PROPERTIES OF FLUIDS
Density  = mass
Volume
Specific weight
Therefore
γ
(kg)
(m3)
γ
=
=
 xg
weight
Unit Volume
N
m3
 = γ
g
or
Specific Volume
Volume occupied by unit mass
=
1

=
m3
kg
Specific Gravity (s)
Sliquid
=
 liquid
 water
Example If the specific weight of water is taken as 9.81 KN/m3 and the
specific gravity of mercury is 13.56, calculate:
(a)
The density of water
(b)
The specific weight of mercury
(c)
The density of mercury
propflu/docs/science/DC/MGS
(a)

(b)
(c)
γ
g
=
Smercury
=
 mercury
 water

=
Smercury
=
13.56 x 1000
=
13560 kg/m3
γ
=
mercury
=
 xg
=
13560 x 9.81
=
133024 N/m3
9.81 x 103
9.81
x
=

1000kg/m3
water
IDEAL FLUID
In an ideal fluid there is no friction or shearing forces within the liquid. Ideal
fluids no not exist in practice, however it is useful to consider fluids as ‘ideal’
in order to develop theories.
Viscosity μ The viscosity of a fluid is a measure of its resistance to shear or
angular deformation. Motor oil has a high viscosity relative to petrol. The
dimensions of absolute viscosity are force per unit area divided by velocity
gradient.
Dimensions of μ
=
Ns
m2
Kinematic Viscosity
Kinematic Viscosity
ν
propflu/docs/science/DC/MGS
=
absolute viscosity
density
=
μ

Units
NS
m2
m3
kg
=
kg m
S2
S
m2
m3
kg
=
m2
S
It is called Kinematic viscosity because there are no forces involved.
The absolute viscosity μ of most fluids is virtually independent of pressure for
most engineering purposes, however the kinematic viscosity ν of gases varies
strongly with pressure because of the change of density.
PRESSURE IN FLUID FLOW
Static Pressure. If the pressure of a fluid is measured with an instrument
which is ‘static with respect to the fluid’, then this is known as static pressure.
In practice, it is measured with a piezometer tube which minimizes the
disturbance of the flow.
Stagnation Pressure (Dynamic Pressure)
1
Flow
Stagnation Point
2
Flow
Applying Bernoulli’s equation to 1 and 2
P1
g
+
C12
2g
+
propflu/docs/science/DC/MGS
h1
=
P2
g
+
C22
2g
+
h2
h1
=
therefore:
h2
cancel
g
P1

+
C12
2
P2
=
P1
terms
c2
=
P2

+
 C12
=
0
2
Example
Calculate the stagnation pressure 5m below the surface of fluid of specific
gravity 1.4 when it is moving at 5 m/s.
Pstagnation
 liquid
=
Pstatic +
 C2
2
=
 gh
+
=
1.4
X
 water
=
1.4
x
1000 kg/m3
=
1400
 C2
2
kg/m3
Using 1 above
Pstagnation
=
(1000 x 9.81 x 5) + 1000 x 52
2
=
49050 + 12500 = 61550 N
m2
propflu/docs/science/DC/MGS
eqn 1
POWER IN A FLUID
Power
or
=
Energy Used
Time
Power
=
Energy
unit weight
of liquid
=
mgh
mg
=
P
There
Also,
P
=
Qγ
=
Qγh
Efficiency
Efficiency
=
Power output
Power input
propflu/docs/science/DC/MGS
Weight
time
mg
s
.
.
= h m g (where m = mass flow rate (kg/s)
.
hmg
=
x
x
hmg
S
.
mg
Therefore
J
s
Example
Water flows along a pipe at 20 m3/s If there is a head loss due to friction of
15m over a 5km length of pipe, calculate the power required to overcome
friction per km of pipe.
Head loss (h) per km =
P
15
5
=
3m
γ = g
=
γQH
=
9810 x 20 x 3
=
=
588600 W
588.6 KW
= 1000 x 9.81
= 9810 N
m3
Example Problem
A turbine located 200m below the water surface at intake has a flow rate of
15m3/s The head loss due to friction in the pipe is 25m. Find the power
produced by the turbine if it is 85% efficient.
Power
=
hmg
Available head h = 200 – 25 = 175m
m
Therefore Power
=
=
Qx 
15 x 1000 = 15000 kg/s
=
=
175 x 15000 x 9.81
25.75 x 106 W = 25.75 MW
propflu/docs/science/DC/MGS
Therefore 25.75 MW is delivered ‘to’ the turbine by the flowing water. If the
turbine is 85% efficient, the output power of the turbine is
Pout
=
Pin
x
=
25.75
=
21.89 M W
efficiency
x
0.85 MW
Example Oil enters a pump at a pressure of 50KN/m2 through a pipe 200mm
diameter. The oil leaves the pump at a pressure of 200KN/m2 through a pipe
150mm diameter. Find the power of the pump if:
- the elevation of the suction and discharge pipes are the same
- the oil has a specific gravity of 0.85
- the flow rate of the oil is 500 l/s
P = 50 kN/m2
200mm
P = 200 kN/m2
Pump
1
 oil
C1
=
Q
A1
2
=
0.85 x 1000
=
850 kg/m3
=
0.5 x 4
π x 0.22
C2 = Q
A2
=
15.92 m/s
C2 = 28.29 m/s
propflu/docs/science/DC/MGS
=
0.5 x 4
π x 0.152
150mm
Applying Bernoulli’s equation, and neglecting changes in potential energy
C12
2g
P1 +
g
+ head due to pump =
50 x 103
+ 15.922
850 x 9.81
2 x 9.81
6.0
+
18.92
+
Power
12.92
Hp
Hp
=
=
=
=
+
+
=
=
Hp
Hp
64.8
45.88m
Qγh
0.5 x 850 x 45.88
19499 W
19.5 KW
propflu/docs/science/DC/MGS
=
P2
g
+
C22
2g
=
200 x 103 + 28.292
850 x 9.81
2 x 9.81
24.0
+
40.80
Water from a reservoir supplies a power station 400m below the surface of
the reservoir. The water jet discharges through a nozzle 300mm diameter at
a velocity of 50m/s. Find the power lost due to friction between the reservoir
and the jet, and the power of the jet.
Applying Bermoulli between the surface of the reservoir (1) and the jet (2).
C12 +
2g
(a) P1 +
g
h1
=
P2
g
+ C22
2g
+
h2 + Losses
Pressures at 1 and 2 are equal (atmospheric).
Velocity at 1 is zero.
h1
=
400m if h2
= 0
Therefore:
C22
2g
h1 =
+
400
=
502
2 x9.81
400
=
127.4 +
Losses
(a)
=
Power
losses
+
losses
losses
272.6m
=
Q γ h
= (C2 A2) (  g) (h)
(h = lost head)
= (50 x π x 0.32 ) (1000 x 9.81) (272.6)
4
= 9.44 x 106 W
= 9.44 MW
propflu/docs/science/DC/MGS
(b)
Power = Q γ h
(h = delivered head)
= 3.53 x 9810 x (400 – 272.6)
= 4.4 x 106 W
= 4.4 MW
Check
Power of jet
=
.
mc2
2
=
Q  c2
2
=
3.53 x 1000 x 502
2
=
4.4 x 106 W
=
4.4 MW
propflu/docs/science/DC/MGS