REVISION TEST 8 (Page 290)
This Revision Test covers the material contained in chapters 22 to 25. The marks available are
shown at the end of each question.
Problem 1. A circular piston exerts a pressure of 150 kPa on a fluid when the force applied to the
piston is 0.5 kN. Calculate the diameter of the piston, correct to the nearest millimetre.
Marks
Pressure =
force
force
, hence, area =
area
pressure
Force = 0.5 kN  0.5 103 N = 500 N, and
pressure = 150 kPa = 150000 Pa = 150000 N/m 2 .
Hence,
area =
500 N
= 0.003 3 m 2
2
150000 N / m
3
 d2
Since the piston is circular, area =
, where d is the diameter of the piston.
4
 d2
= 0.003 3
4
Hence,
area =
from which,
d 2 = 0.003 3 
i.e.
d=
4
= 0.004244

0.004244 = 0.0651 m, i.e. 65.1 mm
3
Hence, the diameter of the piston is 65 mm, correct to the nearest mm.
Total:
6
Problem 2. A tank contains water to a depth of 500 mm. Determine the water pressure (a) at a
depth of 300 mm, and (b) at the base of the tank.
51
© John Bird & Carl Ross Published by Taylor and Francis
Marks
Pressure p at any point in a fluid is given by p = gh Pascal’s, where
 is the density in kg/m 3 , g is the gravitational acceleration in m/s 2
and h is the height of fluid vertically above the point.
(a) At a depth of 300 mm, i.e. 0.30 m.
p = gh = 1000  9.81  0.30 = 2943 Pa = 2.94 kPa
3
(b) At the base of the tank, the vertical height of the water is 500 mm, i.e. 0.50 m.
Hence, p = 1000  9.81  0.50 = 4905 Pa = 4.91 kPa
3
6
Total:
Problem 3. When the atmospheric pressure is 101 kPa, calculate the absolute pressure, to the
nearest kilopascal, at a point on a submarine which is 50 m below the sea water surface. Assume
that the density of sea water is 1030 kg/m 3 .
Marks
The pressure due to the sea, that is, the gauge pressure (p g ) is given by:
p g = gh Pascal’s
i.e.
p g = 1030  9.81  50 = 505215 Pa = 505.22 kPa
3
Absolute pressure = atmospheric pressure + gauge pressure
= (101 + 505.22) kPa = 606.22 kPa
2
i.e. the absolute pressure at a depth of 50 m is 606.22 kPa
Total:
5
52
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Problem 4. A body weighs 2.85 N in air and 2.35 N when completely immersed in water.
Determine (a) the volume of the body, (b) the density of the body, and (c) the relative density of the
body.
Marks
(a) The apparent loss of weight is 2.85 N – 2.35 N = 0.50 N. This is the weight of
water displaced, i.e. Vg, where V is the volume of the body and  is the density of
water,
i.e.
Hence,
0.50 N = V  1000 kg/m 3  9.81 m/s 2 = V  9.81 kN/m 3
V=
2
0.50
m 3 = 5.097  10 5 m 3 or 5.097  10 4 mm 3
3
9.81 10
(b) The density of the body =
2
mass
weight
2.85 N


2
volume
gV
9.81m / s  5.097 10 5 m3
2.85
kg 105
= 9.81
= 5700 kg/m 3 = 5.70 tonne/m 3
3
5.097 m
(c) Relative density =
3
density
density of water
5700 kg / m3
Hence, the relative density of the body =
= 5.70
1000 kg / m3
2
Total:
9
Problem 5. A submarine dives to a depth of 700 m. What is the gauge pressure on its surface, if the
density of seawater is 1020 kg/m 3 and g = 9.81 m/s 2 .
53
© John Bird & Carl Ross Published by Taylor and Francis
Marks
Gauge pressure on surface, p = gh
1
= 1020  9.81  700
2
= 7  10 6 = 7 MPa
2
Total:
5
Problem 6. State the most appropriate fluid flow measuring device for the following applications:
(a) A high accuracy, permanent installation, in an oil pipeline
(b) For high velocity chemical flow, without suffering wear
(c) To detect leakage in water mains
(d) To measure petrol in petrol pumps
(e) To measure the speed of a viscous liquid
Marks
(a) Venturi meter
1
(b) Flow nozzle
1
(c) Pitometer
1
(d) Rotary vane positive displacement meter
1
(e) Electromagnetic flowmeter
1
Total:
5
Problem 7. A storage tank contains water to a depth of 7 m above an outlet pipe, as shown below.
The system is in equilibrium until a valve in the outlet pipe is opened. Determine the initial mass
rate of flow at the exit of the outlet pipe, assuming that losses at the pipe entry = 0.3 v 2 , and losses
at the valve = 0.2 v 2 . The pipe diameter is 0.05 m and the water density, , is 1000 kg/m 3 .
54
© John Bird & Carl Ross Published by Taylor and Francis
Marks
From Bernoulli’s equation:
i.e.
from which,
P1 v12
P v2

 gz1  2  2  gz 2  0.3v2 2  0.2 v2 2
 2

2
4
0 + 0 + 7g = 0 + 0.5 v 2 2 + 0 + 0.5 v 2 2
3
v2  7g  7  9.81 = 8.287 m/s
2
Mass rate of flow at the outlet pipe
= Av 2 = 1000
kg  0.052 2
m

m  8.287
= 16.27 kg/s
3
m
4
s
6
Total:
15
Problem 8. Determine the wind pressure acting on a slender building due to a gale of 150 km/h
that acts perpendicularly to the building. Take the density of air as 1.23 kg/m 3 .
Marks
1
Wind pressure acting on a slender building, p = 0.5v 2
Wind speed, v = 150
Hence,
km
h
m

