Units and dimensions PDF

Title Units and dimensions
Author Thuto Moholo
Course Chemical Engineering Technology 1
Institution Cape Peninsula University of Technology
Pages 31
File Size 1.1 MB
File Type PDF
Total Downloads 88
Total Views 169

Summary

Download Units and dimensions PDF


Description

CET150S

Units and dimensions, Sample variance, Processes and Process variables

S. Shoko Semester 2 2020

1. Units and Dimensions:

Learning Outcomes Statement: At the end of this section you should:  Know the definition of Units and Dimensions  Know how to convert between the various systems of units  Understand the concepts of dimensional consistency and conversion factor, etc 1.2)

Systems of Units

Stream variables require both a number and units in order to have any meaning. Numbers without units aren't very useful. For example, the value 57.6 doesn't mean anything to an engineer, because it doesn't indicate 57.6 of anything in particular. 57.6 moles is quite different from 57.6 L, which is different from 57.6 K. Engineers must specify stream variables very specifically so that designers, technicians, and operators know exactly what stream variable is being specified and what the actual value or setting must be. Specifications in a stream variable must have a unit, a dimension and a value (exceptions will be discussed later).

unit dimension value ¿ ¿ } ¿ } ¿ ¿ What's the difference?¿

Let's try and get a set of definitions we can use. Consider   

110 mg of sodium 24 hands high 5 gal of gasoline Value

Unit

Dimension

110

mg

mass

24

hand

length

5

gal

volume (length3)

Dimensions are our basic concepts of measurements – something that can be measured or derived. For example, a length dimension tells us that the metric (or measured) property is a distance. Likewise, a temperature dimension tells us that the metric is something that we would measure with a thermometer or other temperature-sensitive device. There are four fundamental dimensions. These dimensions, along with their customary abbreviations are the following: 

Mass (M)



Length (L) Time (t) Temperature (T)

 

There are also many derived dimensions. These are multiplicative combinations of the fundamental units. Acceleration, for example is a derived dimension because it has dimensions of L/t2. Similarly, volume, (L3), velocity, density, etc are derived dimensions. Units are the means of expressing the dimensions. Many different units can be used for a single dimension, as inches, miles, centimeters, furlongs, and meters are all units used to measure the dimension length. To maintain consistency, we generally work within a set or system of units. A system of units is an accepted or agreed upon set of units that are consistent, well recognized, and adopted by a group. There are several systems of units in existence today – Table 1 shows the units associated with some of these. Table 1: System SI AES CGS FPS British

Mass (M) Kilogram (kg) Pound mass (lbm) Gram (g) Pound mass (lbm) Slug

Length (L) Meter (m) Foot (ft) Centimeter (cm) Foot (ft) Foot (ft)

Time (t) Second (s) Second (s) Second (s) Second (s) Second (s)

Temperature (T) Kelvins (K) Degrees Fahrenheit (oF) Kelvins (K) Degrees Fahrenheit (oF) Degrees Celsius (oC)

Every system of units has  a set of "basic units" for the dimensions of mass, length, time, absolute temperature, electric

current, luminous intensity, and amount of substance.

 derived units, which are special combinations of units or units used to describe combination

dimensions (energy, force, volume, etc.)  unit multiples, which are multiples or fractions of the basic units used for convenience (years instead of seconds, kilometers instead of meters, etc.). Derived units are formed by combinations of units from within the set. For example mass flow rate would be given in units of kg/s and force would have units of kg*m/s 2. One should not mix units from different systems. For example, one would specify density as either kg/m 3 or lbm/ft3, not as a composite of units from two different systems such as kg/ft3. Table 2 shows some of the derived units that we will for the two systems of units to be used in this course. Table 2: Quantity Volume Force (ma) Pressure (F/A) Energy (Fx) Power (w/t)

Unit is SI m3 N Pa, kPa, MPa J W, kW

Unit in AES ft3 lbf psi Btu, ft-lbf Btu/s, ft-lbf/s

Unit in Other System L, cm3, ml, gas mm Hg, bar, atm cal hp

It is imperative that engineers be conversant with the above units and systems of units and be able to interconvert between units. The next section specifically deals with unit conversions.

