ICS FLUID FLOW
Important Conversions and Equalities:
1 Newton force (N) causes 1 kg mass to accelerate at 1 m/s2
1 dyne force (dyn) causes 1 g mass to accelerate at 1 cm/s2
1 pound force (lbf) causes 1 slug mass to accelerate at 1 ft/s2
1 pound force (lbf) causes 1 lb mass to accelerate at 32.2 ft/s2
1 poundal force (pdl) causes 1 lb mass to accelerate at 1 ft/s2
1 pdl =
1N =
1 lbf =
1 lbmft/s2
1 kgm/s2
1 slugft/s2
1 lbm =
1 kg
1 slug =
0.454 kg
14.594 kg
=
=
7.233 pdl
32.2 pdl
=
=
2.204623 lbm
32.174 lbm
g = 9.81 N/kg or m/s2
g = 981 dyn/g or cm/s2
g = 32.2 lbf/slug or ft/s2
g = 1.0 lbf/lbm
g = 32.2 pdl/lbm or ft/s2
105 dyn
32.2 lbmft/s2
=
=
=
4.448 N
g = 32.2 lbf/slug = 1 lbf/lbm= 32.2 pdl/lbm = 9.81 N/kg = 981 dyn/g = 1037 pdl/slug
The following conversions apply at 3.98 C where water is densest.
H2O = 1

g
kg  1 slug  
1 m3
slug

  1.9403 3


=
1000


3
3
3
3 



cm
m  14.594 kg   3.2808 ft 
ft
981 dyn
 1 g   981 dyn 
 
3 
 cm   g 
cm3
DH2O = g = 
or
 32.2 lbm 
lbm
  62.43 3
 
ft
 1slug 
 1000 kg   9 .807 N 



 m3   kg 

9807 N
m3
.
slug   32 .2 lbf 
62 .43 lbf
2009 pdl
 19403
 62.43 lbm  32 .2 pdl 
or 
 

 

3
3
3

  slug 

  lbm 
ft
ft
ft
ft 3
DH2O = 
1m
1 IMP gal
1 US gal
1 IMP gal H2O
1 US gal H2O
=
=
=
=
=
3.28083 ft
4.5461 L = 1.201 US gal
3.7854 L = 0.8327 IMP gal
4.5461 L  (1 kg/L)  (2.2046 lb/kg) = 10.02 lb/IMP gal
3.7854 L  (1 kg/L)  (2.2046 lb/kg) = 8.345 lb/US gal
H2O = 62.43 lbm/ft3  (1 IMP gal/10.0 lbm) 
H2O = 62.43 lbm/ft3  (1 US gal/8.345 lbm) 
Power =
Also,
work
time
=
Fd
t
 N
Power = p  Q e.g.  2
m
e.g.
6.24 IMP gal/ft3
7.48 US gal/ft3
Nm
J
=
= W
s
s
3
 m  N  m


s
 s 
 lbf
or  2
 ft
or
lbf  ft
s
 ft 3  lbf  ft


 s   s


1 horse power (HP) = 745.7 watts (W) = 550 ftlbfs-1 = 33000 ftlbfmin-1
FLUID FLOW
Page 1 of 51
EQUATIONS FOR FLUID FLOW
When an incompressible fluid flows through a pipe at a steady flow rate, the quantity of
flow past all points in the pipe is constant.
A = d2/4
d
Q
V (A d)

 Av
t
t
The front of this section of fluid moves a distance ‘d’
in time ‘t’, so its velocity, v = d/t.
Volumetric flow rate (Q) is the volume of fluid passing a certain point per unit time.
Volumetric flow rate has units of V/t, e.g., m3/h, L/min, mL/s, ft3/s, gal/h, etc.
Be careful not to confuse V for volume with v for velocity.
It is often easier to calculate cross sectional area as d2/4 rather than r2.
Problem 1: The nozzle of a garden hose has an I.D. of 0.500 in. If water flows through
the hose at 10.0 US gal/min, find the speed at which it leaves the nozzle in ft/s.
Ans: 16.3 ft/s
wall thickness
I.D. = internal diameter
O.D. = outside diameter
OD – 2(thickness) = ID
I.D.
O.D.
FLUID FLOW
Page 2 of 51
THE CONTINUITY EQUATION or STEADY STATE FLOW EQUATION
Consider a steady flow rate of a liquid flowing through a pipe of varying cross-sectional
area, from point 2 to point 1. The flow rate must remain constant even though the
velocity may vary.
2.
1.
Q1, A1, v1
Q 1 = Q2
Q1 = A 1   1
Since Q1 = Q2 ,
  d1 2
or
Q2, A2, v2
A1 < A 2
v1 > v2
and
Q2 = A 2   2
thus:
Q = A1   1 = A2   2
2
d
v2
A
d1
4
 1 

  1
2
2
v1
A2
  d2
d2
 d2
4
2

r
   1

 r2



2
Note that velocity is inversely proportional to cross-sectional area and thus also to the
square of the radius and diameter of the pipe, i.e., the larger the pipe, the lower the
velocity for a given flow rate. The equation of continuity holds for incompressible fluids
(liquids) in any direction of flow, i.e., up, down, back, forth, etc.
Problem 2: Water in a pipe 12.0 cm in diameter flows with a velocity of 8.00 m/s.
Calculate the velocity at a point in the pipe where its diameter is constricted to 3.00 cm.
Ans: 128 m/s.
FLUID FLOW
Page 3 of 51
BERNOULLI’S EQUATION
When a liquid flowing through a pipe enters a region where the pipe diameter is reduced,
its speed increases (because the flow rate is constant). A change in speed involves
acceleration, which means a net force must be acting upon the liquid. This force can only
arise from a difference in pressure between the different parts of the pipe. Obviously the
pressure in the section of pipe where the diameter is larger must be greater since the
liquid increases in speed as it enters the constriction. By similar reasoning, the pressure
must be lower in the constriction where the speed is greater.
Q1, A1, v1
p1
Q2, A2, v2
p2
v1 > v2
p1 < p2
Ignoring friction, change in elevation, heat transfer, chemical reaction, etc. the total
mechanical energy of a liquid remains constant. As a pipe constricts, v (KE) increases at
the expense of pressure, i.e., as velocity increases, pressure must decrease.
Thus we expect a relationship between the pressure in a moving liquid and its speed.
This was derived by Daniel Bernoulli (1700-1782), a Swiss mathematician and physicist
and it is thus called Bernoulli’s equation.
p1 + gz1 + ½12 = p2 + gz2 + ½22
where p1, z1, & 1 are pressure, height above a reference level, and speed of a liquid of
mass density  at point 1 in a body of liquid and p2, z2, & 2 are values of these quantities
at point 2.
According to Bernoulli’s equation, the quantity (p + gz + ½v2) is a constant at all points
in an incompressible liquid with negligible viscosity that undergoes laminar flow, i.e., no
losses due to friction are considered. If viscosity (internal friction) is not negligible, then
the quantity (p + gz + ½v2) decreases in the direction of flow.
FLUID FLOW
Page 4 of 51
DERIVATION OF BERNOULLI’S EQUATION
W = Fd = weight  height (of lifting) = mgh = gravitational PE
Recall:
KE = ½m2
Pressure energy (PV work) = PV
i.e., (F/A)(V) = Fd
3
units;
Pam = (N/m2)(m3) = Nm = J
For a closed system
E1 = E2
Considering mechanical energy (ignoring heat and chemical energy) then:
pressure E + GPE + KE = constant
1.
PV
+ mgz + ½m2 = constant
(Units = Nm or lbfft)
Convert V term to mass using  = m/V  V = m/ we obtain:
 m
mv2
p     mgz 
 cons tan t
2
 
“h” is height of a particle not height of a column of fluid above it so it was changed to “z”
to avoid confusion; reserve “h” for height of a fluid column, i.e., pressure.
Dividing all terms by ‘m’ since it is common, yields a second form of Bernoulli’s eqn.
p
v2

gz

 cons tan t
2.
units are height2/time2 (e.g., m2/s2 or ft2/s2)

2
In equation 1, mass terms can be converted to V terms using m = V yielding:
PV + (V)gz + ½V2 = constant
Canceling the common V term yields a third form of Bernoulli’s eqn.
3.
p + gz + ½2 = constant
By dividing eqn 2. by g
4.
or
p
v2
z
 cons tan t
g
2g
units are pressure (N/m2 or lbf/ft2 or dyn/cm2)
dividing eqn 3. by g we obtain a fourth equation.
units are height (e.g., m. or ft.)
By substituting p = gh into eqn. 4. we obtain
5.
hz
v2
 cons tan t
2g
  gh
v2
z
 cons tan t . Canceling yields:
g
2g
units are height (m, ft, etc.)
Remember, ‘h’ is pressure expressed as height of a column of the working fluid above a
point (units are ft, m, etc.). ‘z’ is the relative elevation of a point or particle also
expressed in m or ft. Do not confuse these!
Since weight density, D = g we can substitute  = D/g in eqn. 3 to obtain:
 D
 D  v2
p    gz   
 cons tan t and canceling ‘g’ in the second term yields:
 g
 g 2
FLUID FLOW
Page 5 of 51
p  Dz 
6.
Dv 2
 cons tan t
2g
units are pressure (N/m2 or lbf/ft2)
Equation 5. is often the simplest form to use. If we now consider this equation in an open
system we must also include terms for energy input from pumps (pump head, hp) and
energy loss due to friction (friction head, hf) and equation 5. becomes:
7.
hp + h1 + z1 + 12/2g = h2 + z2 + 22/2g + hf
where:
hp = pressure head added by pump
h1 = pressure head at point 1
h2 = pressure head at point 2
z1 = elevation of point 1
z2 = elevation of point 2
1 = velocity at 1
2 = velocity at 2
hf = head loss due to friction.
All terms are in height of the working fluid.

