Solved Problems in Fluid Mechanics
Solved Problems in Fluid Mechanics
Solved Problems in Fluid Mechanics
Unit I
1. Similarity - Velocity
2. Similarity - Pressure Drop
3. U - tube Manometer
4. Manometer Connected between Two pipes
Unit II
5. Flow Direction
6. Siphon
7. Capillary Tube Viscosity Measurements
8. Frictional Losses
9. Frictional Losses - with fittings
Unit III
Unit IV
Similarity - Velocity
Water at 20oC (viscosity = 1 cp) flows through a smooth straight pipe A of inside
diameter 4 cm at an average velocity of 50 cm/sec. Oil flows through another pipe B
of inside diameter 10 cm. Assuming similarities, calculate the velocity of oil through
pipe B. Specific gravity of oil is 0.8 and its viscosity is 2 cp.
Data:
A:
Diameter of A = 4 cm
Velocity of fluid in A = 50 cm/sec
Density of fluid in A = 1 g/cc
1
Viscosity of fluid in A = 1 cp
B:
Diameter of B = 10 cm
Density of fluid in B = 0.8 g/cc
Viscosity of fluid in A = 2 cp
Formulae:
Calculations:
NReA = NReB
Therefore,
4 x 50 x 1 / 1 = 10 x VB x 0.8 / 2
200 = 4 VB
VB = 200/4 = 50 cm/sec
A geometrically similar model of an air duct is built 1:30 scale and tested with water
which is 50 times more viscous and 800 times denser than air. When tested under
dynamically similar conditions, the pressure drop is 2.25 atm in the model. Find the
corresponding pressure drop in the full-scale prototype.
Data:
Dimensions of the model = 1/30 times the dimensions of prototype
Viscosity of fluid in the model = 50 times the viscosity of fluid in prototype
Density of fluid in the model = 800 times the density of the fluid in prototype
Formulae:
From 1,
NRem = NRep
Calculations:
NRem = NRep
1 x vm x 800 / 50 = 30 x vp x 1 / 1
vm = 1.875 vp
2
f = 0.079 (Dv/)-0.25
fm/fp = (1 x vm x 800 / 50)-0.25 / (30 x (vm/1.875) x 1 / 1)-0.25 = 16-0.25/16-0.25 =
1
U-tube Manometer
Data:
Manometer reading (h) = 26 mm Hg = 0.026 m Hg
Density of mercury (m) = 13.6 gm/cc
Density of water = 1 gm/cc
Molecular weight of air (MW) = 28.8
Temperature of air = 20 o C = 293 K
R = 8314 J/(kmol.K)
Formulae:
P1 - P2 = p = (m - )gh.
Calculations:
3
centerline of pipe Y. Find the manometer reading as shown by a centimeter scale
attached to it.
Data:
Formula:
P = gh
Principle:
Pressure at the same level in a continuous body of static fluid is equal.
Calculations:
Equating the pressure at the two legs of the manometer at OO':
For the right hand leg pressure at O' = Po' = Py + (1.5 + a) x 800 x 9.812 N/m2
Px + (2.5 + 1.5) x 1590 x 9.812 + a x 13600 x 9.812 = Py + (1.5 + a) x 800 x
9.812
4
Flow Direction - Determination
The pressures at two sections of a horizontal pipe are 0.3 kgf/cm 2 and 0.6 kgf/cm2
and the diameters are 7.5 cm, and 15 cm respectively. Determine the direction of
flow if water flows at a rate of 8.5 kg/sec. State your assumptions.
Data:
P1 = 0.3 kgf/cm2
P2 = 0.6 kgf/cm2
D1 = 7.5 cm
D2 = 15 cm
Mass flow rate = 8.5 kg/sec
Formulae:
Equation of continuity
1A1v1 = 2A2v2
Bernoulli's equation
For the flow direction from 1 to 2,
2 2
p v p v
1 1 z 2 2 z h w q
ρ g 2g 1 ρ g 2g 2
1 2
Calculations:
So to balance the above equation, the quantity h has to have negative values. This
is not possible.
Therefore the flow direction is from the end at which pressure is 0.6 kgf/cm2 and
diameter is 15 mm to the end at which pressure is 0.3 kgf/cm 2 and diameter is 7.5
mm.
5
Flow Rate of Water through Ion-Exchange Column
Figure shows a water softener in which water trickles by gravity over a bed of
spherical ion-exchange resin particles, each 0.05 inch in diameter. The bed has a
porosity of 0.33. Calculate the volumetric flow rate of water.
