2 Units and Dimensions
2.3 CONVERSION OF UNITS
2.3.1 Some Important Derived Quantities and Their Conversion
10 in = 10 in ×m = 0.254 m⎛⎞⎜⎟⎜⎟⎝⎠
Here, the length in metres is multiplied by the conversion factor 2.54 ¥ 10–2 m/in so that the resulting expression has the desired units of metre. The choice of whether the original expression is to be
multiplied or divided by the conversion factor is to be made so as to cancel all the units in the original expression leaving only the desired units. Thus the mass expressed in kilogram is divided by 4.535 924 ¥ 10–1kg/lb to obtain mass in pounds. Thus
50 kg = 50 kg
⎜⎟⎜⎟
⎛⎞=110.23 lb4.535924×10 kg
⎝⎠
2.3.1 Some Important Derived Quantities and Their Conversion
Force: According to Newton’s second law of motion, the force acting on a body is directly
proportional to the time rate of change of momentum. For a body of constant mass, Newton’s law reduces to
F = cma (2.1) where F is the force, m is the mass of the body, a is the acceleration and c is a proportionality
Table 2.7 Conversion factors To convert from atmosphere, standard (atm)
bar (bar)
British thermal unit (Btu)
British thermal unit per pound degree Fahrenheit (Btu lb–1 °F–1)
cubic foot per minute (ft3 min–1) dyne (dyn)
erg (erg) foot (ft)
foot per hour (ft h–1)
gram per cubic centimetre (g cm–3) horsepower (550 ft lbf s–1), (hp) horsepower (metric) inch (in)
mile (based on U.S. survey foot), (mi) mile per hour (mi h–1) poise (P)
pound (avoirdupois) (lb) poundal
pound-force (lbf)
pound-force per square inch, psi, (lbf in–2) pound per cubic foot (lb ft–3) square foot (ft2)
degree Fahrenheit (temperature) (°F) degree Fahrenheit (temperature) (°F) degree Rankine (°R) To Multiply by
5 pascal (Pa) 1.013 25 ¥ 10 pascal (Pa) 1.0 ¥ 105 joule (J) 1.055 056 ¥ 103 joule per kilogram kelvin 4.1868 ¥ 103 (J kg–1 K–1)
joule (J) 4.1868 metre (m) 1.0 ¥ 10–2 pascal second (Pa s) 1.0 ¥ 10–3 metre square per second (m2 s–1) 1.0 ¥ 10–6 cubic metre (m3) 2.831 685 ¥ 10–2 cubic metre per second (m3 s–1) 4.719 474 ¥ 10–4
–5newton (N) 1.0 ¥ 10 joule (J) 1.0 ¥ 10–7 metre (m) 3.048 ¥ 10–1 metre per second (m s–1) 8.466 667 ¥ 10–5 kilogram per cubic metre (kg m–3) 1.0 ¥ 103 watt (W) 7.456 999 ¥ 102 2watt (W) 7.354 988 ¥ 10 metre (m) 2.54 ¥ 10–2 joule (J) 4.1868 ¥ 103 newton (N) 9.806 65 pascal (Pa) 9.806 65 ¥ 104
joule (J) 3.6 ¥ 106 metre (m) 1.609 347 ¥ 103 metre per second (m s–1) 4.4704 ¥ 10–1 pascal second (Pa s) 1.0 ¥ 10–1 kilogram (kg) 4.535 924 ¥ 10–1 newton (N) 1.382 550 ¥ 10–1 newton (N) 4.448 222 pascal (Pa) 6.894 757 ¥ 103 kilogram per cubic metre (kg m–3) 1.601 846 ¥ 101 2) 9.290 304 ¥ 10–2square metre (m
metre square per second (m2 s–1) 1.0 ¥ 10–4 kilogram (kg) 1.0 ¥ 103 pascal (Pa) 1.333 224 ¥ 102 joule (J) 1.0
kelvin (K) T/K = t/°C + 273.15 degree Celsius (°C) t/°C = (t/°F – 32)/1.8 kelvin (K) T/K = (t/°F + 459.67)/1.8 kelvin (K) T/K = (T/°R)/1.8
constant. In the SI system, the constant c is unity and we have F = ma (2.2)
When a body of mass 1 kg is accelerated by 1 m/s2, the force acting on the body is 1 kg m/s2, which is designated as 1 newton or (1 N).