Equations 3, 5 and 7 are most commonly used.
FLUID FLOW
Page 6 of 51
SPECIAL CASES OF BERNOULLI’S EQUATION
In many cases the pressure, height, or speed of a liquid is constant and simplified forms of
Bernoulli’s equation result.
1. Liquid at Rest: When a column of liquid is stationary, equation 3. becomes:
p1 + gz1 = p2 + gz2 
p = (p2 - p1) = g(z1 – z2)
This is simply the pressure-depth equation (p = gh)
Extra problem: The gage pressure of water standing in a vertical pipe is 0.20 atm at a
certain depth. Calculate the gage pressure in the water at a point 2.0 m higher.
(ans. = 2.6” H2O)
2. When p1 = p2: When an open tank is draining, both ends are at atmospheric press.
gz1 + ½12 = gz2 + ½22
patm
If the orifice is small compared to the cross section of
the tank, the liquid level (draining under gravity) will
fall slowly enough that the liquid speed at the top of
the tank (1) is negligible (assume 1 = 0), then:
gz1 = gz2 + ½22
Solving for  we get:

½22 = g(z1 - z2) = gz
v  2gz
(Toricelli’s Theorem)
Note that the speed with which a liquid is discharged under gravity is the same as the
speed of a body falling from rest from height, h, and is independent of the density of the
object. Recall intro physics, the KE of an object at any point all comes from PE.
½m2 = mgh 
v  2gh
The flow rate at which liquid discharges through an orifice can be calculated as follows:
Since Q = A,  Q = A 2gh
(Flow from an orifice)
Problem 3: Calculate the flow rate at which water will leak through a hole 1.0 cm2 in
area at the bottom of a water tank which is filled to a height of 3.0 m. Ans: 0.77 L/s
FLUID FLOW
Page 7 of 51
3. Constant Height: p1 + ½12 = p2 + ½22
Pressure and speed are inversely related.
As  increases (through a constriction),
p decreases.
This relationship is responsible for such
phenomena as lift on an airplane wing,
movement of sailboats at angles to the
wind, venturi action in carburetors,
Bunsen burners, and paint sprayers,
aspirators (nebulizers) in perfume bottles
and AA spectrophotometers.
h = p
h
Extra Problem: Calculate the difference in height (in cm) of H2O in the pressure taps
shown above when the velocity of H2O is 10.0 cm/s in a pipe with diameter of 2.00 cm,
which then constricts to a diameter of 1.00 cm. (ans = 0.765 cm)
Problem 4: The pressure is 25 ft. of water and flow velocity is 100 ft/s in a 2.00 in. I.D.
pipe. The pipe widens to 28 in. I.D. and rises 5.0 ft. from its original level. Calculate the
pressure (in psi) in the elevated and enlarged section of pipe.
Ans: 76 psi.
Problem 5: Water flows in a pipe in the direction of ‘a’ to ‘b’ at a rate of 8.50 m3/min.
The diameter at ‘a’ is 30.4 cm. and the diameter at ‘b’ is 15.2 cm. The pressure at ‘a’ is
1.03  105 N/m2. Calculate the pressure (in Pa) at ‘b’ if the center at ‘b’ is 60.4 cm lower
than the center at ‘a’.
Ans: 8.03  104Pa
FLUID FLOW
Page 8 of 51
LIQUID HEADS IN FLUID FLOW
1. Vertical depth in ft. or m. of any point below the surface of a liquid is called pressure
head or hydrostatic head, i.e., it is a pressure usually expressed in ft. or m. of the liquid.
Recall p = Dh  h = p/D
Extra Problem: Using fresh water as the fluid, calculate the depth of water that would
exert a force of 1.00 atm. Use the equation above and use p and D in British and SI units.
(ans. h = 33.9 ft and h = 10.3 m)
2. A drop in pressure in moving water due to friction is called friction head. It is
directly proportional to the total area of the rubbing surface as well as the roughness of
the surface and the velocity of the fluid. Abrupt changes in cross-sectional area or
direction of flow also causes increased friction head.
3. Pressure in a pipe or channel due to the velocity of liquid particles is called velocity
head. From Toricelli’s Theorem, v  2gh

hv = 2/2g
This indicates that a pressure due to velocity is equivalent to the hydrostatic head, which
would produce such a velocity (at least theoretically, in absence of friction). In real
systems, friction head causes velocity head to be less than pressure head.
In other words, a static fluid under pressure would move with a velocity calculated by
Toricelli’s theorem if a hole (or valve) were opened in the side of a pipe or vessel.
Neglecting friction, all static pressure energy would be converted to velocity.
Problem 6.: Water discharges under gravity from a tank with a very large surface area
through a 6 in. I.D. pipe. The depth of water above the opening is 12.0 in. As water
leaves the pipe it travels a horizontal distance of 3.45 ft. through the air, before reaching a
pond which is 3.68 ft. below.
a) Calculate the actual horizontal velocity of water as it leaves the pipe. Ans: 7.22 ft/s
b) Calculate the theoretical velocity of water in the pipe.
Ans: 8.02 ft/s
c) Calculate the coefficient of velocity (Cv), (actual v/theor. v)
Ans: 0.90
d) Calculate the actual flow rate of water
Ans: 1.42 ft3/s
e) Calculate the hydrostatic head (hh), velocity head (hv), and friction head (hf). Friction
head is simple the head (pressure) lost to friction, i.e., (hf = hh – hv)
Ans: hh = 1 ft. H2O, hv = 0.81 ft H2O, hf = 0.19 ft. H2O
Note: For all Bernoulli equations use ‘h’ and ‘z’ as actual heights (actual dist.). Do not
factor in density to calculate an equivalent height of water but understand that the answer
will be in height of solution (not water). Then apply density to convert to usual pressure
units.
FLUID FLOW
Page 9 of 51
Problem 7: A pump draws a solution (s.g. = 1.84) from a storage tank through a 3.00 in.
I.D. pipe. The velocity in the suction line is 3.00 ft/s. The pump discharges through a
2.00 in. I.D. pipe into an overhead tank. The end of the discharge is 50.0 ft. above the
level of solution in the feed tank. Friction losses are 10.0 ft. of solution.
a) What pressure must the pump develop in psi?
Ans: 48.3 psi.
b) What is the HP of the pump assuming it is 65% efficient?
Ans: 2.86 H.P.
FLUID FLOW
Page 10 of 51
H
10'
E
F
G
50'
A
10'
B
P
D
C
Pressure head in feet (’) at designated points in the process diagram
A
B
C
D
E
F
G
H
I
hp
0’
0’
0’
100’
0’
0’
0’
0’
0’
hh
0’
10’
1’
1’
47’
30’
10’
0’
0’
Z
10’
0’
0’
0’
50’
50’
50’
60’
50’
v2/2g
0’
0’
8’
8’
8’
20’
10’
0’
35’
hf
0’
0’
1’
1’
5’
10’
40’
50’
25’
Total
10’
10’
10’
110’
110’
110’
110’
110’
110’
hh = hydrostatic head (gauge pressure or pressure due weight of liquid above a point or
pressure applied by a pump).
hp = pump head (pressure due to a pump)
Z = elevation
2
v /2g = velocity head (pressure of moving fluid)
hf = head loss due to friction
Point 1
( hh + Z + v2/2g )

=
Point 2
( hh + Z + v2/2g )
The 1st Law of Thermodynamics tells us that energy may change forms but is conserved.
The basic form of Bernoulli’s equation above shows this (in terms of pressure), but only
applies in this form to ideal systems where no pressure energy is lost due to friction and
no pressure energy is added by pumps. Terms for both head loss due to friction and
increase in head due to a pump (pump head) are thus included.
( hh + Z + v2/2g ) + hp
= ( hh + Z + v2/2g ) + hf
Note that hf is really hf , i.e., the increase in head loss due to friction between 2 points.
FLUID FLOW
Page 11 of 51
VISCOSITY
Viscosity is internal friction or resistance to flow. Absolute or dynamic viscosity () is ...

SI units :
shear stress
F/A
F  dx

 =
velocity gradient dV / dx
A  dV
(kg  m  s -2 )(m)
(m 2 )(m  s -1 )

kg
 Pa  s
m s
cgs units :
(g  cm  s -2 )
(cm 2 )(cm  s 1 )

g
 poise (P)
cm  s
10 poise (P) = 1 Pa  s  1 poiseuille (Pl) = 1000 centipoise (cP)
1 cP = 6.72  10 -4
pdl  s
lbm
 6.72 10 -4
ft  s
ft 2
= 2.089 10 -5
slug
lbf  s
= 2.089 10 -5
ft  s
ft 2
Kinematic Viscosity () is absolute viscosity adjusted for density of the fluid.



cgs units :
SI units :
1
poise  g  cm 3  cm 2


 stokes (St)

s
g/cm 3  cm  s  g 
3
2
 kg  m  m






s
 m  s  kg 
m 2 10 4 cm 2

 10 4 stokes  10 6 centistokes (cSt )
s
s
 lbf  s  ft 3   lbf  s  ft 3  ft  ft 2
 lbm  ft 3  ft 2
 
  




British units : 
or 
2 
2 
2 
s
slug
s
ft
ft
lbf

s
 ft  s  lbm 



 

1.076  10-5 ft2/s = 1 centistokes
One advantage of kinematic viscosity is that it eliminates confusion of the British units
for absolute viscosity in lbm vs. lbf.
Viscosity has been covered in detail in physical chemistry so we will only review some
units of viscosity and their conversion factors here. Viscosity is needed to calculate
Reynolds number in the next section.
FLUID FLOW
Page 12 of 51
FLOW OF LIQUIDS
When a liquid flows through a passage such as a pipeline, it will encounter resistance due
to friction and viscosity. If the average velocity of the fluid is very low, the fluid will
flow in parallel lines along the sides of the pipe. In such a case the flow is said to be
laminar, viscous, or streamline. If the velocity is increased beyond a certain critical
value when eddy currents start to form, the laminar flow pattern is changed and the flow
becomes turbulent.
The critical value of the velocity occurs when the Reynolds number (Re) is equal to about
2000. In practice it is found that:
1. If the Re value is less than 2,000, the flow is laminar, viscous, or streamline.
2. If the Re value is between 2,000 and 4,000, the flow is critical, i.e., in a transition
phase, changing from laminar, to turbulent or vice versa.
3. If the Re value is greater than 4,000, the flow is turbulent.
TURBULENT FLOW
LAMINAR FLOW
Velocity
Velocity
r
0
r
distance from center of pipe
r
0
r
distance from center of pipe
COLOUR-BAND METHOD FOR DETERMINING CRITICAL VELOCITY
The critical value may be found by allowing water to flow through a glass tube and
injecting a thin stream of colored liquid (e.g., KMnO4 solution) into the center of the
stream. As long as the velocity is below the critical value, the colour band will remain in
a straight line pattern along the center of the stream. As soon as the velocity exceeds the
critical value, the colored band will be broken up because of turbulence and will mix with
the water.
FLUID FLOW
Page 13 of 51
REYNOLDS NUMBER
The wide variety of pipes and other conduits that must be considered, along with an
equally wide variety of fluid properties, makes it difficult to describe flow conditions in a
generalized manner. However, the ratio of internal forces to viscous forces for a flowing
fluid has been found to be a characteristic of the flow conditions. This ratio, known as
the Reynolds number, is defined by the following equation.
Re 
d v 

or
 d    D

 
   g 
d = diameter (e.g. cm, m, or ft.)
 = absolute viscosity
-1
-1
v = average or bulk velocity (e.g., cms , ms or fts-1 )
g = grav. accel.
-3
-3
-3
-3
 = mass density (gcm , kgm , slug/ft or lbm/ft ) D = weight density (e.g. N/m3, etc.)
Re is dimensionless and  2000 for laminar flows but > 4000 for turbulent flows.
2000 is called the lower critical velocity and 4000 is called the upper critical velocity.
2000 < Re > 4000 is a zone of transition between laminar and turbulent flow.
(cm)(cm  s -1 )(g  cm -3 )
cgs :
 dimensionl ess
( g  cm 1  s 1 )
(ft)(ft  s -1 )(lbm  ft -3 )
British :
(lbm  ft 1  s 1 )