Calculations:
Data:
= 1 cp = 1 x 6.72x10-4 lb/(ft.sec)
= 0.33
= 62.3 lb/ft3
x = 1 ft
g = 32.2 ft/sec2
Formula:
Applying Bernoulli's equation from the top surface of the fluid to the outlet of the
packed bed and ignoring the kinetic-energy term and the pressure drop through the
support screen, which are both small, we find
g(z) = hf
hf = p/
Calculations:
Therefore,
Vs = 32.2 x 1.25 x (0.05/12)2 x 0.333 x 62.3 / ( 150 x (1 x 6.72x10-4) x (1 - 0.33)2
x 1)
= 0.035 ft/sec = 0.011 m/sec.
6
NRem = (0.05/12) x 0.035 x 62.3 / (1 x 6.72x10-4 x (1 - 0.33) ) = 20.2
This is slightly above the value of 10 (up to which the Blake-Kozeny Equation can be
used), for which we can safely use without appreciable error.
Flow in Siphon
A siphon consisting of a 3 cm diameter tube is used to drain water from a tank. The
outlet end of the tube is 2 m below the water surface in the tank. Neglecting
friction, calculate the discharge. If the peak point of the siphon is 1.4 m above the
water surface in the tank, estimate the pressure of fluid at the point of siphon.
Data:
Diameter of tube = 3 cm = 0.03 m
Formula:
Applying Bernoulli's equation for the points 1 and 3, (i.e. comparing the energy
levels for the fluid in the tank surface and at the discharge point of tube)
p1 = 0 N/m2(g)
p3 = 0 N/m2(g)
z1 = 0 m
z3 = -2 m
Since the rate of fall of liquid level in the tank is almost negligible,
v1 = 0 m/sec.
Therefore,
0 + 0 + 0 = 0 + (v32 / 2g) - 2
7
v3 = (2 x 2g)0.5 = 6.265 m/sec
Q = (/4)D2 v = (/4) x 0.032 x 6.265 = 0.00443 m3/sec = 15.94 m3/hr
Applying Bernoulli's equation for the points 1 and 2, (i.e. comparing the energy
levels for the fluid at the tank surface to the peak point of siphon)
p1 = 0 N/m2(g)
z2 = 1.4 m
v2 = v3 = 6.265 m/sec (since the cross sectional area of sections 2 and 3 are the
same)
0 + 0 + 0 = p2 / (g) + 6.2652 / (2g) + 1.4
p2 / (g) = -3.4 m
p2 = -3.4 x 1000 x 9.812 N/m2(g) = -33360.8 N/m2(g)
A capillary tube 0.2 cm in diameter and 10 cm long discharge one liter of a liquid in
ten minutes under a pressure difference of 5 cm mercury. Find the viscosity of the
liquid using the following data:
Data:
Hagen-Poiseuille law (pressure drop for laminar flow is related to the flow
parameters)
4
ΔpD
Q
128μ L
Calculations:
= 15.72 cp
8
Frictional Losses in Pipelines
2.16 m3/h water at 320 K is pumped through a 40 mm I.D. pipe through a length of
150 m in a horizontal direction and up through a vertical height of 12 m. In the pipe
there are fittings equivalent to 260 pipe diameters. What power must be supplied to
the pump if it is 60% efficient? Take the value of fanning friction factor as 0.008.
Water viscosity is 0.65 cp, and density = 1 gm/cc.
Data:
Flow rate (Q) = 2.16 m3/h = 2.16/3600 m3/sec = 0.0006 m3/sec
Diameter of pipe (D) = 40 mm = 0.04 m
Length of pipe in horizontal direction = 150 m
Length of pipe in vertical direction (z) = 12 m
Equivalent length of fittings = 260 pipe diameters
Friction factor (f) = 0.008
Efficiency of pump () = 0.6
Viscosity of fluid () = 0.65 cp = 0.00065 kg/(m.sec)
Density of fluid () = 1 gm/cc = 1000 kg/m3
Formulae:
1. Bernoulli's equation
p1 v12 p 2 v 22
z1 z2 h w q
ρ1g 2g ρ 2 g 2g
3. Power required for pumping = (mass flow rate x head developed by pump)/
= (volumetric flow rate x pressure developed by pump)/
Calculations:
Length of pipe with fittings = 150 + 12 + 260 x 0.04 m = 172.4 m
Velocity = volumetric flow rate/flow cross sectional area
= 0.0006/((/4) x 0.042) = 0.477 m/sec
= 15.69/9.812 = 1.6 m
Pump head(q) = z + h = 12 + 1.6 = 13.6 m
Pressure developed by pump = 13.6 x 1000 x 9.812
= 133443.2 N/m2
9
Equivalent Length of Fittings
Water is pumped from a reservoir to a height of 1000 m from the reservoir level,
through a pipe of 15 cm I.D. at an average velocity of 4 m/s. If the pipeline along
with the fittings is equivalent to 2000 m long and the overall efficiency is 70%, what
is the energy required for pumping? Friction factor f = 0.046 Re-0.2.
