ISRO Mechanical Engineering Fluid Mechanics
Fluid Properties - Bulk modulus
2011. 10. A liquid compressed in cylinder has a volume of 0.04 m3 at 50kg/cm2 and a volume of 0.039 m3 at 150 kg/cm2. The bulk modulus of elasticity of liquid is
a) 400 kg/cm2
b) 40 x 106 kg/cm2
c) 40 x 105 kg/cm2
d) 4000 kg/cm2
Answer
d) 4000 kg/cm2
Explanation
The bulk elastic properties of a material determine how much it will compress under a given amount of external pressure. The ratio of the change in pressure to the fractional volume compression is called the bulk modulus of the material. The reciprocal of the bulk modulus is called the compressibility of the substance.
Bulk Modulus, B = ΔP/(ΔV/V)
Here ΔP = 150-50 =100 kg/cm2
ΔV = 0.04-0.039 = 0.001m3
V = 0.04m3
B = (100/(0.001/0.04)) = 100x0.04/0.001 = 4000 kg/cm2
Fluid Properties - Surface Tension
2011.13 If the surface tension of water-air interface is 0.073N/m, the gauge pressure inside a rain drop of 1mm diameter will bea) 0.146 N/m2
b) 73 N/m2
c) 146 N/m2
d) 292 N/m2
Answer
d) 292 N/m2
Explanation
Gauge pressure inside a bubble = 4T/r
Gauge pressure inside a drop = 2T/r
=2 x 0.073/0.0005 = 292 N/m2
2018.2. A glass tube of 8 mm diameter is immersed in water. Surface tension of water is 0.0075 kg/m and rise of water in the tube due to capillary effect is
a) 7.5 mm
b) 3.75 mm
c) 11.25 mm
d) none of these
Answer
b) 3.75 mm
Explanation
Surface tension of the water tends to pull the water above the level and the weight of water tends to pull it back to its level. For water the angle between the surface tension force and plane of water is 90o.
Force due to surface tension = Weight of water above the level
Force due to surface tension = Surface tension x circumference (as surface tension acts only at the interface) = 0.0075 x π x 0.008 N
Let the rise of the water be h m
Weight of water above the level = Density of water xArea of cross section x height = 1000xπ d2/4 h N
Equating both
0.0075 x π x 0.008 = 1000xπ x 0.008 x 0.008 x h/4
0.0075 = 8h/4
h = 0.0075/2 m = 7.5/2 mm = 3.75 mm
Pressure Measurement
2006. 23. 100m of water column is equal toa) 1000 kN/m2
b) 100 kN/m2
c) 10 kN/m2
d) 1 kN/m2
Answer
a) 1000 kN/m2
Explanation
P = ρgh, ρ = 1000kg/m3, g=10m/s2, h = 100m
P = 1000x10x100 N/m2 = 1000kN/m2
Stream Function & Velocity Potential
2009.2. If the stream function of a two dimensional flow is Ψ = 3xy, then the velocity at a point (2,3) is
a) 7.21 unit
b) 18 unit
c) 10.82 unit
d) 54 unit
Answer
c) 10.82 unit
Explanation
u = ∂ Ψ/∂ y = 3x
v =- ∂ Ψ/∂ x = -3y
At (2,3), u = 3x2 = 6 unit
At (2,3), v = -3x3 = -9 unit
Resultant velocity = √(62+(-9)2)= √(117)=3√(13)=10.81 unit
Answer
Explanation
2018.1. If the stream function of a two dimensional flow is Ψ = 5xy, then the velocity at a point (3,4) is
a) 25 m/s
b) 20 m/s
c) 7 m/s
d) 5 m/s
Answer
a) 25 m/s
Explanation
u = ∂ Ψ/∂ y = 5x
v =- ∂ Ψ/∂ x = -5y
At (3,4) u = 5x3 = 15 m/s
At (3,4), v = -5x4 = -20 m/s
Resultant velocity = √(152+(-20)2)= 25 m/s
Dimensionless Numbers
2018.41. Euler's dimensionless number relates the following
a) inertial force and gravity
b) viscous force and inertial force
c) pressure force and inertial force
d) viscous force and pressure force
Answer
c) pressure force and inertial force
Explanation
inertial and gravity - Froude's number - The Froude's number is a measurement of bulk flow characteristics such as waves, sand bed-forms, flow/depth interactions at a cross section or between boulders.
