Introduction
1.1 m = 3000 kg
p
V= (1.5)2 ¥ 4.2 = 7.422 m3
4
m 3000
r= = = 404.2 kg/m3 Ans.
V 7.422
m3 kg
m = 0.032 ¥ 404.2 ¥ 3 = 12.934 kg/s Ans.
s m
Fig. 1.1
1.2 At H = 10,000 m,
g = 980.6 – 3.086 ¥ 10–6 ¥ 10,000 ¥ 100
= 977.514 cm/s2
At sea-level, H = 0,
\ g = 980.6 cm/s2 = 9.806 m/s2
W = mg = m ¥ 9.806 = 90,000 N
,2 Solution Manual
\ m = 9178.054 kg
Gravity force, W at 10,000 m from sea-level
= 9178.054 ¥ 977.514 ¥ 10–2 = 8,9716.4 N Ans.
90,000 - 89716.4
% difference = ¥ 100 = 0.315% Ans.
90,000
1.3 In MKS (metre-kilogram-second) system of units, the unit of force is kgf
and the unit of mass in kgm. By Newton’s second law of motion, the force
F
F • mf
mf
or, F=
g0
1
where is the constant of proportionality. Here m is the mass and f the
g0
kg m m
acceleration of the body. The unit of g0 is thus . Thus, the weight
kg f s 2
of a body, the unit of which is kgf , is given by
mg
W= ,
g0
g being the acceleration due to gravity.
By Newton’s law of gravitation, the force between two bodies is inversely
proportional to the square of the distance between them, or
1
F• .
r2
mg
At sea-level, W =
g0
Weight of the body at height H is
mg ( d /2) 2
= Fig. 1.3
g0 ( d /2 + H ) 2
=
mg FG
d IJ 2
Proved.
H
g0 d + 2 H K
, Introduction 3
1.4 Radius of the earth = 6340 km
Radius of the orbit of the satellite,
r = 916 + 6340 = 7256 km
28,840 ¥ 103
V= = 8011.1 m/s
3600
mV 2
Now, = mg,
r
m2
b80111. g
s2
2
Fig. 1.4
or g= 3
7256 ¥ 10 m
= 8.8448 m/s2 Ans.
W = mg = 86 ¥ 8.8448 = 760.65 N Ans.
1.5 (a) 90 cm Hg gauge = 90 + 76 = 166 cm Hg abs.
166
\ Pressure = ¥ 101.325 = 221.315 kPa Ans.
76
(b) 40 cm Hg vacuum = 76 – 40 = 36 cm Hg abs.
36
\ Pressure = ¥ 101.325 = 48 kPa Ans.
76
(c) 1.2 mH2O
\ Pressure = hrg
= 1.2 m ¥ 1000 kg/m3 ¥ 9.8 m/s2
= 11,760 Pa = 11.76 kPa Ans.
(d) 3.1 bar = 3.1 ¥ 100 = 310 kPa Ans.
1.6 P = hrg
kg m
= 30 m ¥ 1878 3
¥ 9.65 2
m s
= 543681 Pa = 543.681 kPa Ans.
1.7 z z dp =
1
L
gdH
Now, pL1.4 = 2.3 ¥ 105
Fig. 1.6
( 230000) 1/ 1.4 6759
L= 1/ 1.4
= 0.7143
p p
p = 101325
\
6759
9.81 z
0
p = 101325
p - 0.7143 dp = z
0
H
dH
, 4 Solution Manual
(101325) 0.2857
or, H = 689
0.2857
= 2411.59 ¥ 26.92 = 64929 m
= 64.93 km Ans.
1.8 Making a pressure balance on AB
m
p + 0.03 m ¥ 1000 m ¥ 9.81
s2
Fig. 1.7
= p0 + 0.50 ¥ 13.6 ¥ 1000 ¥ 9.81
p – p0 = (0.50 ¥ 13.6 – 0.03) ¥ 1000 ¥ 9.81
= 66413.7 Pa
= 66.414 kPa Ans.
Fig. 1.8
1.9 0.66 m Hg gauge
= 0.76 – 0.66 = 0.10 m abs. = 10 cm abs.
10
\ Pressure = ¥ 101.325 = 13.33 kPa Ans.