Data:
Diameter of pipe (D) = 15 cm
Average Velocity (v) = 4 m/s
Equivalent Length of pipe with fittings = 2000 m
Efficiency of pump () = 0.7
Formula:
Bernoulli equation:
2 2
p v p v
1 1 z 2 2 z h w q
ρ g 2g 1 ρ g 2g 2
1 2
Frictional losses:
2
2fLv
h
f D
Calculations:
10
Data:
Cv 2(Pa P )
b
V
b 4 ρ
1β
Where
= Db / Da
The pressure difference measured by the manometer
Pa - Pb = (m - )ghm
Power consumption of the meter = volumetric flow rate x permanent pressure loss
Calculations:
The rate of flow of water in a 150 mm diameter pipe is measured with a venturi
meter of 50 mm diameter throat. When the pressure drop over the converging
section is 100 mm of water, the flow rate is 2.7 kg/sec. What is the coefficient of the
meter?
Data:
11
Formulae:
Cv 2(P P )
a b
V
b 4 ρ
1β
Where = Db/Da
Calculations:
Volumetric flow rate = mass flow rate / density = 2.7 / 1000 = 0.0027 m 3/sec
Velocity at the throat = Volumetric flow rate / cross sectional area of throat
= 0.0027 / (Db2/4) = 0.0027 / ( x 0.052/4) = 1.375 m/sec
Therefore,
Orifice Sizing
Data:
Density of brine () = 1.2 x 1000 kg/m3 = 1200 kg/m3
Diameter of pipe (Da) = 10 cm = 0.1 m
Maximum flow rate (Q) = 1200 liters/min = 1.2 m3/min = 0.02 m3/sec
Maximum manometer reading (hm) = 40 cm = 0.4 m of Hg
Density of manometric fluid (m) = 13600 kg/m3
Orifice coefficient (Co) = 0.62
Formulae:
Co 2(Pa P )
b
V
b 4 ρ
1β
Where
= Db/Da
Db = diameter of orifice
Da = diameter of pipe
For the U-tube manometer,
Pa - Pb = (m - )ghm
12
Calculations:
Therefore,
Cross sectional area of orifice = volumetric flow rate / velocity at the orifice
= 0.02 / 5.584 = 0.00358 m2
A Newtonian fluid having a viscosity of 1.23 poise, and a density of 0.893 gm/cm3,
is flowing through a straight, circular pipe having an inside diameter of 5 cm. A pitot
tube is installed on the pipeline with its impact tube located at the center of the pipe
cross section. At a certain flow rate, the pitot tube indicates a reading of 8 cm of
mercury. Determine the volumetric flow rate of the fluid.
Data:
Viscosity of fluid () = 1.23 poise = 0.123 kg/(m.sec)
Density of fluid () = 0.893 gm/cm3 = 893 kg/m3
Diameter of pipe (D) = 5 cm = 0.05 m
Manometer reading (hm) = 8 cm of Hg = 0.08 m of Hg
Formulae:
Calculations:
Since the flow is laminar (NRe < 2100), the average velocity is given by
Vavg = Vo / 2 = 4.73 / 2
= 2.365 m/sec (velocity at the centerline is the maximum velocity)
A rectangular weir 0.75 m high and 1.5 m long is to be used for discharging water
from a tank under a head of 0.5 m. Estimate the discharge (i) when it is used as a
suppressed weir (ii) when it is used as a contract weir. Use Rehbock equation for
estimating Cd in both cases.
13
Data:
Weir height (P) = 0.75 m
Width of weir (B) = 1.5 m
Head (H) = 0.5 m
Formulae:
Rehbock equation:
1 0.08H
C 0.605
d 1000H P
Suppressed weir:
2 3
QC B 2gH 2
d3
Contracted weir:
2 3
QC
d3
B 0.1nH 2gH 2
i. Suppressed weir:
A rotameter calibrated for metering has a scale ranging from 0.014 m 3/min to 0.14
m3/min. It is intended to use this meter for metering a gas of density 1.3 kg/m3
with in a flow range of 0.028 m3/min to 0.28 m3/min. What should be the density of
the new float if the original one has a density of 1900 kg/m 3? Both the floats can be
assumed to have the same volume and shape.
Data:
Old:
Density of float (f)= 1900 kg/m3
Density of gas () = 1.3 kg/m3
Qmin = 0.014 m3/min
New:
Density of gas () = 1.3 kg/m3
Qmin = 0.028 m3/min
14
Formula:
2V (ρ ρ)g
QC A f f
D 2 A 2
ρA 1 2
f
A1
Where
Q = volumetric flow rate
A2 = area of annulus (area between the pipe and the float)
A1 = area of pipe
Af = area of float
Vf = volume of float
CD = rotameter coefficient
Calculations:
From the equation, for the same float area and float volume and the pipe geometry,
Q = k (f - )0.5
Calculate the pressure drop of air flowing at 30 oC and 1 atm pressure through a bed
of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter
and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cP
and the density is 0.001156 gm/cc.