A multiple system is used based on powers of 10: tera- (T) 1012, giga- (G) 109, mega- (M) 106, kilo- (k) 103, centi- (c) 10-2, milli- (m) 10-3, micro- () 10-6, nano- (n) 10-9, pico- (p) 10-12.

1.3)

Converting Units:

Why is it important to know how to convert from one system of units to another? Conversion from one group of units to another is accomplished by multiplication of the original value by conversion factors that multiply by the new units and divide by the previous units. A list of convenient conversion factors is found in Table 3 below:

Table 3:

A convenient way of listing conversion factors is to specify one or more equalities between the same quantity expressed in different units. For example, 1 m3 = 264.17 gal = 35.3145 ft = 106 cm3 The conversion factor is then formed by taking a ratio of two quantities, with the new units on top and the old units on bottom. Because the numerator and denominator are equivalent, multiplying a given quantity by the ratio simply multiplies by unity and therefore does not change the value of the original quantity. However, the units in the denominator of the conversion factor cancel with the original units and leave the answer in the new, desired units. This can often be a multi-step process with many intermediate units and conversion factors. The following steps show the process used in virtually all unit conversions:  

Begin with the number and units to be converted Create a table to the right, containing the necessary conversion factors.



Verify that units cancel.



Perform multiplication and division to yield the new number and its accompanying units. When you apply conversion factors, the old units should cancel out and the new units should remain.

Example 1 illustrates the process and shows how the old units cancel, leaving the answer in the new set of units. Eg1: How many hp are there in 300 J/min? Step 1:

Step 2:

Step 3:

Step 4:

This is the preferred way for engineers to do unit conversions because it makes it clear what has been done and makes the work easy to check and understand. It is not necessary to find a single factor to do the conversion at one swoop - it is perfectly normal to string several along in a row. Example: Convert 5 mph to yds/week 5 mile hr 5 mile 1.0936 yd hr 0.0006214 mile

yds week 24 hr 1 day

7 day 1 week

= 1 478 313. 49 yds/week

When units are raised to powers, such as "square feet" or "cubic meters", the conversion factors for these also need to be raised to the same power as the unit. For example, to go from cubic meters to cubic feet, you use the conversion from meter to foot, but raise everything (numbers and units) to the 3rd power. Eg. convert 1 cm/s2 to km/yr2 Care must be taken in converting dimensional equations to a different set of units!

Example: The vapour pressure p* (in units of mmHg) of water is given as a function of temperature T (in units of oC) by the following equation: ¿

log 10 p =7 . 96681−

1668 . 21 T + 228. 0

To convert this to an equation for vapour pressure (in units of bar) as a function of temperature (in units of K), first write conversion equations for p* and T, as follows

760 mmHg =750.062 p¿ (bar ) p¿ ( mmHg )= p¿ (bar )⋅ 1. 01325 bar o T ( C )= T ( K )−273 .15 Now, substitute these into the original equation and simplify as desired

1.4)

1668 . 21 ¿ log 10 [ 750. 062p (bar )]=7 . 9668− T ( K )−273 .15+228 . 0 1668 . 21 ¿ log 10 p ( bar )= 7. 9668−log 10 (750 .062 )− T ( K )−45 . 2 1668 .21 ¿ 5 .0917− T ( K )−45 .2 Units and Calculations

It is always good practice to attach units to all numbers in an engineering calculation. Doing so attaches physical meaning to the numbers used and reduces the possibility of accidentally inverting part of the calculation. Addition and Subtraction 

Values MAY be added if UNITS are the same.



Values CANNOT be added if DIMENSIONS are different.

Examples: 6 ft + 4 oC = ??

different dimensions: length, temperature - so cannot be added

72 in + 4 in = 76 in

same dimension: length, different units - can add

Multiplication and Division

Values may be combined; units combine in similar fashion. Examples: 12 g  2 ml = 6g/ml 2N3m=6Nm 9 cm  2 cm = 4.5 4.5 is a "dimensionless" quantity (in this case a pure number)

You cannot cancel or lump units unless they are identical. Functions Trigonometric functions can only have angular units (radians, degrees). All other functions and function arguments, including exponentiation, powers, etc., must be dimensionless. Examples: (6 ft)2 = (6 ft)(6 ft) = 36 ft2 makes sense but 6

(2 ft)

is meaningless.