The force unit poundal in FPS system is similar to newton. It is the force that gives an acceleration of 1 ft/s2 when applied to a body having a mass of one pound. 1 poundal = 1 ft lb/s2
In engineering practice, two technical units of force, kilogram force (abbreviated as kgf) in MKS system and the pound force (abbreviated as lbf) were in force for quite a long time. These
are defined by letting c in Eq. (2.1) as equal to 1/gc, where gc is called the Newton’s law conversion factor. That is
F =ma (2.3)gc
The value of gc is 9.806 65 kg m (kgf)–1 s–2 in MKS system and 32.174 lb ft (lbf)–1 s–2 in FPS system.
The numerical value of gc corresponds to the acceleration due to gravity at mean sea level. Although g (acceleration due to gravity) varies from locality to locality, gc does not. The dimension of g is LT–2 whereas the dimension of gc is MLF–1T–2 where F stands for the technical dimension of force.
The force acting on a body under gravitational acceleration is called the weight of the body. Thus weight W is given as
W =ma (2.4)gc
Although g is a local variable, in ordinary calculations g/gc is taken as 1 kgf/kg or 1 lbf/lb, so that the weight of the body in technical force units is numerically equal to the mass of the body.
Pressure: The pressure is defined as the normal component of the force per unit area exerted by the fluid on a real or imaginary boundary. The unit of pressure in the SI system is newton per square meter (N/m2), also called the pascal (Pa). A multiple of pascal, called the bar is also used as a unit of pressure.
1 bar = 105 Pa = 105 N/m2
The pressure exerted by the atmosphere is called the atmospheric pressure and it varies with location and elevation on the earth’s surface. One standard atmospheric pressure abbreviated as atm is used in
all systems of units as an empirical unit of pressure. It is the average pressure exerted by the earth’s atmosphere at sea level. Although actual atmospheric pressure varies from place to place, its value at mean sea level is taken to be 1.013 25 ¥ 105 N/m2 (or 1.013 25 bar).
In engineering practice the pressure is frequently reported as gauge pressure. The absolute pressure is related to the gauge pressure by the following relation:
absolute pressure = gauge pressure + atmospheric pressure
Similarly, pressures below atmospheric (subatmospheric) are reported as vacuum such that absolute pressure = atmospheric pressure – vacuum
Pressure can also be expressed in terms of the height of the column of mercury, which it will support at a temperature of 273.15 K in a standard gravitational field. At standard atmospheric pressure this height is 0.76 m (760 mm or 760 torr) with density of mercury taken as 13.5951 ¥ 103 kg/m3.
1 standard atmosphere (atm) = 1.013 25 bar = 1.013 25 ¥ 105 Pa = 1.013 25 ¥ 105 N/m2 = 760 mm Hg Other important pressure units are the kgf/cm2 in MKS system and lbf/in2 (commonly written as psi) in FPS system.
Work: Energy is expended in the form of work when a force acts through a distance. Thus, dW = F dZ (2.5)
where W is the work done, F is the force acting and Z is the displacement. The unit of work in the SI system is N m (newton-metre) or J (joule).
Energy: Energy is a quantity that can be stored within the system and can be exchanged between the system and the surroundings. The exchange of energy occurs either as heat or as work. Heat and work cannot be stored within the system: they can be treated as energy in transit. Energy stored within the system due to its position above some arbitrary reference plane is referred to as its potential energy (PE). If mass m is at an elevation z above the ground, the potential energy of the mass is
PE = mgz (2.6)
where g is the acceleration due to gravity. The energy possessed by the body by virtue of its motion is called its kinetic energy (KE). If a body of mass m is moving at a velocity u, the kinetic energy of the body is
KE=1 mu2 (2.7)2
The unit of energy in SI system is the joule (J).
