Here are the key steps to solve this problem: 1) Use Bernoulli's equation between points 1 and 2: P1/γ + V12/2g + Z1 = P2/γ + V22/2g + Z2 + HL 2) Given: P1 = 200 kPa, Q = 30 L/sec, HL = 20 kPa 3) Use continuity equation: A1V1 = A2V2 4) Solve for P2 The pressure at point 2 is 180 kPa.

2. FLUID MECHANICS
m kg
ρ=
V m3
V m3
υ=
m kg
W
mg
ρg KN
γ=
=
=
V 1000V 1000 m 3

3. Its specific gravity (relative density) is equal to the ratio
of its density to that of water at standard temperature and pressure.
ρL γ L
SL = =
ρW γW
Its specific gravity (relative density) is equal to the ratio
of its density to that of either air or hydrogen at some specified
temperature and pressure.
ρG γG
SG =
=
ρ ah γ ah
where: At standard condition
3
W = 1000 kg/m
3
W = 9.81 KN/m

4. °F - 32
1.8
°F =1.8°C+ 32
°C =
K
C 273
R
F 460
F
P=
KPa
A
dF
P=
KPa
dA
where: F - normal force, KN
A - area, m2

5. y
P3 A3
A
P1 A1
x
B
C
z
P2 A2
Fx = 0 and Fy = 0
P1A1 – P3A3 sin = 0
P2A2 – P3A3cos = 0
From Figure:
A1 = A3sin
A2 = A3cos
3
4
1
2
Eq. 3 to Eq. 1
P1 = P3
Eq. 4 to Eq. 2
P2 = P3
Therefore:
P1 = P2 = P3

6. Atmospheric pressure: The pressure exerted by the atmosphere.
At sea level condition:
Pa = 101.325 KPa
= .101325 Mpa
= 1.01325Bar
= 760 mm Hg
= 10.33 m H2O
= 1.133 kg/cm2
= 14.7 psi
= 29.921 in Hg
= 33.878 ft H2O

7. Pgage
Atmospheric pressure
Pvacuum
Pabs
Absolute
Zero
Pabs = Pa+ Pgage
Pabs = Pa - Pvacuum
Pabs

8. moving plate
v
v+dv
dx
x
v
Fixed plate
S dv/dx
S = (dv/dx)
S = (v/x)
= S/(v/x)
where:
- absolute or dynamic
viscosity
in Pa-sec
S - shearing stress in Pascal
v - velocity in m/sec
x -distance in meters

9. = /
m2/sec
Ev = - dP/(dV/V)
Where negative sign is used because dV/V is negative for a positive dP.
Ev = dP/(d / )
because -dV/V = d /
where:
Ev - bulk modulus of elasticity, KPa
dV - is the incremental volume change
V - is the original volume
dP - is the incremental pressure change

10. Where:
- surface tension, N/m
- specific weight of liquid, N/m3
r – radius, m
h – capillary rise, m
r
h
Surface Tension of Water
C
0
10
h
0.0742
20
2σ cos θ
γr
0.0756
0.0728
30
0.0712
40
0.0696
60
0.0662
80
0.0626
100
0.0589

11. FREE SURFACE
h1
1•
h
h2
2•
dP = - dh
Note:Negative sign is used because pressure decreases as elevation increases
and pressure increases as elevation decreases.

12. Pressure Head:
P
h
γ
where:
p - pressure in KPa
- specific weight of a fluid, KN/m3
h - pressure head in meters of fluid
MANOMETERS
Manometer is an instrument used in measuring gage pressure in length of some
liquid column.
 Open Type Manometer : It has an atmospheric surface and is capable in
measuring gage pressure.
 Differential Type Manometer : It has no atmospheric surface and is capable in
measuring differences of pressure.

13. Open Type Manometer
Fluid A
Differential Type Manometer
Fluid A
Open
Manometer Fluid
Fluid B
Manometer Fluid

14. Determination of S using a U - Tube
Open
Open
Fluid A
x
y
Fluid B
SAx = SBy

15. Example no. 1
A building in Makati is 84.5 m high above the street level. The required
static pressure of the water line at the top of the building is 2.5 kg/cm2.
What must be the pressure in KPa in the main water located 4.75 m
below the street level. (1120.8 KPa)
Point 1: Main water line, 4.75 m below street level
Point 2: 84.5 m above street level
∆h = h2 – h1 = (84.5 + 4.75) = 89.25 m
P2 = 2.5 kg/cm2 = 245.2 KPa
P2
P1
P1
P1
P1
(h2 h1 )
P2 (h2 h1 )
245.2 9.81(89.25)
1,120.743 KPa

