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LECTURE UNIT. 6
Fluid Flow Continuity Equations
         - deals with fluid flow rates.
Law of Conservation of Mass
         -   mass can neither be created nor destroyed.




For Frictionless Flow
         m1 = m 2
                            .        .
                 but:       m=ρV
                            .
                 and:       V=Av
Hence;
             .        .
         ρ1 V 1 = ρ 2 V 2
      ρ 1 A 1 v 1 = ρ2 A 2 v 2
for frictionless flow t1 = t2, thus ρ1 = ρ2
therefore;
                                 .
         A1 v1 = A2 v2 = Q = V
General Formula:
         Q=Av


Example:
         For the pipe shown above, the following dimensions are given:
                 Fluid is water:
                            D1 = 4 in., D2 = 2 in.
                 v1 = 4 ft/s
Find the
         a. Volume flow rate.
         b. Fluid velocity at point 2.
         c. Mass flow rate.
         d. Weight flow rate.
Considering Flow with Friction




Q = Cc Cv (A1 v1) = Cc Cv (A2 v2)
       Cv = velocity coefficient (due to turbulence)
       Cc = coefficient of contraction (due to vena contrata)
       Cd = Cc x Cv = coefficient of discharge
Q = Cd A1 v1 = Cd A2 v2
Bernoulli’s Equation
       - keeps track of the various forms of energy possessed by fluids.
Law of Conservation of Energy
       -   energy can be either created or destroyed




Assumptions used to derive Bernoulli’s Equation
   1. We have incompressible flow, and thus the specific weight of the fluid is essentially constant.
   2. The fluid is ideal (has no viscosity), and thus there are no energy losses due to friction.
   3. There are no mechanical devices such as pumps or turbines between point 1 and point 2 to add or
       remove energy from the fluid.
   4. The flow is steady.
   5. No heat transferred (added or removed) to or from the fluid between point 1 and point 2 via
       devices called heat exchangers.
Applying conservation of energy:


                                               [Ein = Eout]
                          PEs + KEs + Us + Wfs = PEd + KEd + Ud + Wfd
              Us, Ud = 0, since Internal energy is a function of temperature and knowing that t W = ts = td




Example:

                                         P1
                                                                             P2


                  Q




       Determine pressure at point 2 (P2) for a horizontal pipe with D1 = 2 in., P1 = 20 psig, D2 = 1.5 in.,
and 200 gpm of water.


Torricelli’s Theorem
       The velocity of a free jet pf fluid is ideally equal to the square root of the product of two times the
acceleration of two times the acceleration of gravity and the head producing the jet.



                        v1
     Liquid




              h



                                          v2




Example:
       For the system above, h = 38 ft and the diameter of the side opening is 2 in. Find the: (a) Jet
velocity in ft/s and (b) volume flow rate in gpm.
VENTURI METER

        The venturi meter is a tube that is installed into a pipeline whose flow is to be measured. The venturi tube
consist of a converging section followed by a constant-diameter section (called throat), followed by a diverging section
that returns the tube to its original diameter.
        The maximum recommended included angle of the converging section is 22° and that for the diverging
section is 8° to keep frictional pressure losses to a very small value.

Problems:
   1. Water flows thru a Venturi meter as shown at a rate of 2.12 cfs.




            a. Determine the change in pressure head between 1 and 2.
            b. Determine the velocity at the throat
            c. Determine the discharge coefficient.
    2. For the Venturi meter shown, the deflection of mercury in the differential gage is 14.3 in. Assuming no energy
       is lost between A and B, compute the following




           a. Velocity at B
           b. Velocity at A
           c. Flow of water through the meter.
    3. Installed is a Venturi meter on a pipe 250 mm diameter in which the maximum flow is 125 liters per second
       and the pressure head is 6m. of H2O. Calculate the minimum diameter of the throat to ensure that a negative
       head will be formed in it. (To ensure a negative head at the throat, use zero pressure head).

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006

  • 1. LECTURE UNIT. 6 Fluid Flow Continuity Equations - deals with fluid flow rates. Law of Conservation of Mass - mass can neither be created nor destroyed. For Frictionless Flow m1 = m 2 . . but: m=ρV . and: V=Av Hence; . . ρ1 V 1 = ρ 2 V 2 ρ 1 A 1 v 1 = ρ2 A 2 v 2 for frictionless flow t1 = t2, thus ρ1 = ρ2 therefore; . A1 v1 = A2 v2 = Q = V General Formula: Q=Av Example: For the pipe shown above, the following dimensions are given: Fluid is water: D1 = 4 in., D2 = 2 in. v1 = 4 ft/s Find the a. Volume flow rate. b. Fluid velocity at point 2. c. Mass flow rate. d. Weight flow rate.
  • 2. Considering Flow with Friction Q = Cc Cv (A1 v1) = Cc Cv (A2 v2) Cv = velocity coefficient (due to turbulence) Cc = coefficient of contraction (due to vena contrata) Cd = Cc x Cv = coefficient of discharge Q = Cd A1 v1 = Cd A2 v2 Bernoulli’s Equation - keeps track of the various forms of energy possessed by fluids. Law of Conservation of Energy - energy can be either created or destroyed Assumptions used to derive Bernoulli’s Equation 1. We have incompressible flow, and thus the specific weight of the fluid is essentially constant. 2. The fluid is ideal (has no viscosity), and thus there are no energy losses due to friction. 3. There are no mechanical devices such as pumps or turbines between point 1 and point 2 to add or remove energy from the fluid. 4. The flow is steady. 5. No heat transferred (added or removed) to or from the fluid between point 1 and point 2 via devices called heat exchangers.
  • 3. Applying conservation of energy: [Ein = Eout] PEs + KEs + Us + Wfs = PEd + KEd + Ud + Wfd Us, Ud = 0, since Internal energy is a function of temperature and knowing that t W = ts = td Example: P1 P2 Q Determine pressure at point 2 (P2) for a horizontal pipe with D1 = 2 in., P1 = 20 psig, D2 = 1.5 in., and 200 gpm of water. Torricelli’s Theorem The velocity of a free jet pf fluid is ideally equal to the square root of the product of two times the acceleration of two times the acceleration of gravity and the head producing the jet. v1 Liquid h v2 Example: For the system above, h = 38 ft and the diameter of the side opening is 2 in. Find the: (a) Jet velocity in ft/s and (b) volume flow rate in gpm.
  • 4. VENTURI METER The venturi meter is a tube that is installed into a pipeline whose flow is to be measured. The venturi tube consist of a converging section followed by a constant-diameter section (called throat), followed by a diverging section that returns the tube to its original diameter. The maximum recommended included angle of the converging section is 22° and that for the diverging section is 8° to keep frictional pressure losses to a very small value. Problems: 1. Water flows thru a Venturi meter as shown at a rate of 2.12 cfs. a. Determine the change in pressure head between 1 and 2. b. Determine the velocity at the throat c. Determine the discharge coefficient. 2. For the Venturi meter shown, the deflection of mercury in the differential gage is 14.3 in. Assuming no energy is lost between A and B, compute the following a. Velocity at B b. Velocity at A c. Flow of water through the meter. 3. Installed is a Venturi meter on a pipe 250 mm diameter in which the maximum flow is 125 liters per second and the pressure head is 6m. of H2O. Calculate the minimum diameter of the throat to ensure that a negative head will be formed in it. (To ensure a negative head at the throat, use zero pressure head).