Mechanical Engineering Principles

254 MECHANICAL ENGINEERING PRINCIPLES

wherea 1 =cross-sectional area at section 1,
a 2 =cross-sectional area at section 2,
v 1 =velocity of fluid at section 1, and
v 2 =velocity of fluid at section 2

22.15 Bernoulli’s Equation

Bernoulli’s equation states that for a fluid flowing
through a pipe from section 1 to section 2:

P 1

ρ

+

v 12

2

+gz 1 =

P 2

ρ

+

v^22

2

+g(z 2 +hf)

where ρ=density of the fluid,
P 1 =pressure at section 1,
P 2 =pressure at section 2,
v 1 =velocity at section 1,
v 2 =velocity at section 2,
z 1 =‘height’ of pipe at section 1,
z 2 =‘height’ of pipe at section 2,
hf=friction losses (in m) due to the fluid
flowing from section 1 to section 2,

and g= 9 .81 m/s^2 (assumed)

Problem 2. A storage tank contains oil

whose free surface is 5 m above an outlet

pipe, as shown in Figure 22.12. Determine

the mass rate of flow at the exit of the outlet

pipe, assuming that (a) losses at the

pipe entry= 0. 4 v^2 , and (b) losses at the

valve= 0. 25 v^2.

Free surface

Valve

z 1 = 5 m

v 2

Figure 22.12

Pipe diameter= 0 .04 m, density of oil,

ρ=770 kg/m^3.

Letv 2 =velocity of oil through the outlet pipe.

From Bernoulli’s equation:

P 1

ρ

+

v 12

2

+gz 1 =

P 2

ρ

+

v 22

2

+gz 2

+ 0. 4 v 22 + 0. 25 v^22

i.e. 0+ 0 +g(5m)= 0 +

v 22

2

+ 0 + 0. 65 v^22

(where in the above, the following assumptions have

been made:P 1 =P 2 =atmospheric pressure, and

v 1 is negligible)

Hence, 5 m× 9. 81

m

s^2

=( 0. 5 + 0. 65 )v^22

Rearranging gives: 1. 15 v^22 = 49. 05

m^2

s^2

Hence, v^22 =

49. 05

1. 15

from which, v 2 =

√

49. 05

1. 15

= 6 .531 m/s

Cross-sectional area of pipe

=a 2 =

πd 22

4

=

π× 0. 042

4

= 0 .001257 m^2

Mass rate of flow through the outlet pipe

=ρa 2 v 2

= 770

kg

m^3

× 1. 257 × 10 −^3 m^2 × 6. 531

m

s

= 6 .321 kg/s

Flow through an orifice

Consider the flow of a liquid through a small orifice,

as shown in Figures 22.13(a) and (b), where it can

be seen that the vena contracta (VC) lies just to the

right of the orifice. The cross-sectional area of the

fluid is the smallest here and its decrease in area

from the orifice is measured by the coefficient of

contraction (Cc).

Due to friction losses there will be a loss in velocity

at the orifice; this is measured by the coefficient of

velocity, namelyCv,sothat:

Cd=Cv×Cc=the coefficient of discharge