viscous force and inertial force - Reynolds number - The Reynolds Number can be used to determine if flow is laminar, transient or turbulent.
pressure force and inertial force - Euler number - It expresses the relationship between a local pressure drop caused by a restriction and the kinetic energy per volume of the flow, and is used to characterize energy losses in the flow, where a perfect friction-less flow corresponds to an Euler number of 0.
viscous force and pressure force - No such number
Bernoulli's Equation
2018.40. Speed of an aeroplane is measured by
a) pitot tube
b) hot wire anemometer
c) venturimeter
d) rotameter
Answer
a) pitot tube
Pitot tube measures the velocity flowing liquid or air by measuring the pressure difference between stagnation and static points in flow using Bernoulli's equation.
Hot Wire Anemometer - When an electrically heated wire is placed in a flowing gas stream, heat is transferred from the wire to the gas and hence the temperature of the wire reduces, and due to this, the resistance of the wire also changes. This change in resistance of the wire becomes a measure of flow rate.
Venturimeter - A venturimeter is a device used for measuring the rate of flow of a fluid flowing through a pipe.
Two cross section, first at the inlet and the second one is present at the throat. The difference in the pressure heads of these two sections is used to calculate the rate of flow through venturimeter. As the water enters at the inlet section i.e. in the converging part it converges and reaches to the throat.
The throat has the uniform cross section area and least cross section area in the venturimeter. As the water enters in the throat its velocity gets increases and due to increase in the velocity the pressure drops to the minimum.
Now there is a pressure difference of the fluid at the two sections. At the section 1(i.e. at the inlet) the pressure of the fluid is maximum and the velocity is minimum. And at the section 2 (at the throat) the velocity of the fluid is maximum and the pressure is minimum.
The pressure difference at the two section can be seen in the manometer attached at both the section.
This pressure difference is used to calculate the rate flow of a fluid flowing through a pipe.
Rotameter - A rotameter is a device that measures the volumetric flow rate of fluid in a closed tube.
A rotameter consists of a tapered tube, typically made of glass with a 'float' (a shaped weight, made either of anodized aluminum or a ceramic), inside that is pushed up by the drag force of the flow and pulled down by gravity. The drag force for a given fluid and float cross section is a function of flow speed squared only, see drag equation.
A higher volumetric flow rate through a given area increases flow speed and drag force, so the float will be pushed upwards. However, as the inside of the rotameter is cone shaped (widens), the area around the float through which the medium flows increases, the flow speed and drag force decrease until there is mechanical equilibrium with the float's weight.
Floats are made in many different shapes, with spheres and ellipsoids being the most common. The float may be diagonally grooved and partially colored so that it rotates axially as the fluid passes. This shows if the float is stuck since it will only rotate if it is free. Readings are usually taken at the top of the widest part of the float; the center for an ellipsoid, or the top for a cylinder. Some manufacturers use a different standard.
The "float" must not float in the fluid: it has to have a higher density than the fluid, otherwise it will float to the top even if there is no flow.
The mechanical nature of the measuring principle provides a flow measurement device that does not require any electrical power. If the tube is made of metal, the float position is transferred to an external indicator via a magnetic coupling. This capability has considerably expanded the range of applications for the variable area flowmeter, since the measurement can observed remotely from the process or used for automatic control.
