Page 24 - eEC20510_化工裝置(上)_課本PDF

1 化工裝置（上）

例題 1-3

N dyne kg
༊Ӌ 1 Pa( 2 )= ( 3 )= ( f )=
m cm cm 2
lb
psi( f )f
in 2

5
解 N N 10 dyne 1 m
1 Pa( )=1 × ×( ) 2
m 2 m 2 1 N 100 cm
5
=1 N × 10 dyne × 1 m 2
m 2 1 N 100 cm 2
2
= 10 5 ( dyne )=10 dyne
100 2 cm 2 cm 2

=1 N × 1 kg f ×( 1 m ) =1 N × 1 kg f × 1 m 2
2
m 2 9.8 N 100 cm m 2 9.8 N 100 cm 2
2
-5 kg
= 1 ( kg f )=1.02×10 ( f )
9.8Ò100 2 cm 2 cm 2
2
2
=1 N × 1 lb f × 0.3048 m 2 ×( 1 ft ) ∴ 1 lb =4.45 N
m 2 4.45 N 1 ft 2 12 in f
2
=1 N × 1 lb f × 0.3048 m 2 × 1 ft 2
m 2 4.45 N 1 ft 2 12 in 2
2
-4 lb
= 0.3048 2 ( lb f )=1.45×10 ( f )=1.45×10 psi。
-4
4.45Ò12 2 in 2 in 2

例題 1-4

1 kg €ʮ˘ɢ= N€ˬ཭k lb €ᆼɢk
f
f
dyne€༺ځk

2
2
解 1 kg =1 kg×9.8 m/s =9.8(kggm/s )=9.8 N
f
kggm 2.2 lb 1 ft lb gft
=9.8 × m × ≒ 2.2×32.174( m )
s 2 1 kg 0.3048 m s 2
lb gft
=2.2 lb ∵ 1 lb =32.174 m s 2
f
f
kggm 1000 g 100 cm 5 ggcm
=9.8 × × =9.8×10 ( )
s 2 1 kg 1 m s 2
5
=9.8×10 dyne。

16

ʷʈༀໄ ɪ DI JOEE ɪʹ