Q1. A 200mm diameter pipeline divides into two smaller pipelines, one being 100mm in diameter and the other being 150mm in diameter. If the velocity in the 100mm pipe is 0.3m/s and that in the 150mm is 0.6m/s, calculate the flowrate in litres/s and the velocity in m/s in the 200mm pipe. [ 13 l/s, 0.413m/s]
Solution: Flow Rate = (pi/4)(0.1)2(0.3) + (pi/4)(0.15)2(0.6) = 0.01296 cubicmeters/sec
= 12.96 l/sec. (ans.).
Velocity = (Flow Rate)/(Area) = [0.01296]/[(pi/4)(0.1)2] = 0.4125 m/s.
Q2. If the specific volume of water is 0.001251 m3 /kg. calculate:
The mass and weight of the water when put in a container of diameter 1m and length 2m.
Solution: A. The density of water = 1/ the specific volume of water = 1/0.001251
= 799.361 kg/ m3
B.The mass of the water = (The density of water)(volume of container) =
= (799.361)[(pi/4)(1)2(2)] = (799.361)[1.571] = 1255.63 kg.
The weight of the water = (mass)(acceleration due to gravity) = (1255.63)(9.81)
= 12318 N = 12.318 kN.
Q3. One kilogram of lead is taken to the moon, where the local acceleration of gravity is approximately one – sixth of the earth’s gravity. What is its mass and weight on the moon?
Solution: Mass = 1kg.
Weight = (mass)(acceleration due to gravity) = (1)(9.81/6) = 1.635 N.
Q4. The volume rate of flow is often specified in litters per second and it is necessary to know the flow rate in cubic meters per second. Derive a conversion factor to convert from litters per second to cubic meters per second.
Solution: 1 litre/sec = 1000ml/sec = 1000cm3/sec = 1000(10-2m)3/sec = 1000(10-6)m3/sec
= 10-3 cubic meters per second.
Q5. A mass of 10 kg has a velocity of 12 m/s. Determine the kinetic energy of the mass is N.m.
Solution: kinetic energy of the mass = (½)(mass)(velocity)2 = (½)(10)(12)2 = 720 Nm.
Q6. State the difference between
Solids and fluids
Liquids and gasses
Solution: A. The distance between molecules is less for solids compared to the fluids.
The intermolecular attraction is more for solids compared to the fluids.
B. The distance between molecules is less for liquids compared to the gases.
The intermolecular attraction is more for liquids compared to the gases.
Q7. An object of mass 15kg is raised through a height of 2m and then released. Determine its final velocity.
Solution: v2 – u2 = 2gs. So, v2 – 02 = 2(9.81)(2) = 39.24. So, v = 6.26 m/sec.
Bulk modulus of elastic
Q1. What is meant by the term bulk modulus of fluid?
Solution: Bulk modulus of fluid may be defined as the increase in pressure required for unit decrease in volumetric strain.
Q2. Select the correct statement:
The unit for bulk modulus of elasticity is:
N/m2
N/m3
N/m
Solution: A.
The percent reduction in volume of 10m3 of water when subjected to pressure of 120 kpa and bulk modulus = 2.2 Gpa is:
0.0455
0.00545
5.45
Solution: B.
Q3. Determine the percent reduction in volume of 10 ms of water if the pressure is increased by 3 Mpa. Take bulk modulus= 2.2 Gpa. Ans 0.136
Solution: Percent reduction in volume = [(increase in pressure)/(Bulk Modulus)]X100
= [(3X106)/(2.2X109)]X100 = 0.136.
Q4. A hydraulic fluid has a bulk modulus of 1.7 Gpa. By what percent will its volume change with a pressure increase of 100 Mpa. Ans. 5.88
Solution: Percent reduction in volume = [(increase in pressure)/(Bulk Modulus)]X100
= [(100X106)/(1.7X109)]X100 = 5.88.
Q5. A mass travelling at 10 mm/s is brought to rest by hydraulic buffer 0.04 seconds. If the buffer has a piston length of 400mm and is filled with oil, calculate the rise in pressure of the trapped oil. Take bulk modulus = 2.8 Gpa. Use for change in length dL= (u+v)/2x T. Ans. 14 x 105 Pa
Solution: dL = change in length = (u+v)/2x T = [(10+0)/2]X(0.04) = 0.2 mm.
Volumetric strain = linear strain = (0.2mm)/400mm = 0.0005
Rise in pressure = (Bulk Modulus)(Volumetric strain)
= (2.8X109)(0.0005) = 14 x 105 Pa.
Thermos fluids (viscosity)
Q1. A piston 50 mm is diameter move inside a cylinder of 52 mm internal diameter and 10 cm long. The clearance between the piston and the cylinder is filled with oil. Determine the force required to move the piston at 1 m/s. take μ = 4400 x 10 -4Ns/m2 Ans. 6.9N
Solution: From Newton’s Equation of Viscosity,
Shear Stress = (viscosity)(velocity)/(depth of the oil)
= (4400 x 10 -4)(1)/(0.002/2) = 440 N/m2
Force required to move the piston = (Shear Stress)(Area)
= (440)[(pi)(d)(l)] = (440)[(pi)(0.05)(0.1)] = (440)(0.01571) = 6.9N.
