Introduction
1.1 m = 3000 kg V=
p (1.5)2 ¥ 4.2 = 7.422 m3 4
r=
m 3000 = 404.2 kg/m3 = V 7.422
m = 0.032 ¥ 404.2
m3 kg ¥ 3 = 12.934 kg/s 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
Ans.
Ans.
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 ¥ 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 % difference =
F • mf or,
where
F=
mf g0
1 is the constant of proportionality. Here m is the mass and f the g0
acceleration of the body. The unit of g0 is thus
kg m m . Thus, the weight kg f s 2
of a body, the unit of which is kgf , is given by W=
mg , 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 F•
At sea-level, W =
1 . r2
mg g0
Weight of the body at height H is =
mg ( d /2) 2 g0 ( d /2 + H ) 2
=
mg d g0 d + 2 H
FG H
IJ K
Fig. 1.3
2
Proved.
Introduction
3
1.4 Radius of the earth = 6340 km Radius of the orbit of the satellite, r = 916 + 6340 = 7256 km V=
28,840 ¥ 103 = 8011.1 m/s 3600 mV 2 = mg, r
Now,
b80111. g ms 2
or
g=
2
Fig. 1.4
2
7256 ¥ 103 m
= 8.8448 m/s2 W = mg = 86 ¥ 8.8448 = 760.65 N 1.5 (a) 90 cm Hg gauge = 90 + 76 = 166 cm Hg abs.
Ans. Ans.
166 ¥ 101.325 = 221.315 kPa 76 (b) 40 cm Hg vacuum = 76 – 40 = 36 cm Hg abs. \ Pressure =
\ Pressure =
Ans.
36 ¥ 101.325 = 48 kPa 76
Ans.
(c) 1.2 mH2O \ Pressure = hrg = 1.2 m ¥ 1000 kg/m3 ¥ 9.8 m/s2 = 11,760 Pa = 11.76 kPa (d) 3.1 bar = 3.1 ¥ 100 = 310 kPa
Ans. Ans.
1.6 P = hrg
kg m ¥ 9.65 2 3 m s = 543681 Pa = 543.681 kPa Ans. = 30 m ¥ 1878
1.7
z z
1 gdH L Now, pL1.4 = 2.3 ¥ 105
dp =
L=
( 230000) 1/ 1.4 6759 = 0.7143 1/ 1.4 p p
p = 101325 \
z
6759 p = 101325 p - 0.7143 dp = 9.81 0
z
H
0
dH
Fig. 1.6
4
Solution Manual
(101325) 0.2857 0.2857
or, H = 689
= 2411.59 ¥ 26.92 = 64929 m = 64.93 km
Ans.
1.8 Making a pressure balance on AB p + 0.03 m ¥ 1000 m ¥ 9.81
m s2
= p0 + 0.50 ¥ 13.6 ¥ 1000 ¥ 9.81
Fig. 1.7
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. \ Pressure =
10 ¥ 101.325 = 13.33 kPa 76
Ans.
Temperature
2.1
ts p = lim = 1.36605 Tt pt Æ 0 pt Ts = 1.36605 ¥ 273.16 = 373.15 K = 100°C
Ans.
2.2
Fig. 2.2 WB.P.
50
100
200
300
SB.P.
96.4
193
387
582
1.928
1.93
1.935
1.94
S B.P. WB.P.
The intercept A of
S B.P. at WB.P. = 0 is 1.925. WB.P.
\ Sulphur boiling point = 1.925 ¥ 373.15 = 718.3 K = 445.15 °C
Ans.