1-1 Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2014)

Transcription

1 Air Conditioner Sizing Exercise On an uncomfortable summer day, the air is at 87 o F and 80% relative humidity. A laboratory air conditioner is to deliver 1000 ft 3 /min of air at 55 o F in order to maintain the laboratory at an average temperature of 75 o F and a relative humidity of 40%. A stream of outside air is combined with a recirculated laboratory air stream. The combined stream passes through a refrigeration unit which cools it to a low enough temperature to condense the required amount of moisture, and then reheats the dried air to 55 o F and delivers it into the lab. The cooling duty, Q (Btu/min), is the rate at which the refrigeration unit transfers energy from the air to accomplish the required cooling and condensation. Your task is to calculate Q. (The answer determines how large an air conditioner to order.) (a) What quantities needed for this calculation are given in the process specifications? (Two quantities on the list would be the temperature and relative humidity of the outside air; what are the others?) (b) What else would you need to determine to be able to calculate Q? (c) Outline how you would solve the problem. 1-1

2 UNITS, CONVERSION FACTORS, AND DIMENSIONAL EQUATIONS ( ) What we re going to do now you should have learned in high school or last year. Some of you know it, most don t. Learn it and save hours of calculation time and avoid costly (in grades here, in dollars later) and easily avoidable mistakes. (The Mars probe was lost because of a unit conversion error.) Measured and counted quantities have dimensions (length, time, mass, length/time, ) and units (s, ft, kg, miles/h, ft/s 2, kg/m 3, artichokes, moose/mile 2, ) In arithmetic operations, treat units like algebraic variables: 12 ft 7 ft 5 ft (12 x 7 x 5 x) 12 ft 7 cm??? (12 x 7 y???) 5 ft 3 lb = 15 ft lb (5 x 3 y = 15 xy) m m (3 sec) 9 sec (3 x) 9x 9.0 kg 3 ( dimensionless quantity) 3.0 kg miles x 55 3 h 165 miles 55 3y 165x h y Conversion factor: Ratio of a quantity to its equivalent in different units (= 1). Find ones you don t know in the conversion factor table on the inside front cover of the text. Convert (150.0 ft) to cm: ft cm (150.0)( ) cm 4572 cm ft ( ) Dimensional equations. To convert a quantity from one set of units to another, use a dimensional equation. Start with the quantity you re given & multiply by conversion factors to replace old units with new ones. 2 2 Convert g in/min to ton miles/wk g in ton ft mile min min g in ft h wk 5 ton miles 5 ton miles wk wk 2 Why four significant figures? In multiplication & division, sig. figs. in result = lowest number of sig. figs. in any of the terms being multiplied & divided. Read Sect. 2.5a, & obey the rules lose points if you don t. 1-2

3 PROCESSES AND PROCESS VARIABLES (Felder & Rousseau, Ch. 3) Process Unit Process Unit Inputs Outputs Here we have feed (input) and product (output) from the big box (process). Inside the box (process) you have individual boxes (processes) with their own inputs and outputs. Consider a familiar system : A car What do we want to know about these streams? Temperature, pressure, volumetric flow, mass flow, chemical composition. Some will be given, others we will measure, but many we will calculate. Mass flows we get from mass balance (CHE 205) Heating, cooling requirements we get from energy balance (CHE 205) What happens in the boxes (modeling, designing, sizing, and costing unit operations) the rest of the chemical engineering curriculum. Often we will have to convert from something that is easy to measure to something that is useful. Examples include: Volume to Mass to 1-3