d
Since:  =
Re 


cgs:
( cm)( cm  s 1 )
( cm 2  s 1 )
SI:
or
(m)(m  s -1 )(kg  m -3 )
SI :
= dimensionl ess
(kg  m 1  s 1 )
(ft)(ft  s -1 )(slug  ft -3 )
 dimensionl ess
( slug  s  ft  2 )
( m)( m  s 1 )
( m 2  s 1 )
British:
( ft )( ft  s 1 )
( ft 2  s 1 )
all dimensionless
Extra Problem: Oil (sg = 0.765,  = 1.70 centipoise) is pumped through a 4 inch
Schedule 40 steel pipe at a flow rate of 150 Imperial gallons/minute. Calculate the
Reynolds number (to 3 sig figs) for the fluid under these conditions and state whether the
flow is turbulent, laminar or transitional. (ans. Re = 63,500).
FLUID FLOW
Page 14 of 51
FLUID FLOW MEASUREMENT
Flow has been called the controlling variable in fluid process industries, since by
controlling flow, all other process conditions are controlled. Flow measurement is
necessary for determining production rates and for accounting purposes. Although it is
one of the most frequently measured variables, it is the most difficult to measure
accurately.
A flowmeter is an instrument capable of measuring flow rates in circular pipes. Weirs are
flow measuring devices used in open channels such as ditches and open drains where the
fluid has a free surface.
There are many different kinds of flowmeters. We will consider 5 classes of them.
1. Head meters measure volumetric flow indirectly by means of a detecting element
(primary element, e.g., orifice) that produces a pressure-differential effect, which is
observable on a measuring device (secondary element, e.g., manometer). Head meters
include: orifice plates, venturi tubes, and Pitot tubes.
2. Variable area flowmeters adjust their orifice size (area) as flow rates change, e.g.
rotameters.
3. In positive displacement meters the fluid passes in successive quantities or
displacements, filling and emptying in a cyclic manner, containers of known capacity.
The flow is read by counting revolutions.
The rotary vane meter is one example among many.
Metering pumps that deliver known volumes of liquid include: piston, diaphragm,
gear and peristaltic.
4. Velocity-principle volumetric flowmeters respond to the velocity of the flowing fluid.
There are many meters that operate on this principle. One particularly useful for
corrosives is the magnetic (‘mag’) flow meter, discussed later.
5. Mass flow meters respond directly to the true mass flow rate rather than volumetric
flow rate. In general the density of the fluid coupled with its velocity creates a
measurable momentum (m  v) as it impacts various detecting elements. One example
is the gyroscopic type in which the process fluid is channeled through a spiraling path
within the meter. The centripetal forces are proportional to the flow rate.
FLUID FLOW
Page 15 of 51
FLUID FRICTION
Fluids in motion are subjected to certain resistances, which are assumed to be caused by
friction, i.e., viscosity, which is resistance to sliding between two adjacent layers of the
fluid. This is the result of molecular attraction between particles of the fluid.
It is generally found that for steady viscous or streamline motion of a liquid, i.e., for
velocities less than the critical value, the frictional resistance is:
1. Proportional to the velocity
2. Proportional to the area of surface in contact
3. Independent of the pressure
4. Greatly affected by temperature changes
5. Independent of the nature of the surface in contact
Point 5 infers that, when a liquid is flowing past a surface with a velocity of less than the
critical velocity, a film of stationary liquid is formed over the surface. The resistance to
flow in the rest of the pipe is caused by viscosity only.
When the flow is increased beyond the critical velocity, i.e., when turbulent flow occurs,
frictional resistance is:
1. Proportional to the square of the velocity
2. Proportional to the mass density ()
3. Independent of pressure value
4. Independent of temperature effects which are negligible
5. Directly proportional to the area of surface in contact
6. Dependent on the surface in contact
Point 6 infers that the internal surface of the pipe is an important factor, i.e., the smoother
the surface the lower is the frictional resistance.
FLUID FLOW
Page 16 of 51
HEAD LOSS PER UNIT LENGTH OF PIPE
Osborne Reynolds conducted experiments to measure the drop in pressure (head loss) as a
fluid travels through a pipe. The term “head loss per unit length of pipe” (i) is given by:
h
head loss between two points of pipe
i L 
L
distance between these points
His results showed that i increases with fluid velocity according to: i = kn ,
where  is the average velocity, k is some constant, and n = 1 for Re < 2,000 (which
yields a direct, linear relationship for laminar flow) but n  2 for Re > 4,000 (yielding a
direct, exponential relationship between velocity and head loss for turbulent flow).
overflow
water supply
B
A
v1
D
C
v2
F
E
v3

Explain the relative magnitudes of v1, v2 and v3.

Explain the relative fluid heights in the tubes (pressure taps) above A, B, C, D, E
and F
FLUID FLOW
Page 17 of 51
DISCHARGE THROUGH ORIFICES AND NOZZLES
A flowing fluid that passes through an
orifice or constriction of any kind will
contract in area as in the case of water
from a tap or faucet. The contraction is
caused by the inward motion of the fluid
around the opening as it rushes inwards
to funnel through the opening. The
point of minimum cross-sectional area,
called the vena contracta, is also the
point of highest velocity of the flowing
fluid.
vena contracta
The extent of contraction is called the coefficient of contraction (Cc) which is:
area of jet at vena contracta
Cc 
area of orifice
There is a certain amount of frictional resistance at the sides or edge of the orifice. The
resistance is lowest for sharp-edged orifices. In such cases the Cc approximates 0.65.
A related term is the coefficient of velocity (Cv).
Cv 
actual velocity at vena contracta


theoretical velocity
2gh
The difference between the theoretical velocity (as given by Toricelli's theorem) and the
actual measured velocity is caused by friction at the orifice edge and is very small for
sharp-edged orifices. A typical value for C is approximately 0.97, depending upon the
head, shape, and size of the orifice.
A third coefficient is of considerable importance in fluid flow measurement; the
coefficient of discharge (Cd):
Cd 
actual discharge
Q

theoretical discharge A  2  g  h
Again, owing to resistance, the actual discharge (flow rate) is usually less than theoretical.
FLUID FLOW
Page 18 of 51
FLOW RATE of fluids in pipes is often calculated by installing a venturi tube or orifice plate
and measuring the pressure drop (p) that occurs.
With open manometers mounted above the pipe (Figure 1.), the manometer fluid will be the
same as the fluid flowing through the pipe. Since the manometers in this system are open to
atmosphere, air lies above the manometer fluids, hence p can be calculated directly from the
difference in height (h) of manometer fluid levels, i.e., p = fluidgh.
Figure 1.
air
air
h = ? cm.
h = ? cm.
ORIFICE PLATE
VENTURI
d2
d2
d1
d1
A more practical arrangement is to use enclosed Hg manometers as shown in Figure 2. Since Hg
is very dense (s.g. = 13.6), the manometer can be much smaller. In this case, air is trapped above
the Hg and p is simply Hggh.
air
air
Figure 2.
Hg
Hg
h = ? cm.
h = ? cm.
ORIFICE PLATE
VENTURI
d2
d2
d1
d1
A third arrangement (Figure 3.) places Hg manometers below the pipe. In this case the section of
manometer tube above the Hg will be filled with the process fluid. This creates a ‘bifluid’
manometer and thus p = [.gh(for Hg) - gh(for process fluid)] or p = (Hg - fluid)gh or simply
 =gh.
ORIFICE PLATE
VENTURI
d2
d2
d1
process
fluid
process
fluid
h = ? cm.
Hg
FLUID FLOW
d1
h = ? cm.
Figure 3.
Hg
Page 19 of 51
A SINGLE EQUATION FOR FLOW RATE IN VENTURIS & ORIFICE PLATES:
The equation for calculating fluid flow through a venturi is derived from Bernoulli’s
equation:
h1 + z1 + 12/2g =
h2 + z2 + 22/2g (where 2 is in pipe, 1 is in throat)
eqn 1)
Since elevation is constant, (Z1 = Z2), this term is cancelled from both sides of the
equation and the equation then rearranged to solve for p as p2 – p1 (where p2 > p1).
p = (p2 – p1) = ½ (v12 – v22)
eqn 2)
From A1v1 = A2v2, we know (by rearrangement) that v1 = v2(A2/A1)
After substituting for v1 in eqn 2) we obtain
p = (p2 –p1) = ½ ( [v2(A2/A1)]2 – v22]
After factoring out v22 we obtain:
p = (p2 –p1) = ½ v22 [(A2/A1)2 – 1]
Rearranging to isolate v22 and then taking the square root we obtain the following for v2:
v2
2


 2p 





2
  A2   1 
 A 

 1 

2p


v2 
 d2

 d1
4

  1

Units =
ft
2
s2

ft m cm
, ,
, etc.
s s s
difference of
mass densities of
manometer fluids.
mass density of
process fluid
2p
and since Q = Av, then
Q  A2  v 2  A2 

 d2

 d1
4

  1

Units =
ft 3 m 3
,
, etc.
s
s
eqn 3)
p can be calculated as gh for a single fluid manometer (Fig 1 & 2) or as gh for a
bifluid manometer (Fig 3), where  is the difference in the mass densities of the two
different manometer fluids. See the following two equations
2    g  h
2  g  h
Single fluid:

Q  A2 
 d2

 d1
4

  1

or
Bifluid:
Q  A2 

 d2

 d1
4

  1

 Since the manometer and process fluid are the same fluid
 cancels and the equation is further simplified.
Q  A2 
FLUID FLOW
2  g  h 
 d2

 d1
4

  1

Page 20 of 51
Equation 3), although derived for venturis will also work for orifice plates.
However, because greater frictional loss is incurred across an orifice plate, a
correction factor (called the Coefficient of Discharge), Cd, is included for orifice
plates.
2    g  h
Q  C d  A2 

