Mechanical Engineering Principles

FLUID FLOW 255

(a)

Free surface

h

(b)

Vena contracta

n 2

2

Figure 22.13

Leta=area of orifice.
Due to the vena contracta the equivalent cross-
sectional area=Cca
Now the theoretical velocity at section 2√ =v 2 =
2 gh, but due to friction losses,

v 2 =Cv

√

2 gh

Hence discharge =Cca×Cv

√

2 gh

But Cd=CvCc

Therefore, discharge=Cd×a

√

2 gh

Now try the following exercise

Exercise 115 Further problems on fluid

flow

1. If in the storage tank of worked problem 2
   on page 254, Figure 22.12, z 1 = 8m,
   determine the mass rate of flow from the
   outlet pipe. [7.995 kg/s]


2. If in the storage tank of worked problem 2,
   page 254, Figure 22.12,z 1 =10 m, deter-
   mine the mass rate of flow from the outlet
   pipe. [8.939 kg/s]


3. If in Figure 22.13,h=6m,Cc = 0 .8,
   Cv= 0 .7, determine the values ofCdand
   v 2 .[Cd= 0 .56,v 2 = 6 .08 m/s]


4. If in Figure 22.13,h=10 m,Cc= 0 .75,
   Cv= 0 .65, and the cross-sectional area
   is 1. 5 × 10 −^3 m^2 , determine the discharge
   and the velocityv 2.
   [Cd= 0 .488, 9.10 m/s]

22.16 Impact of a jet on a stationary

plate

The impact of a jet on a plate is of importance in

a number of engineering problems, including the

determination of pressures on buildings subjected

to gusts of wind.

Consider the jet of fluid acting on the flat plate of

Figure 22.14, where it can be seen that the velocity

of the fluid is turned through 90°, or change of

velocity=v.

Now, momentum=mvand asvis constant,

the change of momentum=

dm

dt

×v

v

v

v

F

Figure 22.14

However,

dm

dt

=mass rate of flow=ρav

Therefore, change of momentum=ρav×v

=ρav^2 but from Newton’s second law of motion

(see pages 139 and 144),

F=rate of change of momentum

i.e. F=ρav^2

whereF=resulting normal force on the flat plate.

Pressure=

force

area

=

ρav^2

a

=ρv^2

For wide surfaces, such as garden fences, the pres-

sure can be calculated by the above formula, butfor