4 Consider this simple case study: You mix 100 g of water with 100 ml of methanol into a container at 25 C. Draw and label this process below: How much mass is in the container? We could measure it (on a balance), but you can also calculate it. We need a way to relate volume to mass:. Density: Ratio of mass to volume. Density is a conversion factor between mass and volume (or between mass flow rate & volumetric flow rate of a process stream) M () l = kg/l (= g/cm 3 = lb m /ft 3 ) [Note: M is methanol, and l in M(l) means liquid] Liquid and solid densities vary slightly with temperature, almost not at all with pressure. Gas densities depend heavily on both T and P. Look up a liquid or solid density at one T & P, use at another set of conditions without introducing much error. Specific gravity (SG): Ratio of density of a species to the density of a reference species, usually liquid water at 4 o C and 1 atm. (What about gases? Use Equation of State, covered in Ch. 5). SG where ref =1.000 g/cm 3 = kg/m 3 = lb m /ft 3 (3.1-2) ref Look up SG for liquids and solids at 20 o C & 1 atm in Table B.1, use values to estimate density at any T and P. The following entries come from Table B.1 of F&R. Melting point at 1 atm Boiling point at 1 atm Compound Formula Mol. Wt. SG(20 o /4 o ) T m ( o C) T b ( o C) Methyl alcohol (Methanol) CH 3 OH Water H 2 O o Q: What is the mass of 100 ml of methanol at 25 o C? A: First, methanol must be a liquid at 25 o C because. 1-4

5 Then m MeOH CHE 205 Chemical Process Principles 100 ml g ml Q: That value is probably an overestimate of m methanol. Why? A: (Hint: What probably happens to methanol as T rises, and what did we assume in the calculation?) Q: How much mass is in the original container (water + methanol)? A: Q: What is V H 2O (ft 3 /s) of a stream of liquid water whose molar flow rate is 2.17x10 3 kmol/h? V H O kmol h 3 3 ft s How much volume is in the container? This is tougher because volume is not Non-additivity of liquid volumes. Going back to our case study, let s generically call water A and methanol B and when they are together, it forms a solution T (for total). m A (kg) n A (mol) V A (L) m B (kg) n B (mol) V B (L) m T (kg) n T (mol) V T (L) For the mixture, m T = m A + m B (law of conservation of mass) n T = n A + n B (provided only that A and B don t react and neglecting evaporation) but V T V A + V B (volumes of liquids are not additive, although it s usually close) Compound m(g) Mol. Wt. n(mol) (kg/l) V(mL) Methyl alcohol (Methanol) Water o Mixture 179 MW If you don t have density data for a liquid mixture, assume volume additivity of the components. 1-5

6 How many mols are in the container? Like mass, mols are also additive, so just calculate the mols of water and mols of methanol, and add them! First, why do we care about mols? Molecular weight: Sum of atomic weights of the atoms of a molecule ( 12 C has A.W. = ) Values in table reflect mixture of isotopes found in nature. Mole: Amount of a species whose mass is numerically equal to its molecular weight. 1 g-mole CH 3 OH (or 1 mol CH 3 OH) = g CH 3 OH (= 6.02x10 23 molecules of CH 3 OH) 1 lb-mole CH 3 OH = lb m CH 3 OH 1 kg-mol CH 3 OH (or 1 kmol CH 3 OH) = kg CH 3 OH = 1000 mol CH 3 OH Molecular weight is a conversion factor between mass and moles g CH3OH kg CH3OH lb m CH3OH 1 mol CH OH 1 kmol CH OH 1 lb-mole CH OH etc. Q: How many mols of methanol are there (in the case study)? A: Q: How many mols of water are there (in the case study)? A: Q: How many total mols are there in the container (in the case study)? A: Q: How many lb-mols of water are there (in the case study)? A: 1-6

7 What is the composition of the liquid in the container (case study)? Mass fraction, Mole fraction, Concentration For the liquid mixture in the example above: 79.2 g CH3OH g CH3OH lb m CH3OH kg CH3OH Mass fraction: x M g g lb kg 2.47 mol CH3OH kmol CH3OH lb-mol CH3OH Mole fraction: ym mol mixture kmol lb-mol 2.47 mol CH3OH mol CH3OH Concentration: CM L mixture L m The denominator units of the mass and mole fractions (g, mol) always refer to the total mixture The mixture contains 44.0 wt% CH 3 OH (44% w/w, 44% by mass) and 31.0 mole% CH 3 OH. The molarity of methanol in the mixture is 11.8 The mass fractions of all components of a mixture must add up to So must the mole fractions. If you know all but one, you can calculate the last one by difference (y W = 1 y M = 0.69) Before leaving Chapter 3, be sure you know how to calculate mole fractions from known mass fractions & vice versa; the definition of average molecular weight of a mixture, and how to calculate it. (See Examples and ) Mole and mass fractions are dimensionless, but not unitless. What is the average molecular weight in the container (case study)? The average molecular weight is the sum of the molecular weights of the mixture weighted by their mol fractions. Calculate it below: 1-7