1 J = 1 N m = 1 kg m2/s2
Heat: Heat is that quantity which is transferred between bodies due to the difference in temperatures existing between them. Earlier, this quantity was thought of as a substance called caloric. Heat is now recognized as a form of energy that cannot be stored as such within the system. And like work, it is expressed in J (joule). Two other units used for heat are the calorie and the Btu (the British thermal unit).
1 calorie = 4.1868 J; 1 Btu = 1055.06 J
Power: The power is defined as the time rate of doing work. Its unit in the SI system is J/s, commonly designated as W (watts). In engineering calculations, power is generally expressed as horsepower (hp).
1 metric hp = 75 m kgf s–1 = 735.5 W 1 British hp = 550 ft lbf s–1 = 745.7 W
EXAMPLE 2.1 The flow rate of water through a pipe is reported as 15 cubic feet per minute. Taking density of water as 1 g/cm3, calculate the mass flow rate in kg/s.
Solution Volumetric flow rate = 15 ft3/min
Using the relationship 1ft = 0.3048 m and 1 min = 60 s, the volumetric flow rate can now be expressed in m3/s.
15 ft3 0.304833
Volumetric flow rate =⎛⎞⎜⎟⎜⎟⎝⎠= 7.079 ¥ 10–3 m3/s
min⎛⎞⎜⎟⎝⎠
1g⎛⎞⎜⎟⎜⎟⎝⎠ = 1000 kg/m3Density of water is 1 g/cm3 =10–6 m3 cm
3
⎛⎞⎜⎟⎜⎟⎝⎠
Mass flow rate = Volumetric flow rate ¥ density of water = 7.079 ¥ 10–3 m3/s ¥ 1000 kg/m3 = 7.079 kg/s EXAMPLE 2.2 Determine the conversion factor for poundal to newton
0.3048 m⎛⎞⎛⎞⎜⎟ s2Solution 1 poundal is 1 ft lb/s2 = 1 (ft) ⎜⎟⎝⎠
= 0.1383 kg m= 0.1383 Ns2
The conversion factor is 0.1383 N/poundal.
EXAMPLE 2.3 Express kgf/cm2 and lbf/in2 as N/m2. Solution 1 kgf = 9.806 65 N
1 kgf (kgf) = 1
⎛⎞⎜⎟⎝⎠= 0.981×1052
cm
2 10–4 m2 N/m (cm )
2
⎛⎞⎜⎟⎜⎟⎝⎠
The Newton’s law conversion factor relates lbf to poundal. 1 lbf = 32.174 lb ft . Usings2
this and other conversion factors like 0.453 59 24 kg , 0.3048 m and 0.0254 m, we get thelb ft in following.
lbf (lbf) 32.174 lb ft 0.453 59 24 kg
⎛⎞⎛ ⎞ 0.3048 m ⎜⎟⎜ ⎟lb ×ft1in2 = 1 ⎝⎠⎝ ⎠
(in )
2
(0.0254)
22
m
⎡⎤⎢⎥⎣⎦
= 6894.75 kg m = 6894.75 N/m222 sm
(Table 2.7 gives 6.894 757 ´ 103 as the value for the conversion factor for lbf to pascal.)in2
EXAMPLE 2.4 In the SI system, thermal conductivity has the unit W/(m K). The thermal conductivity of a solid material can be calculated as
k
= xQ
A , where Q is the rate of heat ΔT
transfer, x is the thickness of solid, A is the area of heat transfer and ΔT is the temperature difference across the solid. The following values were obtained experimentally: Q = 10 000 kJ/h, A = 1 m2, x = 100 mm and DT = 800 K.
(a) Calculate the thermal conductivity of the solid in W/(m K).
(b) Express the thermal conductivity in kcal/(h m °C).
(c) If thermal conductivity of a second material is 0.15 Btu/(h ft °F), which one will make a better thermal insulator?
