lect01-unitsdimensions

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Introduction to Chemical Engineering Calculations

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baos

Lecture 1.

Units and Dimensions


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Mathematics and Engineering

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los BaosSLIDE

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In mathematics,

If x = 500 and y = 100, then (x + y) = 600

In engineering,

If x = 500m and y = 100m, then (x + y) = 600m

But,

If x = 500m and y = 100kg, then (x + y) = 600???

Units and Dimensions1


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Why Do We Need Units?

Units are important for effective communication

and standardization of measurements

Image Source: http://lamar.colostate.edu/~hillger/feet-to-meters_1.jpg



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4

The Gimli Glider Incident (23 July 1983)

Image Source: http://upload.wikimedia.org

Pounds vs Kilograms



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The Mars Climate Orbiter Incident (23 September 1999)

Newton vs Pound Force

Image Source: http://www.jpl.nasa.gov/... /climate-orbiter-browse.jpg



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Dimension Symbol

Mass m

Length L

Time

Temperature T

Mole n

Luminosity c

Electric Current I

The 7 Fundamental (Base) Dimensions



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Dimension SI Unit Am. Eng. Unit

Masskilogram

(kg)poundmass

(lbm)

Length meter(m)

foot(ft)

Timesecond

(s)second

(s)

TemperatureKelvin

(K)

Rankine

(0

R)

Molegram mole

(gmol)pound mole

(lbmol)

SI and American Engineering System Units for Fundamental Dimensions



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Dimension Symbol

Area L2

Volume L3

Velocity L/

Acceleration L/2

Force m (L/2)

Pressure m (L/2)/L2 = m/2L

Energy m (L/2) L = m (L2/2)

Secondary (Derived) Dimensions



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Volume m3 ft3

Acceleration m/s ft/s

Force kg m/s2 lbm ft/s2

Pressure kg /(m s2) lbm /(ft s2)

Energy kg (m2/s2) lbm (ft2/s2)

SI and American Engineering System Units for Secondary Dimensions



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Force1kg m/s2

= 1 N32.174 lbm ft/s

2

= 1 lbf

Pressure1 kg /(m s2)= 1 N/m2 = 1

Pa

32.174 lbm /(ft s2)

= 1 lbf/ft2

= (1/144) lbf/in2 (psi)

Energy1 kg (m2/s2)= 1 N m = 1 J

32.174 lbm (ft2/s2)

= 1 ft lbf

Defined Equivalent Units



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Conversion of Units: Single Measurements

The equivalence between two units of the samemeasurement may be defined in terms of a ratio(conversion factor):

New UnitOld Unit = New Unit

Old Unit

1 Old Unit 1=

Old Unit New Unit New Unit



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Conversion of Units: Single Measurements

2 2

2 2 2 2 2

2 2 2

9

2 2

2.2 lbm500 kg =1100 lbm

kg

300 1 cm 300 1 cm 3= =10 mmcm cm 10 mm mm

cm 3600s 24h 365d 1m 1km km1 = 9.95x10

1h 1d 1yr 100cm 1000ms yr



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Conversion of Units: Equations or Formula

Consider the following equation of motion:

D (ft) = 3 t(s) 4

Derive an equivalent equation for distance in meters andtime in minutes.

Step 1. Define new variables D(m) and t(min).

Step 2. Define the old variables in terms of the new variable.



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Conversion of Units: Equations or Formula

3.2808 ftD(ft) = D'(m) x or D = 3.2808D'

1 m

60 st(s) = t'(min) x or t = 60t'

1 min

Step 3. Substitute these equivalence relations into theoriginal equation.

(3.2808D) = 3 (60t) 4

Simplifying,

D (m) = 55t(min) 1.22



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The numerical value of two or more quantities can beadded/subtracted only if the units of the quantitiesare the same.

5 kilograms + 3 meters = no physical meaning

10 feet + 3 meters = has physical meaning

10 feet + 9.84 feet = 19.94 feet


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Operation on Units: Addition and Subtraction



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Multiplication and division can be done on quantitieswith unlike units but the units can only be cancelledor merged if they are identical.

5 kilograms x 3 meters = 15 kg-m

3 m2/60 cm = 0.05 m2/cm

3 m2/0.6 m = 5 m2/m = 5 m


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Operation on Units: Multiplication and Division



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Reynolds Number Calculation

Reynolds number is calculated as:


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Dimensionless Quantities

Dv

Reynolds Number =


where = density of the fluid (kg/m3)D = diameter of pipe (m)v = mean velocity of fluid (m/s)

= dynamic viscosity (kg/m s)

What is the net dimension of Reynolds Number?

1


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Importance of Dimensionless Quantities

Used in arguments of special functions such as exponential,logarithmic, or trigonometric functions.

e20 is possible but e(20ft) is undefined

cos(20) is possible but cos(20 ft) is undefined


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Importance of Dimensionless Quantities

Consider the Arrhenius Equation:


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aE

-RTk = Ae


If Ea is activation in cal/mol and T is temperature in K, whatis the unit of R?

To make the argument of the exponential functiondimensionless, R must have a unit of (cal/mol-K).

1


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Every valid equation must be dimensionally consistent.

Each term in the equation must have the same netdimensions and units as every other term to which it isadded, subtracted, or equated.

A + B = C DE

If A has a dimension of L3, then

1. B must have a dimension of L3

since it is added to A.2. (A + B) has a net dimension of L3.3. (C DE) must have a net dimension of L3

4. C and DE have a dimension of L3.


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Dimensional Consistency



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Example on Dimensional Consistency

The density of a fluid is given by the empirical equation

= 70.5 exp(8.27 x 10-7 P)

where = density in (lbm/ft3) and P = pressure (lbf/in2).

a. What are the units of 70.5 and 8.27x10-7

?

b. Derive a formula for (g/cm3) and P (N/m2)


SLIDE

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1


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= 70.5 exp(8.27 x 10-7 P)

1. Since the exponential part is dimensionless, then 70.5must have the same unit as which is (lbm/ft3).

2. Since the argument of the exponential function must bedimensionless, then 8.27 x 10-7 must have a unit of(in.2/lbf) which is a reciprocal to the unit of P.


SLIDE

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1. Define new variables (g/cm3) and P (N/m2).

2. Express the old variables in terms of the new variables.


SLIDE

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3

3 3 3

2-4

2 2 2

lbm g 1 lbm 28,317 cm = ' = 62.43'

453.593 gft cm 1 ft

lbf N 0.2248 lbf 1 mP = P' = 1.45x10 P'

1 Nin m 39.37 in

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3. Substitute the equivalence relations into the originalequation.


SLIDE

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-7

-7 -4

= 70.5 exp 8.27 x 10 P

62.43' = 70.5 exp 8.27 x 10 1.45 x 10 P'

Simplifying,

-10

3 2

g N' = 1.13 exp 1.20 x 10 P'

cm m

1

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