Orifice
2011.11. A fluid jet is discharging from a 100mm nozzle and the vena contracta formed has a diameter of 90 mm. If the coefficient of velocity is 0.95, then the coefficient of discharge for the nozzle is
a) 0.7695
b) 0.81
c) 0.9025
d) 0.855
Answer
a) 0.7695
Explanation
increases with the head of water. The ratio of a actual discharge through an orifice to the theoretical discharge is known as coefficient of discharge. Coefficient of discharge = Actual discharge/Theoretical dischargea) 0.7695
b) 0.81
c) 0.9025
d) 0.855
Answer
a) 0.7695
Explanation
The ratio of actual velocity of the jet, at vena-contracta, to the theoretical velocity is known as coefficient of velocity.The difference between the velocities is due to friction of the orifice. The value of Coefficient of velocity varies slightly with the different shapes of the edges of the orifice. This value is very small for sharp-edged orifices. For a sharp edged orifice, the value of C\sub>v
Actual discharge = Actual area x Actual velocity
Theoretical discharge = Theoretical area x Theoretical velocity
Given Coefficient of velocity = Actual velocity/Theoretical velocity = 0.95
Theoretical area = π(dtheoretical)2
Actual area = π(dactual)2/r
Coefficient of discharge = Actual discharge/Theoretical discharge
= (Actual Area x Actual velocity)/(Theoretical Area x Theoretical velocity)
= (Actual Area/Theoretical Ara) x (Actual velocity x Theoretical velocity)
= (90/100)2 x 0.95
= 0.9 x 0.9 x 0.95 = 0.7695
Losses in flow through pipes
2011.9 Flow takes place at Reynolds number of 1500 in two different pipes with relative roughness of 0.001 and 0.002. The friction factor
a) Will be higher for the pipe with relative roughness of 0.001
b) Will be higher for the pipe having relative roughness of 0.002
c) Will be same in both the pipes
d) In the two pipes cannot be compared on the basis of data given.
Answer
c) Will be same in both the pipes
Explanation
The friction factor f for a laminar flow (Re < 2320) in a circular pipe is given by the formula f = 64/Re. It is independent of relative roughness for the given Reynolds number
a) 0.025 m
b) 0.05 m
c) 0.8 m
d) 1.2 m
Answer
Explanation
Head loss due to sudden expansion is given by HL = (v1-v2)2/2g
Q = Ρ Av = Ρ A1v1 = Ρ A2v2
v1 = Q/ΡA1 = 0.5/(1000x0.1)=5x10-3 m/s
v2 = Q/ΡA2 = 0.5/(1000x0.5)=1x10-3 m/s
HL = (4x10-3)2/(2x9.81) ~ 0.8 x 10-6 m
Hydro-power Plant
2013.14. The purpose of surge tank in a pipeline is to
a) smoothen the flow of water
b) minimize friction loses in pipe
c) prevent occurrence of hydraulic jump
d) relieve pressure due to water hammer
Answer
d) relieve pressure due to water hammer
Explanation [Fluid Mechanics]
Surge tank is a small storage tank or reservoir required in the hydro power plants for regulating the water flow during load reduction and sudden increase in the load on the hydro generator (water flow transients in penstock) and thus reducing the pressure on the penstock.
When the load on the generator is reduced suddenly (load throw-off), governor closes the turbine gates and thus create an increase in pressure on the penstock. This may result in water " hammering effect" and may require pipe of extra ordinary strength this pressure, otherwise the penstock may burst. To avoid this positive water hammering pressure some means is provided to take the rejected flow of water. This can be achieved by providing small storage reservoir or tank to accommodate this rejected flow. This small storage reservoir is surge tank and located close to the power station. When the turbine gates suddenly opens due to requirement of more water due to increase in the load demand on the generator, water has to rush through the pipe (penstock) and there is a tendency to cause a vacuum in the pipe supplying the water.
The pipe supplying the water must withstand the high pressures caused by the sudden closing of the turbine gates (known as positive water hammer) and there should not be any vacuum in the pipe line when the gate suddenly opens.
The water hammering is defined as the change in the pressure rapidly above or below normal pressure caused by sudden changes in the rate of water flow through the pipe according to the demand of the prime-mover. The water hammer occurs at all the points in the penstock between the forebay or surge tank and the turbine because of the sudden changes in the demand of the water during the load fluctuations.
Surge tank is an open tank which is often used with the pressure conduit of considerable length. The main purpose of providing surge tank is to reduce the distance between the free water surface and turbine thereby reducing the water hammer effect on the penstock and also and also protecting upstream tunnel from high pressure rises. Surge tank also serves as a supply tank to the turbine when the water in the pipe is accelerating during increased load conditions and as storage tank when the water is decelerating during the reduced load conditions. Surge tanks are generally built high enough so that the water cannot overflow even with a full load on the turbine.