Solution 1000 J
ÈØ10 000 kJÉÙ (a)
The rate of heat transfer, Q
10 000 kJ/h 3600 s
ÊÚ 2777.78 J/s h ÈØ
ÉÙ ÊÚ
We have:
x = 100 mm = 0.1 m, A = 1 m2, DT = 800 K Substituting these values in the given equation, we get
xQ 0.1× 2777.78k AT' == ×1 800
JW= 0.347 smK = 0.347 mK
WJ(b) 0.347 mK = 0.347 smK J ⎛⎞⎜⎟⎝⎠cal
= 0.347
⎛⎞⎜⎟⎝⎠h°C = 298.557hm°Cs×m×K× K
(c) The thermal conductivity of the material in the FPS system is WJ0.347 mK = 0.347 smK J ⎛⎞⎜⎟⎝⎠Btu
= 0.347 ⎛⎞⎜⎟⎝⎠hft 1.8°F = 0.2hft°F s×m× ×K×
0.3048 m K
The second material will make a good thermal insulator as its thermal conductivity is 0.15 Btu/(h ft
°F) which is less than the thermal conductivity of the first material calculated above (0.2 Btu/h ft °F).
EXAMPLE 2.5 The weight of an object is 300 N at a location where acceleration due to gravity is 9.81 m/s2.
(a) Determine the mass of the object in kilograms.
(b) Express the mass in the FPS system.
Solution (a) Force is equal to the product of mass and acceleration, i.e.
F = ma
300 = m¥ 39.81 Therefore,
m=300 = 30.58 kg9.81
(b) The conversion factor for pound to kilogram is 4.535 924 ¥ 10–1 kg/lb. Therefore the mass of the object in the FPS system is
30.58 (kg)⎛⎞⎜⎟⎜⎟⎝⎠
= 67.42 lb
EXAMPLE 2.6 The potential energy of a body at a height of 15 m is 2 kJ. If the body is moving at a velocity of 50 m/s what is its kinetic energy? (Take g = 9.8067 m/s2)
Solution Potential energy (PE) = mgz Here z = 15 m, PE = 2000 J.
Therefore, 2000 = m¥ 9.8067 ¥ 15
Thus the mass of the body m = 13.596 kg.
1 2Kinetic energy, KE = 2 mv =
1 (13.596)(50)2 = 16 995 J = 16.995 kJ 2
EXAMPLE 2.7 Nitrogen gas is confined in a cylinder and the pressure of the gas is maintained by a weight placed on the piston. The mass of the piston and the weight together is 100 lb. The acceleration due to gravity is 9.81 m/s2 and the atmospheric pressure is one standard atmosphere. Assuming the piston is frictionless, determine:
(a) The force exerted by the atmosphere, the piston and the weight on the gas in N if the piston diameter is 4 in.
(b) The pressure of the gas in bar and psi.
(c) If the gas is allowed to expand pushing up the piston and the weight by 400 mm, what is the work done by the gas in kJ?
(d) What is the change in the potential energy of the piston and the weight after the expansion in part (c)?
