s. The power required to maintain the flow (in KW) isa)2.25b)2.4Correct answer is between '2.25,2.4'. Can you explain this answer?
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