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```Units and Dimensions
SEPTEMBER 30, 1999
Likely Cause Of Orbiter Loss Found
The peer review preliminary findings
indicate that one team used English units
(e.g., inches, feet and pounds) while the
other used metric units for a key spacecraft
operation.
Mars Climate Orbiter
Units and Dimensions
Quantity = numerical value & units
Example - 100 kg/hr
Dimensions = basic concepts of measurement
Units = quantitatively expressing dimensions
All dimensions of interest can be expressed in terms of:
Mass
Length
Temperature
Time
Currency
What are the dimensions & (SI units) on the
following ?
velocity =
acceleration =
force =
pressure =
energy =
L/t
(m/sec)
L/t2 (m/sec2 )
M*L/t2
(Kg m/ sec2)
M/L*t2 (Kg m / m2 sec2)
M*L2/t2 Kg m2 / sec2)
Typically, coefficients in physical laws (eg, KE = ½ mv2),exponents,
and arguments (log x, sin x, exp x = ex) have no dimensions.
There are special dimensionless numbers used in chemical
engineering; for example:
Reynolds Number
Prandtl Number
N Re 
N Pr 
DV

Cp
k
Dimensional Homogeneity & Dimensionless Numbers
• every added and subtracted term in any equation must have
the same dimensions.
Multiplication & Division of quantities
• creates compound dimensions and units
• must have same dimensions & units
Example
Consider the equation D(ft) = 3t(s) + 4
• What are the dimensions and units of 3 and 4 ?
Convert the equation D(ft) = 3t(s) + 4 to D’(m) = __t’(min) + __
Convert each term then substitute ...
D(ft) = D’(m) * 3.2808 ft / m
& t(s) = t’(min) 60 s / min
Thus, 3.2808D’(m) = 3*[60 t’(min)] + 4
D’(m) = 55t’(min) + 1.22
• What are the dimensions & units of 55 and 1.22 ?
Example
You are traveling at 51 km/hr and increase your speed
by 1 ft/s; what is your new velocity?
Can you add these because they have the same dimensions ?
Dimensional ledger/ equations
• think units first, then numerical values
• break big problem down
km
km
ft  ?? 
V
 51
1
* 
hr
hr
sec  ?? 
10 Minute Problem
An empirical equation for calculating the inside heat transfer coefficient, hi, for
the turbulent flow of liquids in a pipe is given by:
0.023 G 0.8 K 0.67Cp0.33
hi 
D0.2  0.47
where
hi = heat transfer coefficient, Btu/(hr)(ft)2(°F)
G = mass velocity of the liquid, lbm/(hr)(ft)2
K = thermal conductivity of the liquid, Btu/(hr)(ft)(°F)
Cp = heat capacity of the liquid, Btu/(lbm)(°F)
μ = Viscosity of the liquid, lbm/(ft)(hr)
D = inside diameter of the pipe, (ft)
a. Verify if the equation is dimensionally consistent.
b. What will be the value of the constant, given as 0.023, if all the variables
in the equation are inserted in SI units and hi is in SI units.
Extra Practice Problems
Problem Set Handout: I-1 – I-17
Mass, Weight, and Force
Mass: amount of material - mass ≠ weight
Weight: Force that material exerts due to gravity (g) which
changes with location, etc.
Force: (Newton, dyne, or lbf) = mass * acceleration (F = m *a)
Mass = kg (SI), g (CGS), or lbm (English)
kg m
F (N )  2
s
lbm ft
F (lb f )  C * 2
s
32.174 lbm ft / sec 2
gc 
lb f
10 Minute Problem
Momentum (lbf) is equal to mass (lbm / sec) X velocity (ft/sec)
Determine the momentum force transferred to a wall by a stream of
water flowing from a fire hose at 50 ft/sec and 1000 lb/hr.
Extra Practice Problems
Problem Set Handout: I-18 – I-21
Moles, Density and Concentration
Moles
Mole = certain number of entities
6.023 X 1023 molecules
• g-mole = amt of substance whose mass in grams is
equal to the molecular weight of the substance
• similarly kg-mole & lb-mole
• molecular weight (MW) =
mass
mole
• atomic weight - atomic mass .... Inside back cover of textbook
10 Minute Problem
Silver nitrate (lunar caustic) is a white crystalline salt, used in marking
inks, medicine and chemical analysis. How many kilograms of silver
nitrate (AgNO3) are there in :
a. 13.0 lb mol AgNO3.
b. 55.0 g mol AgNO3
Calcium carbonate is a naturally occurring white solid used in the
manufacture of lime and cement. Calculate the number of lb mols of
calcium carbonate in:
a. 50 g mol of CaCO3.
b. 150 kg of CaCO3.
c. 100 lb of CaCO3.
Density, Specific Gravity, API Gravity
Density =  [=] M/L3
→ kg/m3, lbm / ft3, g/cc, etc.
•  ≠ constant → f(T,P)
Specific volume = V = volume / unit mass = –1 [=] L3/M
Specific gravity = sp gr = SG
• For liquids & solids:
=
 A (T , P)
 ref (Tref , Pref )
ref = H2O(liquid) at 4°C & 1 atm
[water= 1 g/cm3 = 1000 kg/m3 = 62.43 lbm/ft3]
• For gases:
ref = air at “standard conditions”
Tabulated Specific Gravities
Example: SG of Ethanol at 140 F
F
 140
48.2
EtOH
SG  40 F 
 0.772
 H 20 62.4
API Gravity (Crude Oil)

