FLUID MECHANICS-PHYSICAL ACCESS CODE
FLUID MECHANICS-PHYSICAL ACCESS CODE
8th Edition
ISBN: 9781264005086
Author: White
Publisher: MCG
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Question
Chapter 5, Problem 5.1P
To determine

The volume flow rate in m3 /h which causes the transition.

Expert Solution & Answer
Check Mark

Answer to Problem 5.1P

The volume flow rate is  0.778m3/h.

Explanation of Solution

Given:

Diameter of the circular tube, d=9cm=0.09m

Reynolds number Re=2300.

Fluid: Kerosene

State: 20°C

Mass density of kerosene ρ=804kg/m3

Concept Used:

Reynolds number is given by,

Re=ρvdμ

Where, ρ is the mass density of the fluid in kg/m3.

v is the velocity of flow in m/s.

d is the diameter of the circular tube in m.

μ is the dynamic viscosity of the fluid in kg/m-s.

Calculation:

From the given data, we know that we can use the Reynolds number to find the velocity of flow from which we can find the volume flow rate.

For kerosene at 20°C, ρ=804kg/m3 and μ=0.00192 kg/m-s

Reynolds number is given by,

Re=ρvdμ

Substituting the known values in the Reynolds number formula, we get

2300=804v0.050.00192

v=0.11m/s

Now, we substitute the value of v=0.11m/s in the volume flow rate equation.

Q=Av=πd24v

π(0.05)24x0.11

=2.1598 x 104 m3/s

Q0.778 m3/h

Thus, the volume flow rate is 0.778 m3/h

Conclusion:

The volume flow rate is 0.778 m3/h.

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Chapter 5 Solutions

FLUID MECHANICS-PHYSICAL ACCESS CODE

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