1000
= 41.667 m/s
h 3600s
km
2
wind pressure, p = 0.5  1.23  41.667 2
= 1068 Pa
2
Total:
5
55
© John Bird & Carl Ross Published by Taylor and Francis
Problem 9. Some gas occupies a volume of 2.0 m 3 in a cylinder at a pressure of 200 kPa. A piston,
sliding in the cylinder, compresses the gas isothermally until the volume is 0.80 m 3 . If the area of
the piston is 240 cm 2 , calculate the force on the piston when the gas is compressed.
Marks
An isothermal process means constant temperature and thus Boyle's law applies,
p 1 V 1 = p 2 V 2 where V 1 = 2.0 m 3 , V 2 = 0.80 m 3 and p 1 = 200 kPa
i.e.
Hence,
(200)(2.0) = p 2 (0.80)
from which,
Pressure =
pressure, p 2 =
200  2.0
= 500 kPa
0.80
3
force
, from which, force = pressure  area
area
2
Hence, force on the piston = (500  10 3 Pa)(240  10 4 m 2 ) = 12.0 kN
Total:
5
Problem 10. Gas at a temperature of 180C has its volume reduced by a quarter in an isobaric
process. Determine the final temperature of the gas.
Marks
Since the process is isobaric it takes place at constant pressure and hence Charles’ law
V1 V2

T1 T2
applies, i.e.
where T 1 = (180 + 273) K = 453 K and V 2 =
3
V
4 1
2
3
V1
V1
4
from which,
453 T2
Hence,
final temperature, T 2 =
3
(453) = 339.75 K = (339.75 - 273)C = 66.75C
4
3
Total:
5
56
© John Bird & Carl Ross Published by Taylor and Francis
Problem 11. Some air at a pressure of 3 bar and at a temperature of 60C occupies a volume of
0.08 m 3 . Calculate the mass of the air, correct to the nearest gram, assuming the characteristic gas
constant for air is 287 J/(kg K).
Marks
pV = mRT
where p = 3 bar = 3  10 Pa, V = 0.08 m ,
5
3
2
T = (60 + 273) K = 333 K, and R = 287 J/(kg K).
Hence
(3 10 5 )(0.08) = m(287)(333)
from which, mass of air,
m=
(3 105 )(0.08)
= 0.251 kg or 251 g
(287)(333)
3
Total:
5
Problem 12. A compressed air cylinder has a volume of 1.0 m 3 and contains air at a temperature of
24C and a pressure of 1.2 MPa. Air is released from the cylinder until the pressure falls to
400 kPa and the temperature is 18C. Calculate (a) the mass of air released from the container, and
(b) the volume it would occupy at S.T.P. Assume the characteristic gas constant for air to be
287 J/(kg K).
Marks
V 1 = 1.0 m 3 (= V 2 ), T 1 = (24 + 273)K = 297 K, p 1 = 1.2 MPa = 1.2  10 6 Pa,
T 2 = (18 + 273)K = 291 K, p 2 = 400 kPa = 400  10 3 Pa, and R = 287 J/(kg K).
(a) Using the characteristic gas equation, p 1 V 1 = m 1 RT 1 , to find the initial mass of air
in the cylinder gives:
from which,
(1.2  10 6 )(1.0) = m 1 (287)(297)
mass m 1 =
(1.2 106 )(1.0)
= 14.08 kg
(287)(297)
3
57
© John Bird & Carl Ross Published by Taylor and Francis
Similarly, using p 2 V 2 = m 2 RT 2 to find the final mass of air in the cylinder gives:
(400  10 3 )(1.0) = m 2 (287)(291)
(400 103 )(1.0)
= 4.79 kg
(287)(291)
3
Mass of air released from cylinder = m 1 - m 2 = 14.08 - 4.79 = 9.29 kg
1
from which,
mass m 2 =
(b) At STP, T = 273 K and p = 101.325 kPa.
Using the characteristic gas equation: pV = mRT
from which,
volume, V =
3
mRT (9.29)(287)(273)

= 7.18 m 3
p
101325
Total:
10
Problem 13. A platinum resistance thermometer has a resistance of 24  at 0C. When measuring
the temperature of an annealing process a resistance value of 68  is recorded. To what tempera-ture does this correspond? Take the temperature coefficient of resistance of platinum as 0.0038/C
Marks
R  = R 0 (1 + ), where R 0 = 24 , R  = 68  and  = 0.0038/C.
Rearranging gives: temperature,  =
1
R  R0
68  24

= 482.5C
R 0
(0.0038)(24)
4
Total:
5
Problem 14. State which temperature-measuring device would be most suitable to measure the
following: (a) materials in a high-temperature furnace in the range 1800C to 3000C
(b) the air in a factory in the range 0C to 35C
(c) an inexpensive method for food processing applications in the range - 20C to - 80C
(d) boiler flue gas in the range 15C to 250C.
58
© John Bird & Carl Ross Published by Taylor and Francis
Marks
(a) Optical pyrometer
1
(b) Mercury-in-glass thermometer
1
(c) Alcohol-in-glass thermometer
1
(d) Copper-constantan thermocouple
1
Total:
4
TOTAL MARKS FOR REVISION TEST 8: 90
59
© John Bird & Carl Ross Published by Taylor and Francis