8 What if you needed to convert between mole and mass fractions? Let s stick with the case study: In either case, you can assume some total mass (or total mole) because mole and mass fractions are independent of the total amount (we call these properties intensive, as discussed later). Q: Assuming 1000 g of liquid in the container, calculate the mole fractions knowing that the mass fraction of methanol is Q: Would your answer change if you assumed 20 g of liquid in the container? Q: Assuming 1000 mole of liquid in the container, calculate the mass fractions knowing that the mole fraction of methanol is

9 What is the force exerted by the liquid (case study) on the bottom of the container? Mass and weight. Go through Sect. 2.4 carefully. Make sure you can do the Test Yourself on p. 13 without looking up the answers. o o Force = x How much force results from gravity acting on the liquid? F = kg x = o For convenience, we define the unit N (1 N = 1 kg m/s 2 ) o Likewise, we define the unit (1 lb f = lb m ft / s 2 ) o This conversion factor is often called g c which converts between natural and derived force units. Don t confuse it with the gravitational constant, g. If you are having trouble getting your units to come out right, you may have forgotten to use g c or. o Note the difference between lb m (a mass unit) and lb f (a force unit). When you use the word pounds be sure to know which you are talking about. Archimedes Principle: The mass of a floating object equals the mass of the fluid displaced by that object. 1-9

10 What is the pressure exerted by the liquid (case study) on the bottom of the container? Fluid Pressure (3.4) A container of a fluid of density (kg/m 3 ) has height h(m) and cross-sectional area A(m 2 ). A uniform pressure P o (N/m 2 ) is exerted on the upper surface of the fluid. P o (N/m 2 ) A(m 2 ) (kg/m 3 ) P(N/m 2 ) h(m) Q: How much does the fluid weigh? (Let g denote the acceleration of gravity.) A: W f = Q: What is the pressure exerted by the fluid on the bottom surface of the generic container? A: Let F o and F be the forces exerted on the top and bottom surfaces. Then F 1 1 P [ F 2 o (N) Wf (N)] [ ] A A (m ) A P P gh o (3.4-1) Q: What is the pressure exerted by the fluid on the bottom surface of the container (case study) assuming the container is a cylinder with a radius of 1 cm? A: Q: What would we need to know about the fluid to determine its height in the container? A: Pressure expressed as a head of fluid. Using Eq , you can express any pressure as a head of a particular fluid with known density () that is, the height (h) of a column of that fluid with a surface pressure of zero (P o = 0). So, for example, a pressure of 14.7 lb f /in 2 can also be expressed as a head of mercury (specific gravity = 13.6). 1-10

11 P h(mm) = g 14.7 lb in ft lb ft / s mm in 1 ft 13.6 lb ft/s 1 lb ft f m m f = 760 mm Hg (= 760 torr) Conversion factors for pressure (including the one just calculated) are given in the inside front cover of the book. Atmospheric pressure (or barometric pressure): The earth s atmosphere can be considered a very tall column of fluid (air). The pressure at the bottom of that column is atmospheric pressure. At sea level, P atm is typically on the order of 760 mm Hg (or 760 torr). By definition, 760 torr = 1 atm. Questions: Would atmospheric pressure at a ski resort be greater or less than atmospheric pressure at sea level? Explain your answer. Suppose you are given the height of the atmosphere. Why can t you use P = P o + gh to calculate atmospheric pressure? Absolute pressure and gauge pressure. The absolute pressure of a fluid is its pressure relative to a perfect vacuum (P = 0). The gauge pressure is its pressure relative to atmospheric pressure; that is, the absolute pressure minus atmospheric pressure (that atmospheric pressure can be any value and is not limited to 1 atm). It is called gauge because many pressure gauges, such as the Bourdon gauge shown on p. 57 of the text, are calibrated to read 0 when P = P atm.) Thus P gauge = P absolute P atm Sometimes you will see lb f /in 2 units written as psia (pounds per square inch absolute) or psig (pounds per square inch gauge). As an analogy, suppose you are 6 tall and your friend is 6 2 tall. You could either say your friend is 6 2 tall, or that he is 2 taller than you (his gauge height compared to you is 2 ). Either way, it doesn t change his actual height, just the way you report it. Gauge pressure is just a convenient way of reporting pressure inside closed containers. Unless you are told otherwise, assume that given pressures are absolute. Work through the Test Yourself on p