16. Example No. 2
A mercury barometer at the ground floor of a high rise hotel in Makati
reads 735 mm Hg. At the same time another barometer at the top of
the hotel reads 590 mmHg. Assuming air density to be constant at
1.22 kg/m3, what is the approximate height of the hotel. (1608 m)
Point 1: Ground floor
1.22(9.81)
KN
For air :
0.012 3
h1 = 0 m
1000
m
P1 = 735 mm Hg = 98 Kpa
kg
1.22 3
P2 - P1 - (h2 - h1 )
Point 2: Roof Top
m
g KN
(P2 - P1 )
h2 = h (height)
h2 - h1
1000 m3
P2 = 590 mm Hg = 78.7 KPa
assumin g :
m
g 9.81
sec2
h2 - h1 h
h 1608.33 meters

17. Example No. 3
The reading on a pressure gage is 1.65 MPa, and the local barometer
reading is 94 KPa. Calculate the absolute pressure that is being measured in kg/cm2. (17.78 kg/cm2)
Example No. 4
A storage tank contains oil with a specific gravity of 0.88 and depth of
20 m. What is the hydrostatic pressure at the bottom of the tank in
kg/cm2. (1.76 kg/cm2)
Example No. 5
A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific
gravity of oil is 0.9, what is the mass of oil in the tank?

18. Forces Acting on Plane Surfaces
Free Surface
hp
S
h
S
S
M
M
y
F
•C.G.
•C.P.
•C.G.
•C.P.
yp
e
N
N
F - total hydrostatic force exerted by the fluid on any plane surface MN
C.G. - center of gravity
C.P. - center of pressure

19. where:
Ig - moment of inertia of any plane surface MN with respect to the axis at its centroids
Ss - statical moment of inertia of any plane surface MN with respect to the axis SS not
lying on its plane
e - perpendicular distance between CG and CP

20. Forces Acting on Curved Surfaces
FV
Free Surface
D
E
Vertical Projection of AB
F
h
C
A
C’
L
C
C.G.
Fh
C.P.
B
B’
B
hp

21. Fh = γhA
A = BC x L
A - area of the vertical projection of AB, m2
L - length of AB perpendicular to the screen, m
FV = γV
V = AABCDEA x L, m3
2
F = Fh + Fv
2

22. Hoop Tension
D
T
F
T
h
1m
D
P= h
F=0
2T = F
T = F/2
S = T/A
A = 1t
T
1
F
2
T
t
1m

23. S = F/2(1t) 3
From figure, on the vertical projection the pressure P;
P = F/A
A = 1D
F = P(1D) 4
substituting eq, 4 to eq. 3
S = P(1D)/2(1t)
PD
S
KPa
2t
where:
S - Bursting Stress KPa
P - pressure, KPa
D -inside diameter, m
t - thickness, m

24. Laws of Buoyancy
Any body partly or wholly submerged in a liquid is subjected
to a buoyant or upward force which is equal to the weight of
the liquid displaced.
1.
where:
W - weight of body, kg, KN
BF - buoyant force, kg, KN
- specific weight, KN/m3
- density, kg/m3
V - volume, m3
Subscript:
B - refers to the body
L - refers to the liquid
s - submerged portion
W
Vs
BF
W = BF
W = BVB KN
BF = LVs KN
W = BF
W = BVB
BF = LVs

25. W
2.
Vs
BF
T
W = BF - T
W = BVB KN
BF = LVs KN
W = BF - T
W = BVB
BF = LVs
where:
W - weight of body, kg, KN
BF - buoyant force, kg, KN
T - external force T, kg, KN
- specific weight, KN/m3
- density, kg/m3
V - volume, m3
Subscript:
B - refers to the body
L - refers to the liquid
s - submerged portion

26. 3.
T
W
Vs
BF
W = BF + T
W = BVB KN
BF = LVs KN
W = BF + T
W = BVB
BF = LVs
where:
W - weight of body, kg, KN
BF - buoyant force, kg, KN
T - external force T, kg, KN
- specific weight, KN/m3
- density, kg/m3
V - volume, m3
Subscript:
B - refers to the body
L - refers to the liquid
s - submerged portion

