Question

4-10M: The kinematic viscosity of a hydraulic oil is 0.0001 m^2/s. If it is flowing in a 30 mm inside diameter pipe at a velocity of 6 m/s, what is the Reynolds number? Is the flow laminar, turbulent, or in transition? 4-14M: For the system in Exercise 4-10M, find the head loss due to friction in units of m (and equivalent pressure drop in bar) for a 100 m length of smooth pipe. The oil has a specific gravity of 0.90. What is the fluid resistance of the pipe in problems 4.10M and 4.14M?

          4-10M: The kinematic viscosity of a hydraulic oil is 0.0001 m^2/s. If it is flowing in a 30 mm inside diameter pipe at a velocity of 6 m/s, what is the Reynolds number? Is the flow laminar, turbulent, or in transition?
4-14M: For the system in Exercise 4-10M, find the head loss due to friction in units of m (and equivalent pressure drop in bar) for a 100 m length of smooth pipe. The oil has a specific gravity of 0.90. What is the fluid resistance of the pipe in problems 4.10M and 4.14M?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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4-10M: The kinematic viscosity of a hydraulic oil is 0.0001 m^2/s. If it is flowing in a 30 mm inside diameter pipe at a velocity of 6 m/s, what is the Reynolds number? Is the flow laminar, turbulent, or in transition? 4-14M: For the system in Exercise 4-10M, find the head loss due to friction in units of m (and equivalent pressure drop in bar) for a 100 m length of smooth pipe. The oil has a specific gravity of 0.90. What is the fluid resistance of the pipe in problems 4.10M and 4.14M?
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Transcript

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00:01 So, velocity of the flow that is v this is equal to q by a this is 120 into 10 raised to the power minus 3 by 60 divided by pi by 4 into 0 .03 to the whole square.
00:18 So, we get v as 2 .83 meter per second.
00:25 Now, the reynolds number that is re this is equal to 2 .83 into 0 .03 divided by 0 .0001.
00:42 So, we get re as 849.
00:46 Now the reynolds number is less than 2000.
00:49 So, friction factor that is f this is equal to 64 by re that is 849.
00:59 So, friction factor is 0 .075.
01:03 Now k factor for fittings involved 4 elbows k is equal to 0 .75 that is 3, 1 globe wall that is k 10 that is 10 by 13.
01:15 So, we say equivalent length for fittings that is lf this is equal to kd by f that is 13 into 0 .03 divided by 0 .075.
01:30 So, we get 5 .2 m and le that is the total equivalent length this is 20 plus 5 .2 that is pipeline length is 20 and 5 .2 is lf.
01:44 So, we get 25 .2 m...
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