 d2

 d1
4

  1

Cd is a dimensionless factor that is empirically
determined, i.e., determined by actual flow rate
measurements. Typical values range from 0.78 to
0.98.
The Orifice Plate:
The orifice plate is exactly what its name implies; a plate with an orifice in it. There
remains a permanent head loss downstream of the orifice; pressure does not fully recover
to the pressure ahead of the orifice. The loss can be attributed to the great amount of
friction (turbulence) induced as the flow lines change abruptly through the orifice.
concentric
orifice plate
eccentric
orifice plate
segmental
orifice plate
Orifice Plate Shapes:
1. Concentric: In this case the hole in the orifice plate is on the same center as the pipe.
This is most common.
2. Eccentric: The orifice plate hole is not centered with the pipe center and is usually
situated tangential to the pipe bottom. This is used for slurries in which suspended
solids would build up around the orifice plate.
3. Segmental: The orifice hole is not circular; but rather a partial circle.
FLUID FLOW
Page 21 of 51
Orifice Faces: .
are usually either sharp edged or
round-edged
sharp-edged
orifice
round-edged
orifice
Pressure taps or manometer connections are most important. In order to obtain valid
results, the downstream pressure tap should be located directly over the vena contracta.
However, the position of the vena contracta and its diameter are not known exactly since
these change depending upon the flow rate.
The 3 generally accepted positions recommended by manufacturer’s and ISA are:
1. taps located right at the flanges on either side of the orifice (integral with the flange)
2. taps 1 D (pipe diameter) upstream and ½ to 1 D downstream (approx. vena contracta
taps)
3. taps 2.5 D upstream and 8 D downstream (gives total pressure loss; there is no further
pressure recovery).
Problem 9: Calculate the flow rate of water flowing through a 12.0 in. diameter pipe
given that when passing through a 6.00 in. diameter orifice plate a pressure differential of
3.50 ft. is measured with a manometer containing a fluid with s.g. = 1.25.
Assume the manometer is below the pipe and assume a Cd of 0.970. Ans: 1.48 ft/s
Obtaining Cd:
1. Data and graphs are available from manufacturers, which relate Cd to factors such as
orifice to pipe diameter and position of pressure taps.
2. Cd can be measured in the lab or process by taking pressure readings with known flow
rates. Test conditions should approximate process conditions.
3. A calibration curve can be prepared in the process, which plots pressure differential
versus flow rates again using known flow rates. Cd is thus unnecessary.
The orifice meter is convenient and inexpensive. It can usually be fabricated on the work
site and sufficient data is readily available in the literature so that calibration is often not
necessary. It yields reasonably good accuracy when properly installed and calculations
are carried out correctly. The orifice must be installed in a location devoid of turbulence
from the piping system and so is usually installed in a long straight section of piping.
Accuracy: The square root relationship between pressure differential and flow makes the
orifice plate impractical for low flow rates. At higher flow rates, e.g., Q > 5 gal/min,
orifice plates give an accuracy of  5%.
FLUID FLOW
Page 22 of 51
Venturi Tubes
This is an accurately machined piece and few shops are equipped to fabricate it. It is
normally purchased.
This design gives a Cd of 0.97 to 0.98. It
gives better accuracy than other pressure
differential meters and creates less permanent
head loss but is more expensive.
Problem 10: Calculate the flow rate through a venturi meter having a pipe diameter of
10.0 cm and throat diameter of 5.0 cm if the difference in the height of water in the
manometer is 6.0 cm. p is measured by pressure taps above the pipe. Ans: 2200 cm3/s
Problem 11: Using the same pipe and venturi dimensions as in problem 10 (above),
calculate the flow rate of water when the difference in height of Hg in a manometer is
0.476 cm. Water is the fluid being measured. The manometers are below the pipe.
Ans: 2200 cm3/s
Comparison of Head Meters
Flow meters, like the orifice plate and venturi meter, that measure pressure drop to
determine flow rate are called head meters. Head meters work best at higher flow rates.
At lower rates, accuracy suffers dramatically.
Orifice plates are inexpensive, easy to install and maintenance free. They are usually
accurate to within  3-5%. However, they will wear with time (causing drift in
readings), have a variable & unknown Cd so they may require on site calibration and they
induce high permanent pressure drops (hf).
Venturis, by comparison, are more accurate and stable over time. They also introduce
only a small pressure drop. However, they are considerably more expensive than orifice
plates.
FLUID FLOW
Page 23 of 51
Pitot Tubes
impact pressure
static pressure
Pitot tubes, also designated stagnation
meters or impact meters, are included in
the general class of Head meters
because flow rates are deduced from a
pressure head created by the impact of a
moving fluid.
Consider a bent glass tube submerged in
a flowing liquid and with its submerged
end pointing directly upstream.
x is a point just outside the tube entrance
y is a point just inside the tube entrance
x
y
Total energy at x = Total energy at y
H + v2/2g = H + h
h = v2/2g

