Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 10 mm. Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.
Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 10 mm. Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.
Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 10 mm. Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.
a)
Expert Solution
To determine
The general expression for Reynolds number in terms of volume flow rate and tube diameter.
Explanation of Solution
Given:
Tube diameter 1(D1) is 10mm.
Tube diameter 2(D2) is 6mm.
Reynolds number 1(Re1) is 1800.
Calculation:
Write the equation for the volume flow rate (Q).
Q=AV¯
Write the equation for the mass flow rate (m˙).
m˙=ρAV¯
Write the equation for the cross sectional area (A).
A=πD24
Calculate the Reynolds number in terms of volume flow rate and tube diameter (Re).
Re=ρDV¯μ=ρDμ(QA)=ρDμ(QπD24)=4QπDρμ
=4QπD(1V¯)=4QπDV¯
Thus, the Reynolds number in terms of volume flow rate and tube diameter is 4QπDV¯.
b)
Expert Solution
To determine
The general expression for Reynolds number in terms of mass flow rate and tube diameter and the Reynolds number for the same flow rate in the section.
Explanation of Solution
Calculate the Reynolds number in terms of mass flow rate and tube diameter (Re).
Re=ρDV¯μ=Dμ(ρV¯AA)=Dμ(m˙πD24)=4m˙πDμ
Thus, the Reynolds number in terms of mass flow rate and tube diameter is 4m˙πDμ.
Calculate the Reynolds number for the same flow rate in the section (Re2).
D1Re1=D2Re2
Re2=D1D2Re1=(10mm)(6mm)(1800)=3000
Thus, the Reynolds number for the same flow rate in the section is 3000.
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Chapter 8 Solutions
Fox and McDonald's Introduction to Fluid Mechanics
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