Qu 1

Water flow through a pipe and the pressure head is indicated by

two manometers 6 m apart. Determine the direction of flow and the
head loss between the two manometers.

1.0 m
3.0 m

.1
1.5 m

.2
 Qu.1

1.0 m
3.0 m
.1
Assume flow is Down
Therefore, hloss acts 1.5 m
upwards HL .2

æ p1 V12 ö æ p2 V22 ö
ç + + z1 ÷ - H L = ç + + z2 ÷
è r g 2g ø è r g 2g ø
 Qu.1

1.0 m
3.0 m
Assume flow is Down .1
Therefore, hloss acts
upwards 1.5 m
HL
.2
æ p1 V12 ö æ p2 V22 ö
ç + + z1 ÷ - H L = ç + + z2 ÷
è r g 2g ø è r g 2g ø
= 1.5 m =0m
V1 = V2 =1m =3m
HLoss = −0.5 m….. –Ve implies HL in opp direction, hence
Flow should be uphill
 Qu.1

1.0 m
3.0 m
.2
1.5 m
HL
.1
Assume flow is Upwards
Therefore, hloss acts Down

æ p V 2 ö æ p V 2 ö
ç 1 + 1 + z1 ÷- H Losses = ç 2 + 2 + z2 ÷
èrg 2g ø èrg 2g ø
 Qu.1

1.0 m
3.0 m
Assume flow is UP .2
Therefore, hloss acts
Down 1.5 m
HL .1
æ p V 2 ö æ p V 2 ö
ç 1 + 1 + z1 ÷- H Losses = ç 2 + 2 + z2 ÷
èrg 2g ø èrg 2g ø
=0m =1m
V1 = V2 =3m =1m
HLoss = + 0.5 m….. +Ve flow uphill is correct
Qu.2 Pump head
Hp (m)
Open tank
Pump curve
Hp = 20 −2000Q2

6m

pump

Flow rate
Q (m3/s)
diameter = 0.07 m

Water is pumped from a tank as shown. The system head loss is

known to be 1.2(V2/2g) where V is the velocity in the pipe. The
pump h-Q curve is given as shown. Determine the flowrate.
Qu.2 Pump head
Hp (m)

.1 Pump curve
Hp = 20 -2000Q2
Open tank

6m

pump

dia = 0.07 m
.2 Flow rate
Q (m3/s)

Apply energy equation between 1 and 2

P1 (V1 ) 2 P2 (V2 ) 2
+ + z1 - H Loss + H pump = + + z2
rg 2 g rg 2 g
Qu.2 P1 (V1 ) 2 P2 (V2 ) 2
+ + z1 - H Loss + H pump = + + z2
rg 2 g rg 2 g
atm V1=0 atm

6m 0m

Pump Head (m)

Hp = 20 -2000Q2
Losses (m)
= 1.2 ( V2/2g)
= 1.2 ( V22/2g)

Q = 0.052 m3/s
 Inlet
Qu 3 P1 = 415 kPa
Q = 4.5 m3/s
dia1 = 1 m

Turbine
3m
Outlet
P2 = 25 cm Hg vacuum
dia2 = 1.2 m

Water is supplied at 4.50 m3/s and 415 kPa to a hydraulic

turbine through a 1.0-m inside diameter inlet pipe. The
turbine discharge pipe has a 1.2-m inside diameter. The
static pressure at section (2), 3 m below the turbine inlet,
is 25 cm Hg vacuum. If the turbine develops 1.9 MW,
determine the power lost between sections (1) and (2).
 Inlet
Qu 3 P1 = 415 kPa
Q = 4.5 m3/s
dia1 = 1 m

Turbine
3m
Outlet
P2 = 25 cm Hg vacuum
dia2 = 1.2 m

Apply energy equation between 1 to 2

P1 (V1 )2 P2 (V2 )2
+ + z1 - H L - H turbine = + + z2
r g 2g r g 2g

r.g.Q.Hturbine (W) = Turbine Power

Power loss = r.g.Q.HL (W)
Qu 3
P1 (V1 )2 P2 (V2 )2
+ + z1 - H L - H turbine = + + z2
r g 2g r g 2g

Power Loss = r.g.Q.HLoss (W)