When load demand on the generator decreases, it leads to rise in the water level in the surge tank. This produces a retarding head and reduces water velocity in the penstock. The reduction in the velocity to the desired levels, makes the water in the tank to rise and fall until oscillations are damped out.
When load demand on the generator increases, governor opens the turbine gates in order to allow more water flow through the penstock to supply the increased load demand thereby creating a negative pressure or vacuum in the penstock. This negative pressure in the penstock creates necessary acceleration force and is objectionable for very long conduits due to difficult turbine regulation. Under this condition additional water flows from the surge tank. As a result the water level in the surge tank falls, an acceleration head is created and flow of water in the penstock increases.
Thus surge tank helps in stabilizing the velocity and pressure in the penstock and protects penstock from water hammering and negative pressure or vacuum.
Turbines
2018.3. An impulse turbine produces 125 hp under a head of 25 metres. By what percentage should the speed be increased for a head of 100 metres.a) 25 %
b) 50 %
c) 75 %
d) 100 %
Answer
d) 100%
Explanation
If a turbine is working under different heads, the behavior of turbine can be easily known from the unit quantities
Nu = N1/√(H1) = N2/√(H2)
N1/√(H1) = N2/√(H2)
N2/N1 = √ (100/25)=2
ie, the speed has to be increased by 100%.
2017.2.44. Which of the following is impulse turbine?
a) Francis turbine
b) Kaplan turbine
c) Pelton turbine
d) none of these
Answer
c) Pelton turbine
Explanation [Fluid Mechanics]
Impulse Turbine - The whole pressure energy is transformed to kinetic energy. Eg. Pelton turbine
Reaction Turbine - All pressure energy is not transformed into kinetic energy. Eg. Francis, Kaplan turbine
Answer
c) Pelton turbine
Explanation [Fluid Mechanics]
Impulse Turbine - The whole pressure energy is transformed to kinetic energy. Eg. Pelton turbine
Reaction Turbine - All pressure energy is not transformed into kinetic energy. Eg. Francis, Kaplan turbine
Centrifugal Pump
2008.38. In centrifugal pumps maximum efficiency is obtained when the blades area) Straight
b) Bent forward
c) Bent backward
d) Radial
Answer
c) Bent backward
Explanation
For fluids in in-compressible regime of operation such as water backward curved vanes are used for maximum efficiency.
For radial and forward bent vanes the power requirement increases monotonically as the discharge increases. But for backward curved the power requirement peaks at maximum efficiency.
2018.6. Efficiency of a centrifugal pump is maximum when the blades are
a) straight
b) forward curved
c) backward curved
d) forward and backward curved
Answer
c) backward curved
Explanation
For fluids in in-compressible regime of operation such as water backward curved vanes are used for maximum efficiency.
For radial and forward bent vanes the power requirement increases monotonically as the discharge increases. But for backward curved the power requirement peaks at maximum efficiency.
2018.4. The specific speed of centrifugal pump, delivering 1000 litres of water per second against a head of 16 metre at 800 rpm
a) 50 rpm
b) 100 rpm
c) 150 rpm
d) 200 rpm
Answer
b) 100 rpm
Explanation
An impeller with a low specific speed has a thin profile (the shrouds are close together) and a large outside diameter relative to the eye diameter. An impeller with a high specific speed has a fat profile (the shrouds are far apart) and has an eye diameter that is closer in size to the impeller outer diameter.
Specific speed Ns = N √(Q)/H3/4
N = 800 rpm
Q = 1000 l/s = 1m3/s
H = 16m
Ns = 800 x 1/(16)3/4
= 800/8 = 100 rpm
Answer
a) 2.0-5.0
Explanation
Dimensionless specific speed range
Pelton wheel - 0.03 to 0.3
Francis turbine - 0.3 to 2
Kaplan turbine - 2 to 5
Hydraulic Turbines
2017.2.13. The specific speed of the Kaplan turbine is
a) 2.0-5.0
b) < 0.3<0 .3="" 0="" p="">
c) 0.3-2.0
d) none of these
Answer
a) 2.0-5.0
Explanation
Dimensionless specific speed range
Pelton wheel - 0.03 to 0.3
Francis turbine - 0.3 to 2
Kaplan turbine - 2 to 5