  2g  h

  C 2g  h
where C is a correction factor called the Pitot tube coefficient. C is  1 when Re > 3,000.
The Pitot tube can also be inserted into closed pressurized systems (pipes and vents). In
this case the tube will measure static head in addition to velocity head. To eliminate the
static head, the Pitot tube is encased with an outer tube, which also contains an opening
but at right angles to the flow rather than in line with the flow. The outer port thus only
measures static head. The two ports are connected to opposite ends of a manometer (or
some other p measuring device) and the differential pressure recorded is only the
velocity head (impact pressure).
Flow is thus measured by the same square law formula previously discussed. Installation
is simple and permanent head loss is negligible. However, measurement is an indication
of velocity at one point only, in contrast to other flow elements that measure the average
velocity of the entire stream. To obtain an average velocity it is necessary to traverse the
conduit both horizontally and vertically with the Pitot and average the results. This factor
plus the difficulty in properly positioning the Pitot, the low differential pressure produced,
and the tendency to pluggage has resulted in the Pitot tube being used in limited
applications, e.g., sampling rather than continuous process measurement.
FLUID FLOW
Page 24 of 51
With proper positioning and suitable flow rates, accuracy can attain  1%. Pitot tubes are
used for measuring gas or liquid flows at up to 2000 psi and 1200 ºF.
As previously discussed, flow velocity is not uniform over the entire cross section of a
pipe but in fact a profile of velocities exists beginning with very low flows at the pipe
wall and maximum velocities at the center.
Problem 12: Calculate the Pitot tube coefficient (C) for the following set of data:
v (ft/s)
1.86
3.92
4.20
7.80
h (in. H2O)
0.76
1.73
3.50
14.40
Ans: 1.03
Head Meters and Automatic Process Control
Although converting pressure differentials to flow can be done with the aid of tables or
calculators, for use in automatic process control suitable transducers and microprocessors
are required. A number of the pressure measuring devices already discussed in the unit
on pressure are used in industry. In many instances, temperature and density sensors
(densitometers) are coupled to a computer and volumetric flow can be converted to mass
flow, which is generally more desirable.
CLASS 1 flow meters are ‘DIFFERENTIAL (HEAD) METERS, just discussed, i.e.,
orifice plates, venturis and Pitot tubes, for example.
FLUID FLOW
Page 25 of 51
CLASS 2 flow meters are VARIABLE AREA METERS, the most familiar example
being rotameters.
Rotameters
Rotameters are classed as variable-area meters.
They consist of a tapered glass tube in which a
suitable "float" is buoyed up to a height that
corresponds to the flow rate of the fluid
moving up the tube. The flow rate is
proportional to the height of the float.
The basic eqn for flow through orifice, nozzle,
and venturi meter is ... Q  A  Cd 2g  h
outlet
metal or glass
‘float’ suspended
in tapered tube
by rising fluid.
In the rotameter, the downward force is the
weight of the float (often steel, glass, etc.) and
ROTAMETER
the upward force is a combination of friction
and impact from the rising fluid. However the
p (or h) is constant at any position of the
inlet
float, the area in the annular space between the
float and the tapered walls changes to keep
upward force constant.
Rotameters are suitable for low or high flow measurement, limited
only by the construction of the meter. Rotameters are used for liquids
and gases. The float can be connected to an armature and its position
thus can provide variable inductance to an electronic transducer as
part of an automated process control loop.
Float height is sensitive to density changes (and thus temperature
changes). Density compensation is accomplished by using a float
whose density is one half that of the fluid. Rotameters are made in a
large variety of materials and in sizes from 1/8 in. to 6 in. nominal
pipe size.
Maximum flow capacities range from 2 cm3/min to 900 gal/min for
water and 70 cm3/min to 1000 SCFM for air. Accuracy is normally
about  2% of full scale. At 10 % flow,  2 % full scale translates to
 20 % of rate, however, by comparison, a head meter would be
useless at  5 to 10 % of full scale at the same flow rates.
OMEGA
ROTAMETER
FLUID FLOW
Page 26 of 51
CLASS 3.
POSITIVE DISPLACEMENT METERS
The positive displacement meter for liquid service is made up of a measuring chamber
and a scaling section between the inlet and outlet connections. It splits the flow of the
liquid into separated measured volumes based on the physical dimensions of the meter
and counts them or ‘totalizes’ them. The measuring chamber follows a fixed path and on
each revolution of the meter is filled at the inlet and discharges at the outlet. The total
quantity of fluid passing through the meter in a given time is the product of the volumes
of the chambers and the number of fillings. Flow is measured by counting the number of
revolutions.
Since the moving components of the meter are in the flowing stream, a pressure drop
occurs. Close mechanical clearances are necessary to minimize leakage from one
measuring chamber to the next. Thus operation is enhanced by the lubricity of the
flowing fluid, and leakage decreases as fluid viscosity increases. Solid particles in the
fluid cause rapid wear and loss of accuracy. A strainer is required upstream from each
positive displacement meter to prevent foreign particles from entering the meter.
Many positive displacement flow meters are used industrially. One type is shown.
Rotary Sliding Vane
vane
rotor
Spring-loaded vanes seal increments of
liquid between the eccentrically mounted
rotor and the housing and transport them
from inlet to outlet. The vanes slide in and
out of the rotor as it rotates. As the liquid
enters the meter, the vanes extend to
enclose the measuring chamber. At the
outlet, the vanes retract and discharge the
volume of liquid.
FLUID FLOW
Page 27 of 51
Class 4
Volumetric Flow Meters
Electromagnetic Meter (Mag Flow Meter) are one of the most important examples
of volumetric flow meters.
The mag flow meter is based on the principle that
electrical currents are induced in a conductor that is
moved through a magnetic field. According to
Faraday’s Law, the voltage produced is proportional to
the intensity of the magnetic field and the velocity of the
conductor.
The flowmeter consists of straight section of pipe (no
pressure drop or clogging from solids), encased with the
windings of an electromagnet. A magnetic field
produced by these windings induces a current in a
secondary circuit, which is completed via two electrodes
that make electrical contact through the flowing liquid.
In order to serve as a conductor, the fluid requires a
minimum conductivity of 50 micromhos/cm (5000
microSiemens/m). All electrical hardware must be well
shielded to prevent interference from stray magnetic flux
from nearby equipment.
Accuracy is as good as  1%. Mag meters are not
affected by dirt or varying viscosity but are suitable for
liquids only.
FLUID FLOW
The pipe is straight through
(unobstructed) and encases
electromagnetic windings
OMEGA MAG
FLOW METER
Page 28 of 51
OPEN CHANNEL FLOWMETERS
Most open channel flow measuring devices are sophisticated adaptations of level
measurements practiced by the Egyptians on the Nile over 4000 years ago. Measurements
are made by the use of a calibrated restriction inserted into the channel, which changes
the level of the flowing liquid in or near the restriction. The shape and dimensions of the
restricting structure establish a known relationship. Flow rates in an open channel are
thus derived from a single measurement of liquid level in the channel.
Weirs
A weir is an open dam with an opening
or notch in the upper edge through which
the liquid flows. The weir is low in cost,
relatively easy to install, and quite
accurate ( 5%) if carefully installed.
However, it normally operates with a
significant loss in head, and accuracy is
very sensitive to the approach velocity of
the liquid. Thus it requires an upstream
weir box or silting chamber that permits
the flowing liquid to attain low velocity
before reaching the weir and serves as a
chamber to settle out debris.
notch
weir
h
The weir must be cleaned regularly to prevent accumulation of solids and sediment in the
upstream side that would otherwise adversely affect accuracy.
Weirs are classified according to the shape of the notch; common types are trapezoidal
(Cipolletti), rectangular, and triangular (V-notch). The V-notch weir is well suited to low
flows (0.01 - 10 cfs). The rectangular weir is used in flow ranges of 0.5 - 300 ft3/s. The
Cipolletti weir has a slightly wider range and a simpler level/ flow relationship but is
slightly less accurate. The top edge of a weir is thin or beveled so that the overflowing
liquid does not contact any part of the weir’s downstream face but is projected past it as a
sheet of water. The water surface downstream must be far enough below the bottom of
the notch so that air moves beneath the sheet of water. The condition is known as free
critical flow and is required for dependable measurements.
The sharp-crested weir can be damaged by floating objects in the flow or even by the
force of large quantities of liquid. It is thus used for clean liquids and relatively small
flows. Broad-crested weirs are much more robust, being constructed of concrete, and are
normally use for larger sizes of open channels or rivers. Broad-crested weirs, in practice,
are usually pre-existing structures such as dams and levees. Discharge coefficients are
obtained by calibrating the weir in place or by model studies.
FLUID FLOW
Page 29 of 51
Flumes
A flume is a specially shaped open
channel flow section which consists
of a converging section to restrict
flow, a throat section and/or a change
in channel slope that results in
increased velocity, and a diverging
section that ensures that the
downstream level is less the level in
the converging section.
A float well, connected to the side of
the flume contains a graduated float
that measures the depth of water
moving through the flume.
float well
throat
size
W
Parshall Flume
(top view)
A flume can measure a higher flow rate than an equally sized weir but with much smaller
head loss than the weir. It is better suited to flows carrying solids or sediment, since its
high velocity makes it self-cleaning. Disadvantages include poorer accuracy ( 10 %)
and a more costly installation.
The flow through the flume can be classified as subcritical, critical, or supercritical by the
Froude number, which is a ratio of inertial forces to gravity forces in the flowing liquid.
Since inertia is related to throat velocity, the flume increases velocity by increasing
channel slope and reducing throat width. Flumes that operate at subcritical velocity
require two level measurements, one at the inlet and one at the throat, to determine flow
rate. Critical and supercritical velocities permit determination of flow by a single level
measurement at the inlet. Thus most flumes are designed to pass flow from subcritical
through critical near the point of measurement.
A flume is said to be submerged downstream when the downstream level rises
sufficiently to reduce the discharge. This results in a reduction in velocity and an increase
in flow depth in the throat. The measurement of submerged flow requires level
measurements at the inlet and at the throat and greatly complicates the determination of
flow rate.
Flow rates are usually determined by taking level measurements and reading flow rates
from nomographs or tables.
FLUID FLOW
Page 30 of 51
CARBON STEEL PIPE DATA
Nominal
Tube Size
(in)
Schedule
Number
O.D.
3/8
40
80
1/2
FLUID FLOW
Wall
Thicknes
s
(in)
Wt./Ft.
(lbs)
Wt.
H2O/ft
(lbs)
0.675
0.091
0.126
0.567
0.738
0.083
0.061
40
80
0.840
0.109
0.147
0.850
1.087
0.132
0.101
3/4
40
80
1.050
0.113
0.154
1.130
1.473
0.230
0.186
1
40
80
1.315
0.133
0.179
1.678
2.171
0.374
0.311
1¼
40
80
1.660
0.140
0.191
2.272
2.996
0.647
0.555
1½
40
80
1.900
0.145
0.200
2.717
3.631
0.882
0.765
2
40
80
2.375
3.652
5.022
2½
40
80
2.875
0.154
0.218
0.203
0.276
5.790
7.660
1.452
1.279
2.072
1.834
3
40
80
3.500
0.216
0.300
7.570
10.250
3.200
2.860
3½
40
80
4.000
0.226
0.318
9.110
12.510
4.280
3.850
4
40
80
4.500
0.237
0.337
10.790
14.980
5.510
4.980
5
40
80
5.563
0.258
0.375
14.620
20.780
8.660
7.870
6
40
80
6.625
0.280
0.432
18.970
28.570
12.510
11.290
8
40
80
8.625
0.322
0.500
28.550
43.390
21.600
19.800
10
40
80
10.750
0.365
0.593
40.480
64.400
34.100
31.100
12
40
80
12.750
0.406
0.687
53.600
88.600
48.500
44.000
(in)
Page 31 of 51
Problems for Fluid Flow
1.
Water if flowing through a pipe with a 6.00 inch ID at a flow velocity of 20.0 ft/s. The pipe
ID reduces to 3.00 in. What is the flow rate in the reduced section?
2.
Oil flows through a 1.00 in ID pipe at a rate of 20.0 gpm (Imp.). A valve placed in the pipe
has a passage in it of 0.750 in diameter. What are the flow velocities in the pipe and
through the valve?
3.
Fluid is introduced into a pump through a 1.50 in suction line and discharged through a
1.00 in port. If the discharge flow rate is 40.0 gpm (Imp.), what are the flow velocities in
the inlet and outlet port?
4.
If the discharge pressure of the pump in question 3 is 1.50  103 psi, calculate the
horsepower required to drive the pump. (Assume inlet pressure to be 0).
5.
A standpipe 50.0 ft high rests on top of a hill, 135 ft above the level of a pumping station.
The pump sits in a pit 10.0 ft. below the floor level. If it delivers 1.00  104 gpm (Imp.) of
water to the standpipe, what is the efficiency of the pump if it is driven by an electric motor
of 6.00  102 H.P.?
6.
The ballast pumps in a submarine submerged at a depth of 1.00 102 m discharge
2.00  104 L/s of sea water. Neglecting losses, determine the horsepower required to
drive the pumps.
7.
The fuel pumps in a ballistic missile deliver 1500 kg/min of liquid fuel (sg = 1.47) to the
rocket engine nozzles at a pressure of 2.0 MPa. What horsepower must be delivered by
the pumps?
8.
Sea water is being pumped up a vertical pipe at a rate of 1.5  102 Imp gal/min.
The cross-sectional area of the pipe at the bottom is 10.0 in2. The pipe narrows to a
cross-sectional area of 4.00 in2 at a point 330.0 ft higher. What is the pressure difference
between these two points (in atmospheres)?
9.
If there was a head loss due to friction of 3.0 ft between the two points in question 8, what
would be the pressure difference?
10.
In the pump shown below, flow at the outlet is 35.0 Imp. gpm
a) What is Q at the inlet?
b) If the fluid is water, the pump is 65% efficient and is driven by a 5.00 H.P. motor.
What is the pressure difference across the pump?
2.00” ID
1.00” ID
P
FLUID FLOW
Page 32 of 51
11. In the diagram below, water is being pumped up from a reservoir. The elevation at point 1
is 30.0 m, at point 2 and 3 is 33.0 m, and at point 4 is 40.0 m. The velocity at point 3 is 6.0
m/s. An electrical motor delivers 25 H.P. to the pump, which is 75.0 % efficient. If the head
loss from 1 to 2 is 0.60 m and the head loss from 3 to 4 is 1.20 m, what is the pressure at
point 4 in m H2O?
4.
1.00” ID
3.00” ID
2.
3.
P
1.
12. A 45.0 cm ID pipeline carries oil ( = 0.700 g/mL) at an average velocity of 3.60 m/s. If the
pipe narrows to 30.0 cm ID, what will be the velocity in the narrow portion?
13. A piping system is constructed of schedule 40 steel pipe. A 3 inch pipe carries water at a
velocity of 50 fps and a 2 inch pipe carries water at a velocity of 25 fps. Both the 2 inch and
the 3 inch pipes flow into a 6 inch pipe (Schedule 40). What is the velocity in the 6 inch
pipe?
14. Crude oil (sg = 0.800) at a rate of 3.00  103 lb/min flows in a channel 3.00 ft wide and 18.0
inches deep. If the flow occupies exactly one half of the depth of the channel, what is the
velocity of the crude oil in ft/s?
15. The city water main supplying a large apartment building has a maximum pressure of 51.5
psia. The water taps on the top floor at the building are 150 ft above the water mains.
Calculate the maximum pressure available at these taps in psig.
16. If the head loss between sections 1 and 2 is 3.50 ft, calculate the pressure (in psig) in the
3 inch pipe. The velocity in the 6 inch pipe is 10.0 ft/s of water.
3.00” ID
2.
6.00” ID, 28.0 psig
18.0’
1.
FLUID FLOW
Page 33 of 51
17. A pump takes water at 50F from a large reservoir and delivers it to the bottom of an open
elevated tank. The level in the tank averages 160 ft above the surface of the reservoir. The
line is 3 inch standard pipe and head loss is 18.6 ft. If the pump delivers 150 gal/min and
has an efficiency of 55%, what horsepower is consumed.
18. At one point A in a pipeline carrying oil (sg = 0.900) the diameter is 6.00 in and the
pressure is 25.0 psig. At another point B, 20.0 ft above A, the diameter is 12.0 in and
pressure is 18.0 psig. If oil flows in the system at a rate of 1.00 ft3/s, is the oil flowing from
A to B or
B to A?
19. An aqueous methanol solution ( = 0.83,  = 0.96 cP) is to be pumped from a storage tank
to a process through a 1¼ inch standard steel pipe. The process end of the line must have
a pressure of 8.0 psig and is 22 ft above the surface of the liquid in the storage tank. The
flow is rated at 20 gpm (Imp) but may be overloaded by 50%. The head loss due to friction
is 300 ft of the methanol solution. Assuming a 60 % pump efficiency, calculate the output
required by the electric motor for this service. Report your answer to 3 sig figs.
20. A standard 2 inch steel pipe is conveying an oil of specific gravity of 0.850 at a rate of 90.0
L/min. At a certain point in the line the pressure is 310 kPa. Calculate the Reynolds
number and pressure at a point in the line that is 61.0 m of pipe further along in the
direction of flow and is 15.0 m lower in elevation. Assume a head loss due to friction of 1.00
m of oil. The oil has a viscosity of 20.0 cP at the temperature of the process.
21. A technologist is investigating a gravity feed that is to supply a vinegar solution to a
processing unit as shown in the diagram. The process requires a flow rate of 10.0 gal/min
(Imp) and a pressure of 10.0 psig at point B, just before the processing unit. The vinegar
has a specific gravity of 1.005 and a viscosity of 1.00 cP. The piping is 3 inch schedule 40
steel pipe. Calculate the velocity and Reynolds number at point B. Assuming the frictional
head loss is 0.11 ft. of fluid, determine the minimum height of vinegar required in the
storage tank.
A
?’
10.0’
10.0’
FLUID FLOW
B
Page 34 of 51
Answers to Problems
1. Q = 3.93 ft3/s
2.
2 = 80.1 ft/s
p = 587 ft/min valve = 1044 ft/min
3. in = 8.69 ft/s
out = 19.6 ft/s
4. 42.0 HP
5. 98.5% effic.
6. 2.70  104 HP
7. 45.6 HP
8. p = 10.1 atm
9. p = 330 ft sea H2O
10. a) Q = 35.0 Imp gal/min
b) 302 ft H2O
11. p = 454 m fresh H2O
12.  = 8.1 m/s
13.  = 15.3 ft/s
14.  = 0.44 ft/s
15. p = -28.2 psig
16. p = 8.6 psig
17. HP = 14.8
18. B  A
19. Re = 7.12  104, hp = 17900 lb/ft2, HP = 4.34
20. Re = 1546 (laminar), p = 426 kPa
21. Re = 1.24 104, depth of fluid in tank must be at least 3.1 ft..
FLUID FLOW
Page 35 of 51
SOLUTIONS TO “PROBLEMS FOR FLUID FLOW’ (pages 32-34)
1.
1.
2.
v1 = 20.0 ft/s
d1= 6.00 in.
v2= ?
d2 = 3.00 in.
v2 = v1 (d1/d2)2 = 80.0 ft/s
Q = A1v1 = (d12/4)v1 = 3.93 ft3/s (3 sig figs)
Q is constant so it is calculated as A1v1 or as A2v2
Convert d1 from inches to feet and include units in calc.
2.
Q = (20.0 Imp g/min)(1 min/60s)(1 ft3/6.24 Imp gal)
1.
2.
= 0.0534 ft3/s
A1 = [(1/12 ft)2/4] = 0.00545 ft2
d1 = 1.00 in
d2 = 0.750 in v1 = Q/A1 = (0.534 ft3/s)/[0.00545 ft2] = 9.8 ft/s = 587 ft/min
Q = 20 Imp gpm
v2 = v1 (d1/d2)2 = (9.8 ft/s)(1 in/0.75 in)2 = 1044 ft/min
Q has 3 sig figs.
3.
Q = (40.0 Imp g/min)(1 min/60s)(1 ft3/6.24 Imp gal)
P
1.
2.
= 0.1068 ft3/s
A1 = [(1.5/12 ft)2/4] = 0.0123 ft2
d1 = 1.50 in
d2 = 1.00 in v1 = Q/A1 = (0.107 ft3/s)/[0.0123 ft2] = 8.70 ft/s
Q = 40.0 Imp gpm
v2 = v1 (d1/d2)2 = 19.6 ft/s
Answers should have 3 sig figs.
4.
p = 1.5103 psi. P = pQ = [1500 lb/in2  (144 in2/1 ft2)]  (0.1068 ft3/s) = 23070 ftlbs-1
HP = (23070 ftlbs-1)(1 HP/ 550 ftlbs-1) = 41.9 HP
This calculation assumes that the pump is 100% efficient (which is never true).
To calculate a realistic value, divide the answer by the efficiency, e.g., 0.95, 0.80, etc.
Q = (1.00E4 Imp g/min)(1 min/60s)(1 ft3/6.24 Imp gal)
= 26.71 ft3/s
p = 195 ft H2O (2116 lb/ft2/33.9 ft H2O) = 12,180 lb/ft2
P = pQ = (12,180 lb/ft2)  (26.71 ft3/s) = 3.25E4 ftlbs-1
HP = (3.25E4 ftlbs-1)(1 HP/ 550 ftlbs-1) = 591 HP
The pump consumes 600 HP, but works at a rate of 591 HP
The % efficiency of the pump = (591 HP/600 HP)  100%
= 98.5% efficient
5.
50 ft
135
ftPumping Station
10 ft
P
FLUID FLOW
Page 36 of 51
6.
The pump must work at a pressure equal to the water
pressure at a depth of 100 m
Sea water is denser than fresh water (s.g. = 1.027)
100 m sea H2O exerts the same pressure as 102.7 m
100m
deep
fresh submarine
7.
102.7 m H2O(101325 Pa / 10.33m H2O) = 1.010E6 Pa
Q = 2.00E4 L/s  (1 m3/103L) = 20 m3/s
P = pQ = (1.01E6 N/m2)(20 m3/s) = 2.02E7 watts
= 2.02E7 watts (1 HP/ 746 W) = 2.70E4 HP
Q = 1500 kg/min (1L/1.47 kg)(1 min/60s)(1m3/103L) = 0.017m3/s
p = 2.0 MPa = 2.0E6 Pa or N/m2
P = pQ = (2.0E6 N/m2)(0.0170m3/s) = 34010 Watts (1 HP/746 W) = 45.6 HP
8.
2.
A1 = 10.00 in2, A2 = 4.00 in2
Q = 1.5E2 gal/min  (1 min/60 s)(1 ft3/6.24 Imp gal) = 0.401 ft3/s
330
A1 = 10 in2 (1ft2/144in2) =0.0694 ft2
3
2
ft
v1 = Q/A1 = (0.401 ft /s)  (0.0694 ft ) = 5.77 ft/s
1.
v2 = v1(A1/A2) = 5.77 ft/s  (10in2/4in2) = 14.4 ft/s
You cannot simply use the pressure depth equation because the
fluid is moving through pipes of changing diameter and this also affects pressure.
Use Bernoulli’s equation to calculate the difference in pressure.
2
h1  z1 
2
v1
v
 h2  z 2  2
2g
2g
(5.77 ft / s ) 2
(14.4 ft / s ) 2

