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Chapter 1
Conversion of Units
Table 1.1 Basic units
Time Length Mass Force Temperature
SI s m kg – K, 0C
CGS s cm g – K, 0C
US Engineering s ft lbm lbf
0R, 0F
Table 1.2 Derived units
SI US Engineering
Force N (1 N = 1 kg m/s2) –
Energy J (1 J = 1 kg m2/s2 ) Btu
Power W (1 W = 1 J/s) HP, PS
Area m2 ft2
Volume m3 (1m3 = 1000 l) ft3
Density kg/m3 lbm/ft
3
Velocity m/s ft/s
Pressure Pa (1 Pa = 1 N/m2)
bar (1 bar = 105 Pa)
torr (1 torr = 1 mmHg)
atm (1 atm = 101325 Pa)
psi=lbf/in
2
Table 1.3 Conversion factors
1 ft = 12 in = 0.3048 m 0F = 32þ1.8* 0C
1 in = 2.54 cm 0C = (0F-32)/1.8
1 US gallon = 3.7854 l 0R= 460 þ 0F
1 lbm = 0.4536 kg K = 273.15 þ
0C
1 lbf = 4.4482 N
1 psi = 6894.76 Pa �0C = �0F/1.8
1 HP =745.7 W �0C = �K
1 Btu = 1055.06 J = 0.25216 kcal �0F = �0R
1kWh = 3600 kJ
S. Yanniotis, Solving Problems in Food Engineering.
Ó Springer 2008
1
Examples
Example 1.1
Convert 100 Btu/h ft2oF to kW/m2oC
Solution
100
Btu
h ft2 8F
¼100
Btu
h ft2 8F
�
1055:06 J
1Btu
�
1 kJ
1000 J
�
1 h
3600 s
1ft2
0:3048mð Þ2
�
1:8 8F
18C
�
1 kW
1kJ=s
¼ 0:5678
kW
m2 8C
Example 1.2
Convert 100 lb mol/h ft2 to kg mol/s m2
Solution
100
lbmol
h ft2
¼ 100
lbmol
h ft2
�
0:4536 kgmol
lbmol
�
1 h
3600 s
�
1 ft2
0:3048mð Þ2
¼ 0:1356
kgmol
sm2
Example 1.3
Convert 0.5 lbf s/ft
2 to Pas
Solution
0:5
lbf s
ft2
¼ 0:5
lbf s
ft2
�
4:4482N
lbf
�
1 ft2
0:3048mð Þ2
1 Pa
1 N=m2ð Þ
¼ 23:94 Pa s
Exercises
Exercise 1.1
Make the following conversions:
1) 10 ft lbf/lbm to J/kg, 2) 0.5 Btu/lbm
oF to J/kgoC, 3) 32.174 lbmft/lbfs2 to kgm/
Ns2, 4) 1000 lbmft /s2 to N, 5) 10 kcal/min ft oF to W/mK, 6) 30 psia to atm,
7) 0.002 kg/ms to lbmft s, 8) 5 lb mol/h ft
2mol frac to kg mol/s m2 mol frac,
9) 1.987 Btu/lbmol oR to cal/gmol K, 10) 10.731 ft3lbf/in
2lbmol oR to J/kgmolK
2 1 Conversion of Units
Solution
1) 10
ft lbf
lbm
¼ 10
ft lbf
lbm
�
::::::::::::::m
ft
�
:::::::::::::::N
1 lbf
�
::::::::::::::lbm
0:4536 kg
�
:::::::::::::J
mN
¼ 29:89
J
kg
2) 0:5
Btu
lbm8F
¼ 0:5
Btu
lbm8F
�
1055:06 J
::::::::::
�
:::::::::::::
:::::::::::::
�
1:88F
18C
¼ 2094:4
J
kg8C
3) 32:174
lbm ft
lbf s2
¼ 32:174
lbm ft
lbf s2
�
:::::::::::::::
:::::::::::::::lbm
�
::::::::::::::::::m
1 ft
�
:::::::::::::::
4:4482N
¼ 1
kgm
Ns2
4) 1000
lbm ft
s2
¼ 1000
lbm ft
s2
�
0:4536 kg
::::::::::::::::
�
::::::::::::::::
1 ft
�
1N
1 kgm=s2
¼ 138:3N
5) 10
kcal
min ft oF
¼ 10
kcal
min ft oF
�
1055:06 J
0:252 kcal
�
:::::::::min
60 s
�
::::::::::::ft
:::::::::::m
�
:::::::::::8F
::::::::::K
�
:::::::::W
::::::::::J=s
¼ 4121
W
mK
6) 30 psia ¼ 30
lbf
in2
�
::::::::::::::::in2
:::::::::::::::::m2
�
::::::::::::::::::N
::::::::::::::::::lbf
�
:::::::::::::::::Pa
::::::::::::::N=m2
�
:::::::::::::::::atm
:::::::::::::::::Pa
¼ 2:04 atm
7) 0:002
kg
m s
¼ 0:002
kg
m s
�
::::::::::::::lbm
:::::::::::::::kg
�
:::::::::::::::m
::::::::::::::::::ft
¼ 0:0013
lbm
ft s
8) 5
lbmol
h ft2mol frac
¼ 5
lbmol
h ft2mol frac
�
::::::::::::::::kgmol
::::::::::::::: lb mol
�
:::::::::::::::::h
::::::::::::::::::s
�
:::::::::::::::::ft2
:::::::::::::::::::m2
¼ 6:78� 10ÿ3
kgmol
s m2mol frac
9) 1:987
Btu
lbmol 8R
¼ 1:987
Btu
lbmol 8R
�
::::::::::::::cal
::::::::::::::Btu
�
::::::::::::::::lbmol
::::::::::::::::gmol
¼
�
::::::::::::::8R
::::::::::::K
¼ 1:987
cal
gmol K
10) 10:731
ft3 lbf
in2 lbmol8R
¼ 10:731
ft3 lbf
in2 lbmol8R
�
::::::::::::::::m3
:::::::::::::::::::ft3
�
::::::::::::::::N
::::::::::::::::::lbf
�
::::::::::::::::in2
:::::::::::::::::::m2
�
:::::::::::::lbmol
:::::::::::::kgmol
�
1:88R
K
¼ 8314
J
kgmol K
Exercises 3
Exercise 1.2
Make the following conversions:
251oF to oC
(Ans. 121.7 oC)
500oR to K
(Ans. 277.6 K)
0.04 lbm/in
3 to kg/m3
(Ans. 1107.2 kg/m3)
12000 Btu/h to W
(Ans. 3516.9 W )
32.174 ft/s2 to m/s2
(Ans. 9.807 m/s2 )
0.01 ft2/h to m2/s
(Ans. 2.58x10-7 m2/s)
0.8 cal/goC to J/kgK
(Ans. 3347.3 J/kgK)
20000 kg m/s2 m2 to psi
(Ans. 2.9 psi)
0.3 Btu/lbm
oF to J/kgK
(Ans. 1256 J/kgK)
1000 ft3/(h ft2 psi/ft) to
cm3/(s cm2 Pa/cm)
(Ans. 0.0374 cm3/(s cm2 Pa/cm )
4 1 Conversion of Units