Power developed = 1.9MW

Multiply by rgQ and re-arrange

rQ(V1 - V2 )
2 2
rgQH Loss = Q( P1 - P 2 ) + + rgQ( z1 - z 2 ) - 1.9 x106
2
Inlet Outlet
P1 = 415 kPa P2 = 25 cm Hg vacuum
Q = 4.5 m3/s dia2 = 1.2 m
Power Loss = 290 kW dia1 = 1 m z1 – z2 = 3m

V1 = 5.73 m/s V2 = 3.98 m/s

Qu 4 Section 1 Section 2

Flow 8 cm dia
5 cm dia

orifice
An orifice meter is used to measure the water flow rate along a horizontal section
of a piping system. A 5-cm-diameter orifice plate is inserted into the 8-cm-
diameter pipe. If the water flowrate through the pipe is 0.03 m3/s, determine the
pressure difference indicated by a manometer attached to the flow meter.
Kinematic viscosity of water is given as 1.12×10−6 m2/s.
2 ( p1 - p2 )
Apply BE between 1 to 2 Q = Ao r (1-  4 )
æ p1 V12 ö æ p2 V2 2 ö This is ideal flow ; no Losses
ç + + z ÷ = ç + + z ÷
ç rg 2 g 1 ÷ ç rg 2 g 2 ÷
è ø è ø For actual flow :
2 ( p1 - p2 )
Qactual = Co Ao r (1-  4 )
Co is flow coef for orifice
Qu 4 orifice

Flow
8 cm dia .1 .2
5 cm dia

At the orifice, velocity increases and the corresponding pressure

reduces.
2 ( p1 - p2 )
Applying B.E between 1 to 2 Q = Co Ao r (1-  4 )
where  = 0.50/0.80 = 0.625

Find P1 –P2
Qu 4
2 ( p1 - p 2 )
To find P1 –P2 Q = Co Ao r (1-  4 )

Q is given
Ao is area at orifice
 = d2/d1 = 0.50/0.80 = 0.625
Need to find Co

Given : Q = 0.03 m3/s…

from Q = A.V1 dia1 = 8cm
V1 = 5.97 m/s

VD (5.97)(0.08)
Re = = = 4. 26 x10 5

 1.12 x10 -6
Qu 4
From figure, with Re = 4.26×105 and  = 0.625

Co = 0.608

2 ( p1 - p2 )
Q = Co Ao r (1-  4 )

Q = 0.03 m3/s
Co = 0.608
r = 1000 kg/m3
 = 0.625

Hence, P1 –P2 = 267.6 kPa

Qu 5
A 50-mm-diameter nozzle meter is installed at the end of an
80-mm-diameter pipe through which air flows, as shown in Fig.
3. A manometer attached to the static pressure tap just
upstream from the nozzle indicates a pressure of 7.3 mm of
water. Determine the flow rate, Q .

Pressure at point 1

Tutorial 5
Qu 5
Flow across nozzle is :  = 50/80 = 0.625
2( p1 - p 2 ) p1 – p2 = gwaterh
Q = Cn An r (1-  4 ) p1 – p2 = (9810)7.3/1000
p1 – p2 = 71.5 Pa
Q = 0.0230Cn
dnozzle = 50 mm

Cn depends on  and Re No Hence …Anozzle =

Make an initial Guess or Estimation of Cn

Qu 5 Q = 0.0230Cn
Make an initial Guess or Estimate of Cn

1) 1st estimate Cn = 0.97…gives… Q = 0.0223 m3/s

From Q = 0.0223 m3/s…. then from V = Q/Area …. V= 4.44 m/s

VD(4.44)(0.08)
Re = = = 2.43x10 4

 1.46x10 -5

 = 0.625 found earlier … and Re = 2.43 x 104

With  and Re, we can obtain Cn from Figure

Cn = 0.963…. ( initial guess was Cn ~ 0.97 )
Qu 5

 = 0.625
Re No = 24300
Qu 5 Q = 0.0230Cn
Cn = 0.963
2) Cn = 0.963…gives… Q = 0.0221 m3/s

From Q = 0.0221 m3/s…. then V = Q/Area …. V= 4.40 m/s

VD (4.40)(0.08)
Re = = = 2.41x10 4

 1.46x10 -5

With  and Re = 2.41x104

we can obtain next Cn from Figure to be close
to 0.963 … Thus, Q = 0.0221 m3/s
Orifice Nozzle