12 Q: The absolute pressure of a gas on a sunny day is 3.67 atm. A low pressure storm comes up. If the conditions in the gas cylinder remain unchanged, does the absolute pressure of the gas increase, decrease, or remain the same? What about the gauge pressure? Manometers are useful and simple instruments for measuring pressure visually. Sketch and label a general manometer (see Fig ) What is the pressure at point (a)? What is the pressure at point (b)? If you equate the pressures at (a) and (b), you get the general manometer equation: What happens if fluid 1 and 2 are the same? What happens if one of the ends of manometer is sealed? 1-12

13 What happens if fluid 1 or 2 is a gas at moderate pressure? What happens if both fluids are gases? What if P 1 and P 2 are expressed in units of heads of manometer fluid (i.e. mm Hg or m H 2 O)? Example: A cylindrical tank is filled with methylethyl ketone (MEK) up to a height of 2.78 ft. The tank is open to the atmosphere and has a diameter of 4.00 feet. An open-end mercury manometer is attached to the side of the tank. Assume the atmospheric pressure is 1 atm. a) What is the absolute pressure on the bottom surface of the tank (in psia)? 0 b) What is the gauge pressure on the bottom surface of the tank (in psig)? c) What is the gauge pressure on the bottom surface of the tank (in ft MEK)? d) Suppose the distance d 1 is equal to one foot. What reading would you expect on the manometer (h, in mm Hg)? 1-13

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15 Temperature (3.5) CHE 205 Chemical Process Principles To convert a temperature in one unit to an equivalent unit: T(K) = T( o C) T( o R) = T( o F) T( o R) = 1.8 T(K) T( o F) = 1.8 T( o C) + 32 To convert one temperature interval to another temperature interval, use the following conversion factors: o o o o 1.8 F 1.8 R 1 F 1 C,,, o o 1C 1K 1R 1K Before leaving Chapter 3, be sure you know how a manometer works, and how to derive and use Eqs. (3.4-5) (3.4-7). the difference between a temperature of 20 o C and a temperature interval of 20 o C; how to convert both temperatures and temperature intervals among the four most common temperature scales. For an excellent practice problem and a good head start on Chapter 4, see Problem 3.29 in the Student Workbook. 1-15

16 We won t cover this in class since it should be review, but you may find it helpful: PROCESS DATA REPRESENTATION AND ANALYSIS (2.7) The operation of a chemical plant is based on the measurement of process variables temperatures, pressures, flowrates, concentrations, etc. To a process engineer, a process is defined as the set of all those variables. Sometimes possible to measure directly, but more often we must relate one variable to another that is easier to measure. We might do some kind of calibration experiment from which we can develop an equation relating one variable to another. Example: Set values of C A (say, a concentration of a salt A in an aqueous solution, mg/ml), measure corresponding values of something that s easy to measure [say, (μs/cm) the electrical conductivity of the solution]. (S = Siemen = 1/Ohm.) Solution: Known amounts of A and H 2 O C A (mg A/L) Conductivity meter C A κ Plot C A vs. κ (calibration plot). Then, measure κ for solutions with unknown salt concentration, use calibration plot to determine C A. Possibilities: (a) (b) (c) C A C A C A κ κ κ If (a) or (b), our life is easy and we have a straightforward linear relationship. C A = aκ + b which is much more convenient than a curve for converting κ to C A calculator, spreadsheet, computer If it s (c) but we straight-line it anyway, when do we run into trouble? When we get outside the range of our data. Interpolation versus extrapolation. Better find a nonlinear function that fits the curvature of the data. (Extrapolation still risky.) 1-16