76
1.1 m = 3000 kg
p
V= (1.5)2 ¥ 4.2 = 7.422 m3
4
m 3000
r= = = 404.2 kg/m3 Ans.
V 7.422
m3 kg
m = 0.032 ¥ 404.2 ¥ 3 = 12.934 kg/s Ans.
s m
Fig. 1.1
1.2 At H = 10,000 m,
g = 980.6 – 3.086 ¥ 10–6 ¥ 10,000 ¥ 100
= 977.514 cm/s2
At sea-level, H = 0,
\ g = 980.6 cm/s2 = 9.806 m/s2
W = mg = m ¥ 9.806 = 90,000 N
,2 Solution Manual
\ m = 9178.054 kg
Gravity force, W at 10,000 m from sea-level
= 9178.054 ¥ 977.514 ¥ 10–2 = 8,9716.4 N Ans.
90,000 - 89716.4
% difference = ¥ 100 = 0.315% Ans.
90,000
1.3 In MKS (metre-kilogram-second) system of units, the unit of force is kgf
and the unit of mass in kgm. By Newton’s second law of motion, the force
F
F • mf
mf
or, F=
g0
1
where is the constant of proportionality. Here m is the mass and f the
g0
kg m m
acceleration of the body. The unit of g0 is thus . Thus, the weight
kg f s 2
of a body, the unit of which is kgf , is given by
mg
W= ,
g0
g being the acceleration due to gravity.
By Newton’s law of gravitation, the force between two bodies is inversely
proportional to the square of the distance between them, or
1
F• .
r2
mg
At sea-level, W =
g0
Weight of the body at height H is
mg ( d /2) 2
= Fig. 1.3
g0 ( d /2 + H ) 2
=
mg FG
d IJ 2
Proved.
H
g0 d + 2 H K
, Introduction 3
1.4 Radius of the earth = 6340 km
Radius of the orbit of the satellite,
r = 916 + 6340 = 7256 km
28,840 ¥ 103
V= = 8011.1 m/s
3600
mV 2
Now, = mg,
r
m2
b80111. g
s2
2
Fig. 1.4
or g= 3
7256 ¥ 10 m
= 8.8448 m/s2 Ans.
W = mg = 86 ¥ 8.8448 = 760.65 N Ans.
1.5 (a) 90 cm Hg gauge = 90 + 76 = 166 cm Hg abs.
166
\ Pressure = ¥ 101.325 = 221.315 kPa Ans.
76
(b) 40 cm Hg vacuum = 76 – 40 = 36 cm Hg abs.
36
\ Pressure = ¥ 101.325 = 48 kPa Ans.
76
(c) 1.2 mH2O
\ Pressure = hrg
= 1.2 m ¥ 1000 kg/m3 ¥ 9.8 m/s2
= 11,760 Pa = 11.76 kPa Ans.
(d) 3.1 bar = 3.1 ¥ 100 = 310 kPa Ans.
1.6 P = hrg
kg m
= 30 m ¥ 1878 3
¥ 9.65 2
m s
= 543681 Pa = 543.681 kPa Ans.
1.7 z z dp =
1
L
gdH
Now, pL1.4 = 2.3 ¥ 105
Fig. 1.6
( 230000) 1/ 1.4 6759
L= 1/ 1.4
= 0.7143
p p
p = 101325
\
6759
9.81 z
0
p = 101325
p - 0.7143 dp = z
0
H
dH
, 4 Solution Manual
(101325) 0.2857
or, H = 689
0.2857
= 2411.59 ¥ 26.92 = 64929 m
= 64.93 km Ans.
1.8 Making a pressure balance on AB
m
p + 0.03 m ¥ 1000 m ¥ 9.81
s2
Fig. 1.7
= p0 + 0.50 ¥ 13.6 ¥ 1000 ¥ 9.81
p – p0 = (0.50 ¥ 13.6 – 0.03) ¥ 1000 ¥ 9.81
= 66413.7 Pa
= 66.414 kPa Ans.
Fig. 1.8
1.9 0.66 m Hg gauge
= 0.76 – 0.66 = 0.10 m abs. = 10 cm abs.
10
\ Pressure = ¥ 101.325 = 13.33 kPa Ans.
76
