Venturi Inserts
Posted on | April 8, 2009 | Comments Off
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Venturi Inserts
The venturi flowmeter shown below is used to measure the flow rate of water in a solar collector system.?
The venturi flowmeter shown below is used to measure the flow rate of water in a solar collector system. The flowmeter is inserted in a pipe with diameter 1.9 cm; at the venturi of the flowmeter the diameter is reduced to 0.65 cm. The manometer tube contains oil with density 0.76 times that of water. If the difference in oil levels on the two sides of the manometer tube is 1.4 cm, what is the volume flow rate?
This is, of course, a Bernoulli's principle problem.
Bernoulli's equation states that P + ρv^2/2 + ρgh remains constant.
P is indicated by manometer height: P = ρgh; thus if we lump P into ρgh the equation becomes ρv^2/2 + ρgh = constant, or
ρv1^2/2 + ρgh1 = ρv2^2/2 + ρgh2 (1)
We solve for v1 to find the volume flow rate.
From (1), v1^2-v2^2 = -2g∆h where ∆h=h1-h2
Because of continuity (constancy of volume flow rate due to incompressibility), v2/v1 = area ratio A1/A2, thus
v1^2-v2^2 = v1^2(1-(A1/A2)^2)
v1^2 = -2g∆h/(1-(A1/A2)^2) = 2g∆h/((A1/A2)^2-1)
v1 = sqrt(2g∆h/((A1/A2)^2-1))
Then volume flow rate dV/dt = v1A1
To keep things simple we equate the 1.4 cm deltaH of oil to the equivalent value for water = 0.76*0.014 = 0.01064 m.
The answer is 1.5258E-5 m^3/s.
If you check this using the calculator in ref. 1, you will need to input 1.3214E-5 m^3/s (13.214 cm^3/s) because they use a factor of 4/3 to multiply the v^2 terms, which is explained in ref. 2.
Converted Spirax TD42 to venturi orifice trap
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