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posted by  KPL on 11/1/2009 11:30:23 AM  |  status: Closed  |  Earned Karma: 25

Fluid Energy Equation

Course Textbook Chapter Problem Needs by
Fluid Mechanics Fundamentals of Fluid Mechanics (6th) by Young, Munson, Okiishi 5 94P 11/4/2009 at 9:00:00 AM
Question Details:
A horizontal venturi flow meter consists of a converging-diverging conduit.  The diameters of cross section (1)= 6 in and (2) = 4 in.  The velocity and static pressure are uniformily distributed at cross sections (1) and (2).  Determine in the volume flowrate (ft^3/s) through the meter if p1-p2=3 psi, the flowing fluid is oil (density = 56 lb*m/ft^3), and the loss per unit mass from (1) to (2) is negligibly small.
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posted by Anonymous on 11/6/2009 3:02:06 PM  |  status: Live
Clarification Details:
what about the fact that density is given in units of lbm/ft^3? Shouldnt there be some multiplication factor of (lbf*s^2)/(lbm*32.2ft)?

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posted by Root_Locus(MNK) on 11/2/2009 6:46:45 AM  |  status: Live
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Response Details:
Applying bernoulli's theorem between point  1 and 2
P1/ρ + V12/2 + gZ1 = P2/ρ + V22/2 + gZ2  ................................(1)
Given, Z1 = Z2
also flow rate Q = (πd12/4)V1 = (πd22/4)V2
 so V1 = 4Q/πd12  =>  V12/2 = 8Q2/π2d14
and V2 = 4Q/πd22  =>  V22/2 = 8Q2/π2d24
from (1),
P1/ρ + 8Q2/π2d14 = P2/ρ + 8Q2/π2d24 
or, (P1-P2)/ρ = (8Q2/π2)[1/d24 - 1/d14 ] ..................(2)
Now, P1-P2 = 3 Psi = 3*122 lb/ft2              {12 in = 1 ft}
          ρ = 56 lbm/ft3
         d1 = 6 in = 1/2 ft
         d2 = 4 in = 1/3 ft
putting in (2)
 3*122/56 = (8Q2/π2)[34 - 24]
or, (8Q2/π2) = (3*144)/(56*65) = 0.118681319
or, Q = π*√(0.118681319/8)
     Q = 0.3826 ft3/s
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posted by new_sylar on 11/2/2009 8:04:07 AM  |  status: Live
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Response Details:
Given conditions  : the tube is horizontal so no change in potential Energy due to g
                                change in pressure = 3psi
                                density of liquid =56 lbm/ft3
                                 diameters d(1) =6 d(2)=4

So adjusted equation is
P1+ ρV12/2  = P2 +ρ V22/2   =>      ΔP =ρ/2(V(1)^2-V(2)^2)
where volume flow rate is V= Q/A =4Q/π*d*d applying

3 =( 56/12*12)(8Q2/π2)[34 - 24]

Which gives Q/π =0.121
which gives Q =0.381Ft^3/s
-------Ford :p
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