Sin (/2 ft) is never defined

1.5)

Dimensional Homogeneity

Every valid equation must be "dimensionally homogeneous" (dimensionally consistent). By dimensionally consistent, we mean that an equality, signified by the equals sign, requires not only that the value be identical but that the units be the same on both sides of the equation. All additive terms must have the same dimension. Consider an equation from physics that describes the position of a moving object:

x = x o + ν t+

a t2 2

length [=] length + velocity*time + acceleration*time2 [=] length + (length/time)*time + (length/time2)*time2 [=] length + length + length so the equation is dimensionally homogeneous.

Eg. Suppose that it is known that the composition C varies with time t in the following manner:

C = 0.03 exp(-2.00t) where C [=] (has units of) kg/L and t [=] s. What are the units associated with 0.03 and 2.00? Because the argument of an exponential function must be dimensionless, 2.00t must be unitless. Therefore 2.00 must have units of s-1.

Eg. Now suppose that during a particular experiment the data are obtained in units of C [=] lbm/ft3 and t [=] hr. Change the above equation so that the data can be directly plugged into it in these units. What we will do is write a new expression for the variables C and t in the previous units in terms of new variables C and t in the new units. Thus, t[s] = t [hr]·(3600 s/hr) = 3600·t C[kg/L] = C[lbm/ft3]·(453.59 g/lbm)(1 kg/1000 g)(35.3145 ft3/1000 L) = 0.0160·C So in terms of the new variables we have 0.0160C = 0.03 exp[-(2.0)(3600)t], or finally, C = 1.9 exp(-7200t) When you come across a formula and units for the variables are specified, you need to be aware that the constants may also have units. For instance, the formula for liquid flow through a pipe can be approximated by:

F = 36 .5



ΔP sg

where F is in gal/min, P is the pressure difference in psi between the entrance and exit of the pipe, and sg is the specific gravity of the flowing fluid. If the answer comes out in gal min-1, but we didn't put gallons in anywhere, there have to be some unit conversions hidden in the equation. If you want to use the equation in any way except with the exact given units, you need to examine the equation to determine the hidden units. For the above equation, the constant comes out to be



gal 1 min psiflow equation with different units, say m3/s and Pa? If you only how would you use the above 36 .5

have to do it one time, it isn't hard to convert Pa to psi, plug the psi into the equation to get galmin-1, and then convert the galmin-1 to m3/s. But you may be doing an experiment or problem where you need to use the dimensional equation repeatedly. In this case, it makes

(

36 5

gal

√ )( in2

1 m3

)(

1 min

)√

1 m2



0 . 22481 lb f

= 2 77 x 10−5

m3



1

sense to transform the dimensional equation to the new set of units. The constant (36.5) has all the units attached that make the equation work. All you have to do is convert those attached units:

Therefore the above equation can be written as:

F' = 2. 77 x 10−5

1.6)



Δ P' sg

Dimensionless Quantities

When a quantity is dimensionless, it means one of two things. First, it may just be a number like we get when counting. Example Pi is a dimensionless number representing the ratio of the circumference of a circle to its diameter Second are combinations of variables where all the dimension/units have "cancelled out" so that the net term has no dimension. These are often called "dimensionless groups" or "dimensionless numbers" and often have special names and meanings. Most of these have been found using techniques of "dimensional analysis" (a way of examining physical phenomena by looking at the dimensions that occur in the problem without considering any numbers). Probably the most common dimensionless group used in chemical engineering is the "Reynolds Number", given by

Re =

Dν ρ μ

where D 

[length] [length/time]

the diameter of the pipe the velocity of the fluid

 

[mass/length2] the density of the fluid [mass/(length*time)] the viscosity of the fluid

This describes the ratio of inertial forces to viscous forces (or convective momentum transport to molecular momentum transport) in a flowing fluid. It thus serves to indicate the degree of turbulence. Low Reynolds numbers mean the fluid flows in "lamina" (layers), while high values mean the flow has many turbulent eddies.