h

330
'

2
2  (32.2 ft / s 2 )
2  (32.2 ft / s 2 )
h1  0'0.52'  h2  330 '3.22'
h1  0'
1 atm


h1  h2  333 .2  0.52  333 ft sea water  
  10.1 atm
33.0
ft
sea
water


Note the pressure is lower at the higher elevation. The pressure is 333 ft sea water less at
an elevation 330 ft higher. The additional 3 feet pressure loss is due to the increased fluid
velocity in the smaller pipe.
9.
If 3 ft of fluid pressure were lost due to friction of the fluid traveling between 1 and 2, there
would be 3 ft less fluid pressure difference between these points. The difference in
pressure between point 1 and 2 would then be only 330 ft sea water or 10.0 atm.
FLUID FLOW
Page 37 of 51
10.
a)
b)
2.00”
ID
The inlet and outlet flows must be constant
since liquids are relatively non compressible
fluids. Q = 35.0 gpm
1.00” ID
P
P = 5.00 HP(550 ftlbs-1/1 HP)  0.65 efficiency = 1788 ftlbs-1
Q = (35 Imp gal/min)  (1 min/60 s)(1 ft3/6.24 Imp gal) = 0.0935 ft3/s
p = P/Q = (1788 ftlbs-1) / (0.0935 ft3/s) = 19120 lb/ft2  (33.9 ft. H2O/2116 lb/ft2)
= 306 ft H2O
A1 = [(2”/12 ft)2/4] = 0.0218 ft2
v1 = Q/A1 = (0.0935 ft3/s) / (0.0218 ft2) = 4.29 ft/s
v2 = v1 (d1/d2)2 = 17.1 ft/s
2
h p  h1  z1 
2
v1
v
 h2  z 2  2
2g
2g
(4.29 ft / s ) 2
(17.1 ft / s ) 2

h

0
'

2
2  (32.2 ft / s 2 )
2  (32.2 ft / s 2 )
306 .1  h1  0'0.29'  h2  0'4.56'
306 .1' h1  0'
h2  h1  306 .4  4.56  302 ft fresh water
The pressure added by the pump is the term ‘hp’ in Bernoulli’s equation.
Since no change in elevation is stated we assume the elevation is unchanged so z1
and z2 are both zero and can be left out of the equation if you wish.
11.
3.00” ID
4.
1.00” ID
hf = 1.20 m
2.
P
3.
z = 10 m
hf = 0.60 m
1.
d2 = 3.00 in d3 = 1.00 in
v2 = ? m/s
v3 = 6.0 m/s
FLUID FLOW
v2 = v3 (d3/d2)2 = 0.667 m/s
d3 = 1 in = 2.54 cm
= 0.0254 m
A3 = [(0.0254 m)2/4]
= 5.067E-4 m2
Q = A3v3
= (5.067E-4 m2) (6.0 m/s)
=0.00304 m3/s
P = 25 HP  0.75 effic.
= 18.75 HP  (746 W/1
HP)
= 13990 Watts or Nm/s
Page 38 of 51
p = P/Q = (1.4E4 Nms-1)/(0.00304 m3/s) = 4.60E6 N/m2 (10.3 m/101325 Pa) = 468 m H2O=hp
2
2
v
v
h p  h1  z1  1  h4  z 4  4  h f
The elevation (z1) and pressure head
2g
2g
(h1) at 1 are both zero since the pump is
(0.667 m / s ) 2
(6.0m / s ) 2

h

10
m

 hf
4
2  (9.81m / s 2 )
2  (9.81m / s 2 )
468 m  0.023 m  h2  10 m  1.83m  h f
468 m  0  0 
h2  (456 .2m fresh water - h f )  (456.2m - 1.8m)  454m fresh water
drawing from the surface of the
reservoir.
The head losses due to friction, hf,
(1.2m + 0.6m H2O) total 1.8m. The
pressure at 4 (from Bernoulli’s eqn) is
decreased by this amount.
12.
The density is irrelevant for the calculation required.
Velocity is inversely proportional to the square of the diameter
of the pipe.
v2 = v1 (d1/d2)2 = 8.1 m/s
13.
You must use the pipe tables to determine the
correct internal diameters.
1
2
ID = OD – 2(wall thickness)
d1 = 3.50” – 2(0.216”) = 3.068”
d2 = 2.375” – 2(0.154”) = 2.067”
d3 = 6.625” – 2(0.280”) = 6.065”
A1 = [(3.068”/12 ft)2/4] = 0.0513 ft2
A2 = [(2.067”/12 ft)2/4] = 0.0233 ft2
A3 = [(6.065”/12 ft)2/4] = 0.201 ft2
Q3 = Q1 + Q2 = A1v1 + A2v2 = (0.0513ft2  50 ft/s) + (0.0233ft2  25 ft/s)
= (2.565 ft3/s + 0.5825 ft3/s) = 3.148 ft3/s
v3 = Q3/A3 = (3.148 ft3/s)  (0.201 ft2) = 15.7 ft/s
The flow rate in pipe 3 must be the sum of the flow rates in pipes 1 and 2
3
14.
This is a simple calculation of fluid velocity using Q = Av.
The carrier is a rectangular-shaped trench rather than a cylindrical pipe.
The cross sectional area of flow is 3.00 ft  18”/12 ft  ½ (because the trench is half-full).
A works out to 2.25 ft2. Convert Q to units of ft3/s and you can calculate velocity in ft/s.
Doil = DH2O  0.80 = (62.43 lb/ft3)  0.8 = 49.94 lb/ft3
Qoil = 3000 lb/min  (1 ft3/49.94 lb)  (1 min/60s) = 1.00 ft3/s
voil = Q/A = (1.00 ft3/s)  2.25 ft2 = 0.44 ft/s
15.
Calculate the height of water that exerts the same pressure as supplied
by the water main. Compare this to the height to the top floor of the
building. Don’t forget to work in psi gage not absolute.
When a person on the top floor opens the tap, atmospheric pressure
will push against the water flowing from the tap.
pg = 5.5 psia – 14.7 psi = 36.8 psig
150 ft
36.8 psig  (33.9 ft H2O/ 14.7 psi) = 85 ft!
This is not enough pressure to reach the top.
p = 51.5 psia
The maximum height at which water will flow from the
tap is 85 ft.
FLUID FLOW
Page 39 of 51
Tall apartments buildings must supplement city water
presssure by adding a pump. In this case the building will need an additional 28 psi.
because 150 ft H2O  (14.7 psi/33.9 ft) = 65.0 psi
65.0 psi – 36.8 psi = 28.2 psi.
FLUID FLOW
Page 40 of 51
16.
3.00”
ID
2.
This is another problem where Bernoulli’s eqn is needed because both
elevation and velocity are changing.
v2 = v1 (d1/d2)2 = 40.0 ft/s
6.00” ID, 28.0 psig
1.
2
h1  z1 
18.0
’
hf = 3.50 ft
2
v1
v
 h2  z 2  2  h f
2g
2g
(10 ft / s ) 2
(40 ft / s ) 2

h

18
'