Data:
Mass flow rate of Air = 60 kg/min = 1 kg/sec
Density of Air () = 0.001156 gm/cc = 1.156 kg/m3
Viscosity of Air () = 0.0182 cP = 0.0182 x 10-3 kg/(m.sec)
NRePM = DpVo/((1 - ) )
For laminar flow (i.e. NRePM < 10) pressure drop is given by Blake-Kozeny equation:
15
2 2 3
ΔpΦ s Dp ε
150
2
LV0μ(1 ε)
For turbulent flow (i.e. NRePM > 1000) pressure drop is given by Burke-Plummer
equation:
3
ΔpΦ s D p ε
1.75
2
Lρ Vo (1 ε)
3
ΔpΦ sDp ε 150(1 ε)
1.75
2 Φ sDp Voρ/μ
Lρ V (1 ε)
o
Calculations:
Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m 3/sec
i.e. p x 0.0125 x 0.383 / ( 2.5 x 1.156 x 0.7052 x ( 1 - 0.38 ) ) = 150 / 903 + 1.75
where p is the pressure drop in kgf/cm2, Z is the bed height in meter, is the
density in g/cc, Dp is nominal diameter of Berl saddles in cm, V s is the superficial
linear velocity in m/sec.
16
Data:
Formula:
Ideal gas law:
PV = nRT
Formula given,
Calculations:
= Mn/V = MP/(RT)
= 29 x 7 x 1.01325 x 105 / (8314 x (273 + 127) )
= 6.185 kg/m3
= 6.185 x 10-3 g/cc
A regenerative heater is packed with a bed of 6 mm cubes. The cubes are poured
into the cylindrical shell of the regenerator to a depth of 3.5 m such that the bed
porosity was 0.44. If air flows through this bed entering at 25 oC and 7 atm abs and
leaving at 200oC, calculate the pressure drop across the bed when the flow rate is
500 kg/hr per square meter of empty bed cross section. Assume average viscosity
as 0.025 cP and density as 6.8 kg/m3.
Data:
Mass flow rate of Air / unit area = 500 kg/(hr.m2) = 0.139 kg/(sec.m2)
Density of Air () = 6.8 kg/m3
Viscosity of Air ()= 0.025 cP = 0.025 x 10-3 kg/(m.sec)
Bed porosity () = 0.44
Length of bed (L) = 3.5 m
Diameter of particles (Dp)= 6 mm = 0.006 m
17
Formulae:
Sphericity ( s) = 6 Vp / (DpSp)
Vp = volume of particle = Dp3 (for cube)
Sp = surface area of particle = 6 x Dp2 (for cube)
NRePM = DpVo/( (1 - ) )
For laminar flow (i.e. NRePM < 10) pressure drop is given by Blake-Kozeny equation:
2 2 3
ΔpΦ s Dp ε
150
2
LV0μ(1 ε)
For turbulent flow (i.e. NRePM > 1000) pressure drop is given by Burke-Plummer
equation:
3
ΔpΦ s D p ε
1.75
2
Lρ Vo (1 ε)
3
ΔpΦ sDp ε 150(1 ε)
1.75
2 Φ sDp Voρ/μ
Lρ V (1 ε)
o
Calculations:
i.e. p x 0.006 x 0.443 / ( 3.5 x 6.8 x 0.02042 x ( 1 - 0.44 ) ) = 150 / 59.45 + 1.75
p x 0.0712= 4.273
18
same temperature. The pressure at the bottom of the bed is 30 lb f/in.2 abs. The bed
porosity is 0.40. If the gas has average properties similar to propane and the time
of contact (based on superficial velocity of gas) between the gas and the catalyst is
10 s, what is the inlet pressure?
Data:
Diameter of tower (D) = 20 ft = 20 x 0.3038 m = 6.096 m
Height of tower (L) = 50 ft = 15.24 m
Diameter of packing (Dp) = 1 inch = 2.54 cm = 0.0254 m
Temperature of gas (T) = 500oF = (500 - 32) x 5/9 oC = 260oC
= (273 + 260) K = 533 K
Pressure at the bottom = 30 lbf/in.2 abs = 30/14.7 atm(a) = 2.04
atm(a) = 206785.7 N/m2(a)
Bed porosity () = 0.4
Time of contact = 10 sec
Molecular weight of gas = 44 (molecular weight of propane (C3H8) )
Formulae:
Density of gas
= PM/(RT)
Ergun equation:
Calculations:
For these NRePM Burke-Plummer equation can be used.(i.e. Turbulent part of the
Ergun's equation)
p = 46937.5 N/m2
Pressure at the inlet of the column = 206785.7 + 46937.5 = 253723.2 N/m 2(a) =
36.81 lbf/in2(a)
19