Solution (a) One standard atmosphere = 1.013 25¥ 105 N/m2 The area of the piston is
pp2–22 = 8.1073 ¥ 10–3 m2D×= × ×10 )
44
Force exerted by the atmosphere = pressure ¥ area
= 1.013 25 ¥ 105¥ (8.1073 ¥ 10–3) = 821.47 N Force exerted by the piston and weight = m¥ g
= (100 ¥ 0.4536) ¥ 9.81 = 444.98 N Total force acting on the gas is 821.47 + 444.98 = 1266.45 N
1266.45(b) Pressure = force/area =(8.1073×10–3)
= 1.562 ¥ 105 N/m2 (= 1.562 bar)
From Table 2.7 the conversion factor for converting psi to pascal (Pa) is 6.894 757 ¥ 103 Pa . Therefore, pressure in psi ispsi
1.562 ×105 6.894 757×10–3 = 22.65 psi
(b) Work done = force ¥ displacement = 1266.45 ¥ (400 ¥ 10–3) = 506.58 J = 0.507 kJ (c) Change in the potential energy,
(PE)
= mgDz = 100 ¥ 0.4536 ¥ 9.81 ¥ (400 ¥ 10–3) = 177.99 J 2.3.2 Conversion of Empirical Equations
Many equations used by chemical engineers are correlations of experimental results by empirical means without regard to dimensional homogeneity. These include empirical equations available in literature for estimating thermodynamic properties, transport coefficients and pressure drops for flow of fluids through various geometries. When these equations are to be used for estimation, care should be taken to stick to the units specified for each of the variables. For example, the rate of heat loss from a cylindrical pipe to the ambient air is correlated by the equation
q
7.38 DT1.25 =×D0.25
P
where q is the rate of heat loss in W/m2, DT is the temperature difference in K, and DP is the diameter of the pipe in mm. The dimension of q on the right-hand side is not the same as the dimensions on the left-hand side. Such equations are also known as dimensional equation. The equation as given above will give the correct result only if we adhere to the units specified. If, for example, we want to use the FPS units in the calculation of heat loss, we have to convert the equation into the equivalent FPS form.
In the following paragraphs, we illustrate through some examples the conversion of dimensional equations into their equivalent forms in other systems of units.
EXAMPLE 2.8 For the absorption of sulphur dioxide in water in towers packed with wood grids, the following empirical equation is proposed for evaluating mass transfer coefficient:
k
G = 6.7 ×10–4 ⎡⎤+0.8
ST
⎢⎥⎣⎦
where kG is the mass transfer coefficient in lb-mol/(h ft2 atm), G is the superficial mass velocity in lb/(ft2 h), dS represents clearance between grids in feet, dG represents the height of grids in feet, and dT is the thickness of the grid slats in feet. What is the corresponding equation if kG is given in kmol/(m2 h atm), G is in kg/(m2 h) and the lengths are given in m?
Solution Suppose that the mass velocity is G¢ in kg/(m2 h), and d¢S, d¢ and d¢T are theG clearance between grids, height of grids and thickness of grid, respectively, all measured in m. Then
GG =¢
kg11b ʈ22 lbʈm 0.2048G mh Á˜Ë¯Ë¯
= ¢ 22 2ft h
dd= ⎛⎞ = 3.2808d¢ ft⎜⎟⎝⎠
Since G and d are in the units specified for the given equation, we can substitute these and get the following result.
k
G
=
6.7×10–4 0.2048GddST⎡⎤0.8 ⎢⎥×+
(3.2808dd′′ )⎣⎦d′
SG
k
G = 9.2374×10–5 ′′ ′⎡⎤0.8
ST+
0.4 0.2 ⎢⎥ dd′′⎣⎦SG
where kG is given in lb-mol/(ft2 h atm). Since
lb-mol ⎛⎞⎛⎞⎜⎟2 kmol,1hft atm =1× × ⎛⎞
⎜⎟ ⎜⎟= 4.88252mhatm⎝⎠ ⎝⎠
the right-hand-side of the above result is to be multiplied by 4.8825 so that the mass transfer coefficient is expressed in kmol/(m2.h.atm). Therefore, the desired equation is 4.5102
× 10
–4 ′′⎡⎤0.8 k
ST
G0.4 0.2′= ⎢⎥
ddSG ′′⎣⎦
where k
′
G is in kmol/(m2 h atm), G′is in kg/(m2 h), and the clearances and thickness are in metres.