 3.5'
2
2  (32.2 ft / s 2 )
2  (32.2 ft / s 2 )
64.6'0'1.55'  h2  18'24.8'3.5'
64.6'0'
 14.7 psi 
h2  19.85 ft fresh water  
  8.6 psig
 33.9 ft 
Since you are given p1 = 28 psig, the changes in pressure due elevation and velocity do
not change the fact that we are calculating gage pressure. If the pressure at p1 were
given as 28.0 psia, then the pressure at p2 would be 8.6 psia.
17.
Q = (150 Imp gal/min)  (1 min/60 s)(1 ft3/6.24 Imp gal) = 0.401 ft3/s
p = 160 ft + 18.6 ft = 178.6 ft  ( 2116 lb/ft2/33.9 ft H2O)
patm
= 11,150 lb/ft2
18.6 ft is head loss due to friction, (hf).
P = pQ = (11,150 lb/ft2) (0.401 ft3/s) = 4470 ftlbs-1
= 4470 ftlbs-1  (1 HP/550 ftlbs-1)
= 8.13 HP required.
P consumed by pump = 8.13 HP/0.55 = 14.8 HP
At a temperature of 50°F, the s.g. of water is
very close to standard value of 1.0. The limited
acccuracy of the data in the question does not
warrant factoring in the density. It wouldn’t make
P
any difference in this problem.
FLUID FLOW
160 ft
Page 41 of 51
18.
A tricky problem. Calculate the total energy of the
oil at both A and B. Since it is flowing, there must be
some energy loss due to friction (hf). If energy at A
B
is less than at B, then flow must be from B to A, since
energy would be lost when the oil flows from B to A.
20 ft
Get it?
Oil has s.g. = 0.90, so the height of oil that would
exert a pressure equivalent to 33.9 ft of H2O will be
A
33.9 ft/0.90 = 37.67 ft oil.
It’s best to work in height of process fluid in Bernoulli’s equation and only convert to other
units at the end, if necessary.
v1 = Q/A1 = (1.0 ft3s)/(0.1964 ft2) = 5.09 ft/s [It is assumed that you can calculate A1]
v2 = v1 (d1/d2)2 = 1.27 ft/s
p1 = 25 psig(37.67 ft oil/14.7 psig) = 64.1 ft oil
p2 = 18 psig(37.67 ft oil/14.7 psig) = 46.1 ft oil
2
h1  z1 
2
v1
 Energy at A
2g
h2  z 2 
(5.09 ft / s ) 2
 64.5' oil
2  (32.2 ft / s 2 )
46.1'20'
64.1'0'
v2
 Energy at B
2g
(1.27 ft / s ) 2
 66.1' oil
2  (32.2 ft / s 2 )
Note that the Energy (as pressure) at B is greater than at A by 1.6 ft of oil. Since the
pressure is greater at B, the flow must be from B to A, i.e., from high pressure to low.
19.
The viscosity is irrelvant but the density is important.
to process
The pump must deliver 8 psig greater than the hydrostatic
p = 8.0 psig
head (h1) of the methanol in the tank.
patm
Calculate what height of methanol this 8 psig is
22 ft
equivalent to.
Methanol has s.g. = 0.83, so the height of methanol that
would exert a pressure equivalent to 33.9 ft of H2O will
be 33.9 ft/0.83 = 40.8 ft methanol.
p2 = 8.0 psig(40.8 ft methanol/14.7 psig) = 22.2 ft methanol.
The pump must supply enough power to raise the
P
methanol 22.0 ft and deliver it at a pressure equivalent to
22.2 ft. The pump must also supply power to increase
the velocity of the methanol from rest (0 ft/s) in the tank to some velocity on the output
side of the pump. Calculate this velocity and then combine all terms in Bernoulli’s
equation.
d2 = OD – 2(thickness) = 1.660” -2(0.140”) = 1.38”
(Standard 1 ¼ “ pipe means Schedule 40 1 ¼ “ pipe. Use the pipe tables to get ID)
A2 = [(1.38”/12 ft)2/4] = 0.01039 ft2.
The flow must be multiplied by 1.5 to allow for 50% overload mentioned in the problem
Q = (20  1.5 Imp gal/min)  (1 min/60 s)(1 ft3/6.24 Imp gal) = 0.0801 ft3/s
v2 = Q/A2 = (0.0801 ft3/s)(0.01039 ft2) = 7.71 ft/s
You should now be able to solve Bernoulli’s equation
FLUID FLOW
Page 42 of 51
2
2
v
v
h p  h1  z1  1  h2  z 2  2  h f
2g
2g
(7.71 ft / s ) 2
 300 '
2  (32.2 ft / s 2 )
hp  0'0'0'  22.2'22'0.92  300 '
hp  345 ' methanol
hp  0'0'0'  22.2'22'
h1 is set to 0’ because we know it is
22.2’ of methanol less than h2.
z1 is set to 0’ because we know it is 22
ft less than z2
hf is given in the problem as 300’
methanol
345 ft methanol is equivalent to:
345 ft (0.83/1) = 286 ft H2O (2116 lb/ft/33.9 ft H2O) =17880 lb/ft2
P = pQ = (17880 lb/ft2)(0.0801 ft3/s) = 1432 ftlbs-1 (1 HP/550 ftlbs-1) = 2.60 HP
Finally, we must factor in the efficiency of the pump.
We will need to use a pump that draws 2.60 HP/0.60 = 4.34 HP
20.
For ‘standard’ (Schedule 40) 2” steel pipe:
p = 310 kPa
1.
d = 2.375” – 2(0.154”) = 2.067” ID. (0.0254 in/m) = 0.0525 m
3
3
3
Q = 90 L/min  (1 m /10 L)(1 min/60 s) = 0.0015 m /s
A = [(0.0525 m)2/4] = 0.00216 m2
v = Q/A = (0.0015 m3/s)  (0.00216 m2) = 0.693 m/s
15.0 m
We don’t need velocity for Bernoulli’s equation because the
pipe does not change size (velcoity is constant), but we will
2.
need fluid velocity to calculate Reynolds number.
Convert the pressure at 1 from kPa to meters of oil (s.g. = 0.850)
310 kPA (10.3 m H2O/101.325 kPa) = 31.51 m H2O  (1/0.850) = 37.1 m oil
2
2
v
v
h1  z1  1  h2  z 2  2  h f
2g
2g
37.1m  15m  0m  h2  0m  0m  1m
The mass density of oil = 1000 kg/m3  0.850
h2  51.1 m oil
51.1 m oil (0.850/1) = 43.4 m H2O (101.3 kPa/10.3 mH2O) = 426 kPa
The pressure at elevation 1 is 426 kPa greater than at elevation 1.
d v 
(0.0525m)  (0.693m/s)  (850kg/m 3 )
= 1546 = Laminar flow.

0.020 k g  m 1  s 1
Be careful with viscosity units. 20.0 cP given = 0.20 Poise = 0.020 Pas.
The units of Pa.s are kgm-1s-1 and are required to cancel all units.
Reynolds number is dimensionless.
Be sure to include units in your calculations. One mark is assigned to each question for
units. Even if you get the correct answer, but leave out units, you will still lose a mark.
You will make fewer errors if you include units (and you’ll get more marks from the
instructor as well).
Re 
FLUID FLOW

Page 43 of 51
21.
3” Sch. 40 steel pipe has an ID of 3.068 in or 0.2557 ft.
A = 0.0513 ft2
Q = 10.0 gal/min  (1 min/60 s)(1 ft3/6.24 Imp gal) = 0.0267 ft3/s
v = Q/A = 0.520 ft/s
Pressure at B = 10 psig (33.9 ft H2O/14.7 psig) = 23.1 ft H2O (1/1.005) = 23.0 ft vinegar
2
2
v
v
h1  z1  1  h2  z 2  2  h f
2g
2g
(0.52 ft / s ) 2
 .011'
2  (32.2 ft / s 2 )
0' z1 '0'  23.0  0'0.004 '.011'
0' z1 '0'  23.0  0'
z1  23.1 ft vinegar
You can see from the diagram that the bottom of the tank is 20 ft above point B, so we
only need an additional 3.1 ft of vinegar to attain the 23.1 ft height of vinegar. So the tank
need only be filled to a level of 3.1 ft to supply the required pressure at B.
Re 
d v 


(0.256ft)  (0.520ft/s )  (1.950slug /ft 3 )
 12420  turbulent
2.089  10 5 slug  ft 1  s 1
The mass density of the vinegar is 1.9404 slug/ft3 (for water)  1.005 = 1.950 slug/ft3
1.00 cP = 2.089E-5 slugft-1s-1 Be careful. You must use a consistent set of units in
calculating Re.  is mass density so be sure to use either slug/ft3 or lbm/ft3.
’lbm’ is a mass unit called ‘pounds mass’. It is numerically equal to ‘lbf’, (1 lbm = 1 lbf),
but they are not the same. ‘lbf’ is a unit of force. If you use lbm units for  in calculating
Re, then use a  = 6.74E-4 lbmft-1s-1, as shown below.
Re 
d v 


(0.256ft)  (0.520ft/s )  (62.17lbm/ ft 3 )
 12420  turbulent
6.72  10 4 lbm  ft 1  s 1
You will need to convert the mass density from slug/ft3 to lbm/ft3.
Note that 1 slug = 32.17 lbm, so 1.950 slug/ft3 = 61.17 lbm/ft3.
FLUID FLOW
Page 44 of 51
FLUID FLOW TEST SUMMARY
1. You are responsible for all problems, those in the body of the notes and the problem set at
the end of the notes.
You will be required to read a pipe table (the ones in the notes) to get data to solve
problems, e.g., determine ID for flow calc., or other such info. If you do the Pipes, Fittings
and Valves self study, you should be prepared for this.
2. There will be a few short answer questions on the following topics
a) the various types of flow meters described in the notes.
b) Understand laminar and turbulent flow and critical velocity and the corresponding
Reynolds numbers
c) Understand factors that affect fluid friction in laminar and turbulent flow. You don’t have
to memorize them all, but given certain factors, you should be able to describe their
effects on fluid flow and frictional resistance.
d) You should have a conceptual understanding of the three coefficients of velocity,
contraction and discharge. The formulas are given in the test aids, but what do they
mean/how are they applied?
e) Sketch the cross section of a venturi and/or orifice plate in a pipe and the pressure
changes that occur and read the comparison of only these two meters. Don’t forget
the calculations re: these meters.
From the labs:
Lab #2: Bomb Calorimetry
 Perform calculations like those required for the lab report. You will not be required to
calculate rate corrections for temperature or a corrected temperature rise but you may
be required to use a given corrected T in the calculation of the heat equivalent of a
calorimeter or the heat of combustion of a sample. You should be able to explain
purpose and effect of rate corrections. You may also omit the acid correction
calculations since you did not do this but you are responsible for corrections for the
combustion of the ignition wire. You are also responsible for calculations involving Cl
determination of a sample given data on dilution, aliquots and titration with AgNO3 or
measurement via specific ion electrode, including logarithmic regression analysis of data
(emf vs. ppm Cl-). Equations will not be provided. You can derive them from the units
or purchase them for marks on the test.
Lab #10 Heat Transfer