Q2. The viscosity of a liquid
Is independent of temperature
Increases with increase in temperature
Decrease with increases in temperature
Solution: C.
Q3. The unit of viscosity in the SI system is
Nm/ S2
Ns/m2
m2/s
Solution: B.
Q4. The units of kinematic viscosity in the SI system is
N.m/S2
Ns/m2
m2/s
Solution: C.
Q5. A body weighing 2.5 N and 5cm X 5cm X5cm in size and lubricated with oil.
( μ = 814 x 10-4Ns/m2) slides down a surface making an angle of 45 with the horizontal. Determine the speed of the body if the thickness of the lubricating film is 1mm. Ans. 8.69 m/s
Solution: Shear Force = Tangential component of weight = 2.5cos450 = 1.7678 N.
Shear Stress = (Shear Force)/(Area) = (1.7678)/[(0.05)(0.05)] = 707.11 N/m2
Shear Stress = (viscosity)(velocity)/(depth of the oil).
So, velocity = (Shear Stress)(depth of the oil)/(viscosity)
= (707.11) (0.001)/(814 x 10-4) = 8.69 m/s.
Pressure and pressure measurement
Q1. Calculate the pressure in the ocean at a depth of 2000 m assuming that salt water is incompressible with a constant density of 1002 kg/m3.
Solution: Pressure = (mass density)(acceleration due to gravity)(depth)
= (1002)(9.81)(2000) = 19660000 Pa = 19.66MPa.
Q2. What will be (a) the gauge pressure (b) the absolute pressure of water at the depth of 12m below the free surface. Take the density of water 1000 kg/m3and the atmospheric pressure 101 KN/m2
Solution: gauge Pressure = (mass density)(acceleration due to gravity)(depth)
= (1000)(9.81)(12) = 117.72 kPa.
Absolute Pressure = 117.72 kPa + 101 kPa = 208.72 kPa.
Q3. What depth of oil specific gravity 0.8 will produce a pressure of 120 KN/m2. What would be the corresponding depth of water?
Solution: oil depth = (pressure)/(oil specific weight)
= (120 KN/m2)/(0.8x9.81 KN/m3) = 15.29m.
water depth = (pressure)/(water specific weight)
= (120 KN/m2)/(9.81 KN/m3) = 12.23m.
Q4. At what depth below the free surface of oil having a density of 600 kg/m3 will the pressure be equal to 1 bar.
Solution: 1 bar = 100000 N/m2
oil depth = (pressure)/(oil specific weight)
= (100000N/m2)/(600x9.81 N/m3) = 16.99m.
Q5. What would be the pressure in Pa if the equivalent head is measured as 400 mm of (a) mercury of specific gravity 13.6 (b) water (c) oil of specific weight 7.9 KN/m3 (d) a liquid of density 520 kg/m3
Solution: (a) 400 mm of mercury = (0.4m)(13,6x9.81kN/m3) = 53.37 kPa.
(b) 400 mm of water = (0.4m)(9.81kN/m3) = 3.3924 kPa.
(c) 400 mm of oil = (0.4m)(7.9N/m3) = 3.16 kPa.
(d) 400 mm of liquid = (0.4m)(520x9.81N/m3) = 2.04 kPa.
Q6. The pressure head in a gas main at point 120m above sea level is equivalent to 180mm of water. Assuming that the densities of air and gas remain constant and equal to 1.202 kg/m3 and 0.561 kg/m3 respectively. What will be the pressure head in mm of water at sea level.
Solution: 120 m of gas = 120(0.561)/(1.202) = 56.007 m of water = 56007 mm of water.
So, the pressure head in mm of water at sea level = 56007 + 180 = 56187 mm.
Q7. A manometer connected to a pipe in which a fluid is flowing indicating a negative gauge pressure head of 50mm of mercury. What is the absolute pressure in the pipe in Pa if the atmospheric pressure is 1 bar.
Solution: 1bar = 100 kPa; 50 mm of mercury = (0.050)(13.6)(9.81) = 6.6708 kPa.
absolute pressure in the pipe = 100 kPa - 6.6708 kPa = 93.33 kPa.
Q8. In fig Q8 fluid P is water and fluid Q is mercury. If the specific gravity of mercury is 13.6 times that of water and the atmospheric pressure is 101.3 Kpa what is the absolute pressure at A when h1 = 15cm and h2 = 30 cm
Solution: absolute pressure at A = (0.15)(9.81) + (0.3)(13.6)(9.81) + 101.3 = 142.8 kPa.
Q9. A U-tube manometer shown in fig Q9 measures the difference between two point A and B in a liquid of density 1000 kg/m3. The U-tube contains mercury of specific gravity 13.6. calculate the difference of pressure if a = 2m, b = 1m and h = 0.75m.
Solution: the difference of pressure = (0.25)(9.81) + (0.75(13.6x9.81) – (2)(9.81)
= 82.895 kPa.
Q10. The top of an inverted U-tube manometer is filled with oil of specific gravity 0.98 and the remainder of the tube with water of specific gravity 1.01. find the pressure difference in Pa between two points at the same level at the base of the legs when the difference of water level is 75mm.
Solution: Pressure difference = (0.075)(1.01 - 0.98)(9.81) = 0.02207 kPa = 22.07 Pa.