141.5
API 
  fluid, 60 F
sp.gr.
  water , 60 F




 131.5
Example
The density of a liquid is 1500 kg/m3 at 20°C.
•
What is the specific gravity 20°C/4°C of this
material ?
•
What is the API gravity of the liquid ?
•
What volume (ft3) does 140 lbm of this material
occupy at 20°C ?
Composition
• Mole fraction =
moles (n) of A
total (nT ) moles
mass (m) of A
• Mass fraction =
total ( mT ) mass
• Volume fraction (gas) ????
n RT
V
P
Example
A liquefied mixture of n-butane, n-pentane and n-hexane has
the following composition in weight percent.
n - C4H10 = 50 %
n - C5H12 = 30 %
n - C6H14 = 20 %
Calculate the weight fraction, mol fraction and mol percent of
each component and also the average molecular weight of the
mixture.
10 Minute Problem
A mixture of gases is analyzed and found to have the
following composition (volume percent). How much will
3 lb mol of this gas weigh ?
CO2
CO
CH4
H2
N2
12.0
6.0
27.3
9.9
44.8
Total
100.0
Concentration
Concentration = quantity of A / volume
kg / m3
kg mol / m3
lb / ft3
g/L
lb mol / ft3
g /cc
Example
A solution of HNO3 in water has a specific gravity of 1.10 at
25 C. The concentration of HNO3 is 15 g/L.
What is the mole fraction of HNO3 in the solution ?
What is the ppm (wt) of HNO3 in the solution ?
10 Minute Problem
The 1993 Environmental Protection Agency (EPA) regulation contains
standards for 84 chemicals and minerals in drinking water. According to the
EPA one of the most prevalent of the listed contaminants is naturally
occurring antimony. The maximum contaminant level for antimony and nickel
has been set at 0.006 mg/L and 0.1 mg/L respectively.
A laboratory analysis of your household drinking water shows the antimony
concentration to be 4 ppb (wt) (parts per billion) and that of nickel to be 60
ppb (wt).
Determine if the drinking water is safe with respect to the antimony and nickel
levels. Assume density of water to be 1.00 g/cm3
Extra Practice Problems
Problem Set Handout: I-22 – I-44
Temperature
Temperature - average kinetic
energy of molecules.
Relative
Fahrenheit (°F)
Celsius (°C)
Absolute
Rankin ( °R )
Kelvin (°K)
Conversions
T (°K) = T (°C ) + 273.15
T (°R) = T (°F ) + 459.67
T (°R) = T (°K ) * 1.8
Example
DT ≠ T - conversions approaches are different
Given the following equation:
  (1.096  0.00086 T )e0.000953P
Where:  [=] gm / cm3, T [=] °C, P [=] atm
A) Determine the units on the three constants
B) Convert the constants to accurately reflect the following
revised set of units:
 [=] lbm / ft3, T [=] °R, P [=] psi
Extra Practice Problems
Problem Set Handout: I-45 – I-51
Pressure
Pressure is defined as the amount of force exerted on a unit area of
a substance:
force N
P
 2  Pa
area m
force lb f
P
 2
area
ft
Direction of fluid pressure on
boundaries
Furnace duct
Pipe or tube
Heat exchanger
Pressure is a Normal Force
(acts perpendicular to surfaces)
It is also called a Surface Force
Dam
Units for Pressure
Unit
1 pascal (Pa)
Definition or
Relationship
1 kg m-1 s-2
1 bar
1 x 105 Pa
1 atmosphere (atm)
101,325 Pa
1 torr
1 / 760 atm
760 mm Hg
1 atm
14.696 pounds per sq.
in. (psi)
1 atm
Standard Atmosphere
1 Atmosphere
33.91 ft of water (ft H20)
14.696 psi (lbf / in2)
29.92 in Hg
760 mm Hg
1.013 X 105 Pascal (Pa)
101.3 kPa
Pressure distribution for a fluid at rest
Let’s determine the pressure
distribution in a fluid at rest
in which the only body force
acting is due to gravity
The sum of the forces acting
on the fluid must equal zero
Pressure distribution for a fluid at rest
A force balance in the z direction gives:
F
z
 0  PS z  PS z Dz  SDzg
Pz  Dz  Pz
  g
Dz
For an infinitesimal element (Dz0)
dP
  g
dz
Incompressible fluid
Liquids are incompressible i.e. their density is assumed to
be constant:
P2  P1  g ( z2  z1 )
When we have a liquid with a free surface the pressure P at any
depth below the free surface is:
P   g h  Po
Po is the pressure at the
free surface (Po=Patm)
By using gauge pressures we can simply write:
P  gh
Measurement of Pressure Differences
Apply the basic equation of static fluids
to both legs of manometer, realizing that
P2=P3.
P2  Pa  b g ( Z m  Rm )
P3  Pb  b g ( Z m )   a gRm
Pa  Pb  g Rm (  a  b )
Example
A U-tube manometer is used to determine the pressure drop across an orifice
meter. The liquid flowing in the pipe line is a sulfuric acid solution having a
specific gravity (60°/60°) of 1.250. The manometer liquid is mercury, with a
specific gravity (60°/60°) of 13.56. The manometer reading is 5.35 inches, and
all parts of the system are at a temperature of 60°F.
What is the pressure drop across the orifice meter in psi ?
10 Minute Problem
The barometric pressure is 720 mm Hg. The density of the oil is 0.80
g/cm3 . The density of mercury is 13.56 g/cm3 The pressure gauge
(PG) reads 33.1 psig. What is the pressure in kPa of the gas ?
3 in
Gas
12 in
20 in
24 in
3 in
PG
16 in
Extra Practice Problems
Problem Set Handout: I-52 – I-63
```