An alternative to the above technique is to first determine the units for the constant appearing in the original equation, in the present case it is 6.7 ´ 104and convert the units into the desired system by multiplying with appropriate conversion factors. The constant gets a new numerical value after this manipulation. For example, let the constant be 6.7 ´ 104 (lb-mol)a (lb)b(ft)c(atm)d(h)e. Now equate the exponents of the units on both sides and we get
lb-mol (lb mol) (lb) (ft)ab c d e(h)1ÈØÈ Ølb 0.8 (atm) ÉÙÉ Ù ft
20.62
h atm ÊÚÊ Ú or
(lb-mol)1 (ft)2 (h)1 (atm)1 = (lb-mol)a (lb)b + 0.8 (ft)c 2.2 (atm)d (h)e 0.8 Thus we get, a = 1, b = 0.8, c
= 0.2, d = 1, e = 0.2. And the constant is 6.7 ´ 104 (lb-mol)1(lb)0.8(ft)0.2(h)0.2(atm)1. 6.7 ´ 104 (lb mol)(lb)0.8(ft)0.2(h)0.2(atm)1
= 6.7 ´ 104
⎛⎞⎜⎟⎝⎠
×lb×
0.4536 kg0.8 0.3048 m0.2
⎛⎞⎜⎟⎝⎠
×ft×
⎛⎞⎜⎟⎝⎠
–0.2 –1 ×(h) ×(atm)
= 4.51 ´ 104 (kmol)(kg)0.8 (m)0.2 (h)0.2 (atm)1
We can use the given equation with the constant 6.7 ´ 104 replaced by 4.51 ´ 104 where kG is given in kmol/(m2 h atm), G is in kg/(m2 h) and the distances are given in m.
EXAMPLE 2.9 The effective heat capacity of a mixture of gases is given by CP = 7.13 + 0.577 ´ 103 t + 0.0248 ´ 106 t2
where CP is in Btu/(lb-mol °F) and t is in °F.
(a) What are the units of the constants in the equation?
(b) Change the equation into the form in which CPis given in kJ/(kmol K) and temperature is in K.
Solution (a) Since the unit of CP on the left-hand-side is Btu/(lb-mol °F), each term on the right-hand side of the equation has the same units. These terms are:
First term = 7.13 Btu/(lb-mol °F)
Second term = 0.577 ¥ 10–3 t Btu/(lb-mol °F) Third term = 0.0248 ¥ 10–6 t2 Btu/(lb-mol °F)
The constant 7.13 has the same unit as heat capacity, i.e.
Btu .lb-mol °F
Now, the second term 0.577 Btu can be written as¥ 10–3tlb-mol °F
Btu ×(°F)0.577 ¥ 10–3 lb-mol (°F)2
which means that the units of the constant 0.577 ¥ 10–3 is Btu .lb-mol (°F)2
Similarly, the third term 0.0248 ¥ 10–6t2 Btu can be written aslb-mol °F Btu ×(°F)220.0248 ¥ 10–6
lb-mol (°F)3
Therefore, the units of the constant 0.0248 ¥ 10–6 is Btu .lb-mol (°F)3 The units of the constant are respectively
Btu , Btu and Btu .lb-mol °F lb-mol (°F)2 lb-mol (°F)3
(b) Suppose that the temperature of the gas is T K, then the equivalent Fahrenheit temperature is t = (T – 273.15) ¥ 1.8 + 32.0 = 1.8T – 459.67
Substituting this in the equation for CP, we get the heat capacity in Btu/(lb-mol °F) when T is in K.
CP = 7.13 + 0.577 ¥ 10–3 (1.8T – 459.67) + 0.0248 ¥ 10–6 (1.8T – 459.67)2 = 6.87 + 9.976 ¥ 10–4T + 0.0804 ¥ 10–6T2
The heat capacity given by the above equation is in Btu/(lb-mol °F). Since 1.055 056 ×103 J
Btu Btu ×
1 Btu1 lb-mol °F = 0.453 5924 ×103 mol°F × Klb-mol × 1lb-mol 1.8°F
JkJ= 4.1868 mol K = 4.1868 kmol K
the above equation is to be multiplied by 4.1868 to obtain the heat capacity in kJ/kmol K when the temperature is in K. The resulting equation is
CP = 28.763 + 4.763 ¥ 10–3T + 0.3366 ¥ 10–6T2
This equation evaluates heat capacity in kJ/kmol K when temperature T is in K.