Expect a calculation involving the overall heat transfer coefficient.
The test is closed book but you will be given test aids (attached).
FLUID FLOW
Page 45 of 51
TEST AIDS FOR FLUID FLOW TEST
pabs = gh + patm
or
This pressure-depth equation:
pabs = Dh + patm
p = gh = Dh
The standard density for water is 1.00 g/cm3 which occurs at 3.98 ºC
D4º = 62.4 lb/ft3 = 9810 N/m3
and
4º = 1.00 g/cm3 = 1000 kg/m3 = 1.94 slug/ft3
Atmospheric Pressure at sea level is:
2
5
2
6
2
2
14.70 lb/in = 2116 lb/ft = 1.01325  10 N/m (Pa) = 1.01325  10 dyn/cm
= 101.325 kPa = 1 atm = 760 mmHg. (Torr) = 76 cm Hg = 29.92 inHg
= 33.90 ft fresh water (~ 33 ft. sea water) = 1.01325 bar = 1013.25 mb
p
F
A
sg =
Ds
s
=
Dw
w
=
m
v
D

v
Recall units of force:
A Newton (N) is a force that will cause a 1 kg mass to accelerate at 1 m/s2.
The force of gravity on earth (g) is 9.81 N/kg.
Gravity causes a 1 kg mass to accelerate (fall) at a rate of 9.81 m/s2.
A pound of force (lbf) will cause a mass of 1 slug to accelerate at 1 ft/s2.
The force of gravity on earth (g) is 32.2 lb/slug.
Gravity causes a 1 slug mass to accelerate (fall) at a rate of 32.2 ft/s2.
Conversions:
force-force:
mass-mass:
mass-force:
1 in. = 2.54 cm
FLUID FLOW
1 lb = 4.45 N
1 slug = 14.6 kg
1 kg = 9.81 N
1 N = 105 dynes
1 kg = 2.205 lb
1 slug = 32.2 lb
1 m = 3.2808 ft
g/cm3
H2 O
1.00
Hg
13.596
air
0.00129
Page 46 of 51
ICS FLUID FLOW
Important Conversions and Equalities:
1 Newton force (N) causes 1 kg mass to accelerate at 1 m/s2
1 dyne force (dyn) causes 1 g mass to accelerate at 1 cm/s2
1 pound force (lbf) causes 1 slug mass to accelerate at 1 ft/s2
1 pound force (lbf) causes 1 lb mass to accelerate at 32.2 ft/s2
1 poundal force (pdl) causes 1 lb mass to accelerate at 1 ft/s2
1 pdl
1N
1 lbf
1 lbmft/s2
1 kgm/s2
1 slugft/s2
=
=
=
1 lbm =
1 kg
1 slug =
=
=
7.233 pdl
32.2 pdl
=
=
=
=
2.204623 lbm
32.174 lbm
g = 9.81 N/kg or m/s2
g = 981 dyn/g or cm/s2
g = 32.2 lbf/slug or ft/s2
g = 1.0 lbf/lbm
g = 32.2 pdl/lbm or ft/s2
105 dyn
32.2 lbmft/s2
=
4.448 N
0.454 kg
14.594 kg
g = 32.2 lbf/slug = 1 lbf/lbm= 32.2 pdl/lbm = 9.81 N/kg = 981 dyn/g = 1037 pdl/slug
The following conversions apply at 3.98 C where water is densest.
H2O = 1

g
kg  1 slug  
1 m3
slug

  1.9403 3


=
1000


3
3
3
3 



cm
m  14.594 kg   3.2808 ft 
ft
981 dyn
 1 g   981 dyn 
 
3 
 cm   g 
cm3
DH2O = g = 
or
 32.2 lbm 
lbm
  62.43 3
 
ft
 1slug 
 1000 kg   9 .807 N 



 m3   kg 

9807 N
m3
.
slug   32 .2 lbf 
62 .43 lbf
2009 pdl
 19403
 62.43 lbm  32 .2 pdl 
or 
 

 

3
3
3

  slug 

  lbm 
ft
ft
ft
ft 3
DH2O = 
1m
1 IMP gal
1 US gal
1 IMP gal H2O =
1 US gal H2O
=
3.28083 ft
=
4.5461 L = 1.201 US gal
=
3.7854 L = 0.8327 IMP gal
4.5461 L  (1 kg/L)  (2.2046 lb/kg) = 10.02 lb/IMP gal
=
3.7854 L  (1 kg/L)  (2.2046 lb/kg) = 8.345 lb/US gal
H2O = 62.43 lbm/ft3  (1 IMP gal/10.0 lbm)  6.24 IMP gal/ft3
H2O = 62.43 lbm/ft3  (1 US gal/8.345 lbm)  7.48 US gal/ft3
Power =
Also,
work
time
=
Fd
t
e.g.
Nm
J
=
= W
s
s
or
lbf  ft
s
3
 lbf  ft 3  lbf  ft
 N  m  N  m
 
 
Power = P  Q e.g.  2 
or  2 
s
s
 m  s 
 ft  s 
1 horse power (HP) = 745.7 watts (W) = 550 ftlbfs-1 = 33000 ftlbfmin-1
FLUID FLOW
Page 47 of 51
  d1 2
2
2
 d1 
 r1 
v2
A1
d1
4


 





2
d 
v1
A2
  d22
d2
 2
 r2 
4
V ( A  v ) t
Q 
 A v
t
t
2
The Continuity Equation:
Since Q1 = Q2 ,
thus:
Q = A1  1 = A2  2
Bernoulli’s Equation-the most common forms:
p + gz + ½2 = constant
v2
hz
 cons tan t
2g
p  Dz 
=
units are height (m, ft, etc.)
Dv 2
 cons tan t
2g
hp + h1 + z1 + 12/2g
h2 + z2 + 22/2g + hf
units are pressure (N/m2 or lbf/ft2 or dyn/cm2)
units are pressure (N/m2 or lbf/ft2)
hp = pressure head added by pump
h1 = pressure head at point 1
h2 = pressure head at point 2
z1 = elevation of point 1
z2 = elevation of point 2
1 = velocity at 1
2 = velocity at 2
hf = head loss due to friction.
All terms are in height of the working fluid
LIQUID HEADS IN FLUID FLOW
1. Vertical depth in ft. or m. of any point below the surface of a liquid is called pressure head
or hydrostatic head, i.e., it is a pressure usually expressed in ft. or m. of the liquid.
Recall p = Dh  h = p/D
2. A drop in pressure in moving water due to friction is called friction head. It is directly
proportional to the total area of the rubbing surface as well as the roughness of the surface and
the velocity of the fluid. Abrupt changes in cross-sectional area or direction of flow also causes
increased friction head.
3. Pressure in a pipe or channel due to the velocity of liquid particles is called velocity head.
From Toricelli’s Theorem, v  2gh

hv = 2/2g
FLUID FLOW
Page 48 of 51
SPECIAL CASES OF BERNOULLI’S EQUATION
In many cases the pressure, height, or speed of a liquid is constant and simplified forms of
Bernoulli’s equation result.
1. Liquid at Rest: When a column of liquid is stationary, equation 3. becomes:
p1 + gz1 = p2 + gz2  p = (p2 - p1) = g(z1 – z2)
This is simply the pressure-depth equation (p = gh)
2. When p1 = p2: When an open tank is draining, both ends are at atmospheric press.
gz1 + ½12 = gz2 + ½22
If the orifice is small compared to the cross section of the tank, the liquid level (draining under
gravity) will fall slowly enough that the liquid speed at the top of the tank (1) is negligible
(assume 1 = 0), then:
gz1 = gz2 + ½22  ½22 = g(z1 - z2) = gz
Solving for  we get: v  2gz
(Toricelli’s Theorem)
Note that the speed with which a liquid is discharged under gravity is the same as the speed of
a body falling from rest from height, h, and is independent of the density of the object. Recall
intro physics, the KE of an object at any point all comes from PE.
½m2 = mgh 
v  2gh
The flow rate at which liquid discharges through an orifice can be calculated as follows:
Since Q = A,  Q = A 2gh
(Flow from an orifice)
3. Constant Height:
p1 + ½12 = p2 + ½22
Reynolds number, is defined by the following equation.
Re 
d v 

or
 d    D

 
   g 
 = absolute viscosity
d = diameter (e.g. cm, m, or ft.)
-1
-1
v = average or bulk velocity (e.g., cms , ms or fts-1 ) g = grav. accel.
-3
-3
 = mass density (gcm , kgm , slug/ft-3 or lbm/ft-3 ) D = weight density (e.g. N/m3, etc.)
(cm)(cm  s -1 )(g  cm -3 )
cgs :
 dimensionl ess
( g  cm 1  s 1 )
British :
FLUID FLOW
(ft)(ft  s -1 )(lbm  ft -3 )
(lbm  ft 1  s 1 )
or
(m)(m  s -1 )(kg  m -3 )
SI :
= dimensionl ess
(kg  m 1  s 1 )
(ft)(ft  s -1 )(slug  ft -3 )
 dimensionl ess
( slug  s  ft  2 )
Page 49 of 51
The term “head loss per unit length of pipe” (i) is given by:
i
h L head loss between two points of pipe

L
distance between these points
coefficient of contraction (Cc) which is:
Cc 
area of jet at vena contracta
area of orifice
coefficient of velocity (Cv).
Cv 
actual velocity at vena contracta


theoretical velocity
2gh
coefficient of discharge (Cd):
Cd 
actual discharge
Q

theoretical discharge A  2  g  h
Bernoulli’s Equation for an Orifice
2    g  h
Q  C d  A2 
f
4
 d2 
   1
 d1 
Q = volume flow rate
g = grav. accel.
h = pressure head
Cd = coeff. of discharge
A1 = area of orifice
A2= area of pipe
 = difference in mass densities of 2 manometer fluids
f = mass density of process fluid (fluid in pipe)
Recall from our study of manometers that for a bifluid manometer p is not just gh but rather
()gh. For a single fluid manometer, use p = gh
Bernoulli’s Equation for a Venturi
The same equation is used for flow calculations in a venturi but without Cd. Venturis have lower
pressure drops than orifice plates (minimal frictional losses) resulting in flow calculations that
are close to the true values as obtained by actual flow measurements.
You must include units in your calculations and you must use a consistent set of units.
FLUID FLOW
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VISCOSITY
Viscosity is internal friction or resistance to flow. Absolute or dynamic viscosity () is ...

SI units :
shear stress
F/A
F  dx

 =
velocity gradient dV / dx
A  dV
(kg  m  s -2 )(m)
kg

 Pa  s
2
-1
m s
(m )(m  s )
cgs units :
(g  cm  s -2 )
1
(cm )(cm  s )
2

g
 poise (P)
cm  s
10 poise (P) = 1 Pa  s  1 poiseuille (Pl) = 1000 centipoise (cP)
1 cP = 6.72  10 -4
pdl  s
lbm
 6.72 10 -4
ft  s
ft 2
= 2.089 10 -5
slug
lbf  s
= 2.089 10 -5
ft  s
ft 2
Heat Transfer
q
m  c  T
t
FLUID FLOW
q  U 0  A0  Tm
Tm 
T2  T1
ln (T2 / T1 )
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