Eg:

where

    

CP [=] Btu/(lbm F)  [=] lbm/(hr*ft) k [=] Btu/(hr*ft F) D [=] ft G [=] lbm/(hr*ft2)

What are the units of h, knowing that h/(CPG) is unitless?

Note that you do not need to work through all of the equations to get the units of h. Since each individual group must be unitless or dimensionless, all that we need to do is work from a single group containing h. Thus, h [=] CPG [=] [Btu/(lbmF)][lbm/(hr*ft2)] [=] Btu/(hr*ft2F)

Sample Variance

Sample variance is a way of estimating data during experiments. Suppose we carry out a chemical reaction of the form A _ Products, starting with pure A in the reactor and keeping the reactor temperature constant at 45°C. After two minutes we draw a sample from the reactor and analyze it to determine X, the percentage of the A fed that has reacted.

In theory X should have a unique value; however, in a real reactor X is a random variable, changing in an unpredictable manner from one run to another at the same experimental conditions. The values of X obtained after 10 successive runs might be as follows:

Why don’t we get the same value of X in each run? There are several reasons.



It is impossible to replicate experimental conditions exactly in successive experiments. If the temperature in the reactor varies by as little as 0.1 degree from one run to another, it could be enough to change the measured value of X.



Even if conditions were identical in two runs, we could not possibly draw our sample at exactly t . 2:000 . . . minutes both times, and a difference of a second could make a measurable difference in X.



Variations in sampling and chemical analysis procedures invariably introduce scatter in measured values.

We might ask two questions about the system at this point.

1. What is the true value of X?

2. How can we estimate the true value of X?

We use Sample mean, range, sample variance and standard deviation to estimate data from experiments

The more a measured value (Xj) deviates from the mean, either positively or negatively, the greater the value of .Xj _ X.2 and hence the greater the value of the sample variance and sample standard deviation

For typical random variables, roughly two-thirds of all measured values fall within one standard deviation of the mean; about 88% fall within two standard deviations; and about 99% fall within three standard deviations.

Validating Results After your experiments, there are 3 ways of validating results. Among approaches you can use to validate a quantitative problem solution are 1. Back substitution 2. order-of-magnitude estimation, 3. The test of reasonableness.

Processes and Process Variables

1.1)

Introduction to terms:

Process Streams Chemical engineers are involved in designing plants which convert raw materials into useful products. Such plants commonly consist of several operating units, such as reactors, pumps, columns, and mixers, connected together by piping. The simplified flow of material from one unit to another is commonly shown in a process flow diagram, or PFD, such as the one shown in Figure 1 for the processing of crude oil to produce kerosene. PFDs can be highly simplified, showing piping as lines and process units as boxes or slightly more artistic like those in Figure 1. In this course, we will not be interested in what happens within the process unit, only how the process variables change across the unit. The term process stream refers to the flow of a particular material between operating units. Each unit will have input or feed streams as well as output or product streams. Figure 1:

Figure 1 represents a petroleum refinery which separates crude oil into useful components. The raw crude, which enters at the point labeled "CRUDE FEED" passes through a combination of pumps, heat exchangers, heaters and distillation columns and finally exits having been transformed into the desired products. The "CRUDE TOWER" is an example of one of the

many operating units within the refinery. It is a distillation column which separates the crude oil feedstock into output streams of various fuels as labeled on the diagram. Stream Variables Although a PFD contains detailed information regarding the physical flow of material through the plant, there is additional information associated with each stream that is used to design the process. These include the flow rates and compositions of the streams as well as the conditions (temperature and pressure) of the stream. These properties define the state of the stream. The properties of the process stream, such as flow rate, temperature, pressure, composition, density, etc., are called stream variables. On a complex PFD like that shown in Figure 1, stream variables are usually listed in a table that accompanies the PFD. For simpler processes, the stream variables are often entered right on the PFD as shown in Figure 2.

100 kg/h 40% ethanol 60% water

Process Unit

50 kg/h 60% ethanol 40% water

50 kg/h 20% ethanol 80% water

Figure 2: A convention that we will use here, though it is not universally used, is to place the stream flow rate above the stream and to put the compositions of the stream below the line. Other p...


Similar Free PDFs