27. 4.
W
T
Vs
BF
VB = Vs
W = BF + T
W = BVB KN
BF = LVs KN
W = BF + T
W = BVB
BF = LVs

28. 5.
W
Vs
BF
T
VB = Vs
W = BF - T
W = BVB KN
BF = LVs KN
W = BF - T
W = BVB
BF = LVs

29. Energy and Head
Bernoullis Energy equation:
2
HL = U - Q
Z2
1
z1
Reference Datum (Datum Line)

30. 1. Without Energy head added or given up by the fluid (No work done
bythe system or on the system:
P1 v12
P2 v 2 2
+
+ Z1 =
+
+ Z2 + H L
γ 2g
γ 2g
2. With Energy head added to the Fluid: (Work done on the system
P1 v12
P2 v 2 2
+
+ Z1 + h t = +
+ Z2 + H L
γ 2g
γ 2g
3. With Energy head added given up by the Fluid: (Work done by the
system)
P1 v12
P2 v 2 2
+
+ Z1 + =
+
+ Z2 + H L + h
γ 2g
γ 2g
Where:
P – pressure, KPa
v – velocity in m/sec
Z – elevation, meters
+ if above datum
- if below datum
- specific weight, KN/m3
g – gravitational acceleration
m/sec2
H – head loss, meters

31. APPLICATION OF THE BERNOULLI'S ENERGY THEOREM
Nozzle
Base
Tip
Q
Jet
P1
v12
2g
Z1
P2
v22
2g
Z2
2
v
1 2
2g
HL
1
2
Cv
Q
Av m3 /sec
where: Cv - velocity coefficient
HL

32. Venturi Meter
B. Considering Head loss
P1
γ
Q'
Q'
1
2
2
Meter Coefficient
Manometer
A. Without considering Head loss
2
2
v1
P v2
Z1 2
2g
γ 2g
A1v1 A2 v 2
actual flow
2
P1 v 1
P2 v 2
Z1
γ 2g
γ 2g
Q A1v1 A2 v 2
Q theoretica flow
l
Z2
Q'
C
Q
Z 2 HL

36. 2
Upper
Reservoir
Suction Gauge Discharge Gauge
Gate Valve
1
Lower
Reservoir
Gate
Valve

37. Ht
P2
P1
γ
v2
2
v1
2g
Q = Asvs = Advd m3/sec
WP = Q Ht KW
BP
2πTN
KW
60,000
2
Z2
Z1
HL
meters

38. ηP
WP
x 100%
BP
ηm
BP
x 100%
MP
ηC
ηC
WP
x 100%
MP
ηP ηm

39. MP
MP
where:
EI(cosθ)
KW
1000
3 EI(cosθ)
KW
1000
P - pressure in KPa
T - brake torque, N-m
v - velocity, m/sec
N - no. of RPM
- specific weight of liquid, KN/m3
WP - fluid power, KW
Z - elevation, meters
BP - brake power, KW
g - gravitational acceleration, m/sec2
MP - power input to
HL - total head loss, meters
motor, KW
E - energy, Volts
I - current, amperes
(cos ) - power factor

40. HYDRO ELECTRIC POWER PLANT
1
Headrace
Penstock
turbine
2
Tailrace
Y – Gross Head

41. 1
Headrace
Penstock
Generator
Y – Gross Head
B
Draft Tube
ZB
2
B – turbine inlet
Tailrace

42. Fundamental Equations
1. Net Effective Head
A.
Impulse Type
h = Y – HL
Y = Z1 – Z 2
Y – Gross Head, meters
Where:
Z1 – head water elevation, m
Z2 – tail water elevation, m
B. Reaction Type
h = Y – HL
Y = Z1 –Z2
h
PB
2
vB
2g
ZB
meters
Where:
PB – Pressure at turbine inlet, KPa
vB – velocity at inlet, m/sec
ZB – turbine setting, m
- specific weight of water, KN/m3

43. 2. Water Power (Fluid Power)
FP = Q h KW
Where:
Q – discharge, m3/sec
3. Brake or Shaft Power
BP
2 TN
KW
60,000
Where:
T – Brake torque, N-m
N – number of RPM
4. Turbine Efficiency
BP
e
x 100%
FP
e eh evem
Where:
eh – hydraulic efficiency
ev – volumetric efficiency
em – mechanical efficiency

44. 5. Generator Efficency
g
g
Generator Output
x 100%
Brake or Shaft power
GP
x 100%
BP
6. Generator Speed
N
120f
RPM
n
Where:
N – speed, RPM
f – frequency in cps or Hertz
n – no. of generator poles (usually divisible by four)

45. Pump-Storage Hydroelectric power plant: During power generation the turbine-pump acts
as a turbine and during off-peak period it acts as a pump, pumping water from the
lower pool (tailrace) back to the upper pool (headrace).
Turbine-Pump

46. A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the
same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and
the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the
liquid is oil with S = 0.80. (174.2 KPa)
300 mm
100 mm
1
2

47. A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm
water main. In a differential gage partly filled with mercury (the remainder of the tube
being filled with water) and connected with the meter at the inlet and at the
throat, what would be the difference in level of the mercury columns if the discharge is
150 L/sec? Neglect loss of head. (h=273 mm)

48. The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PB is 15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in
the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m3/sec)
PA
1.2 m
PB