17 Suppose we are lucky and the data appears linear. Plot C A vs. κ, draw line through data. Two points on line are (κ = 10.0, C A = 5.00) & (κ = 310.0, C A = 205). Note: we re not picking two of the data points and drawing a line through them --- we are drawing a line through the data and picking two points on the line we have drawn. C A 100 (κ 1 = 10.0, C A1 = 5.00) (κ 2 = 310.0, C A2 = ) κ Derive an expression for C A (κ). (See Eqs , 2.7-3, and on p. 24 of Felder text book) Aqueous solutions are drawn from two process vessels and analyzed. Their electrical conductivities are found to be κ a = 200 μs/cm and κ b = 550 μs/cm. Estimate the salt concentrations in both vessels. Which of your two estimates would you have least confidence in, and why? Speculate on whether your value is an underestimate or an overestimate, and explain your reasoning. Suppose the salt concentration of the first solution is determined by chemical analysis, and it is found to be significantly different from the value you determined. Think of at least five possible explanations for the discrepancy. 1-17

18 NONLINEAR DATA (2.7c,d) CHE 205 Chemical Process Principles Data is not always linear, but sometimes we can plot it in a different way to make it linear. Linear plots are convenient for fitting (to get slope & intercept), interpolation, and ease of use. The key concept is to get a general expression: (What you plot on vertical axis) = SLOPE x (What you plot on horizontal axis) + INTERCEPT Let s try y = ax 3 + b. A plot of y vs. x is non-linear, but a plot of y vs. x 3 should be a straight line with slope a & intercept b. (Why?) x y x x Slope = y Intercept = x 3 Equation: Example Sometimes we have to rearrange an equation to get it in a linear form. Suppose we have nonlinear (x,y) data, and we have reason to believe that they are related through the equation sin ynx my where n and m are constants (adjustable parameters). Questions: (1) Does that relationship fit our data? (2) If it does, what are the best values of n and m? Task: Rearrange the equation into a form f(x,y) = ag(x,y) + b, plot f vs. g. If linear, original equation is valid. Determine a and b from plot, use to calculate m and n. How about (sin y my) = nx, plot (sin y my) vs. x, if linear then slope = n? Won t work, because Solution: sin y x sin y x x y y y sin y sin y nxmy n m y y f(x,y) g(x,y) y x / y Conclusion: Function is correct, n = slope of line, m = intercept Work through Test Yourself on p. 27. For additional practice, work through Problem 2.31 in the Student Workbook and check your solutions in the back. 1-18

19 SWITCHING VARIABLES IN EQUATIONS and PLOTS Suppose you have a function shown below, where V (cm 3 ) and t (min): V = 2t 3 What are the units of 2 and 3? How would you plot this to make it linear? What would be the slope and intercept? Suppose the equation represents the volume of a balloon V (cm 3 ) as you blow it up over time (min). Some data are below: t (min) T (sec) V (cm 3 ) If you wanted to get an expression in terms of seconds instead of minutes, the temptation would be to do the following: V = 2 (t ) 3 Why is this NOT correct? Instead, define a new variable T (sec), instead of t (min). These are related by T = t * 60 (sec/min). If we plug this into the original equation, we get: Another option is to convert the units of the coefficients, e.g. convert all minutes in all the coefficients into seconds. 1-19

20 EXPONENTIAL AND POWER LAW FUNCTIONS (Section 2.7d) 1-20

21 Example. A toxic waste product from a chemical process, A, is treated in a holding tank with a bacterial agent that causes it to decompose. Effluent from plant High C A Bacteria Tank Discharge to sewer Low C A The initial concentration of A in the tank is C A0 = 10.0 mol A/liter. Samples are frequently drawn from the reactor and analyzed for A, leading to the following data: mol A t(min) C A L CA(mol/L) t (min) We want to determine an expression for C A (t) so we can determine the holding time required for the A to fall below its safe value of mol/l. Curve looks like an exponential decay, so let s try CA bt ae (expect b to be negative). bt C ae ln( C ) ln abt A A Plot ln C vs. t. If plot is linear, assumption is correct, slope = b, intercept = ln a A Rectangular plot Semilog plot t(min) mol A C A ln C A L ln(ca) [(t 1 =0, ln(c A1 )=2.303)] t or CA (t 1 =0, C A1 =10.0) t [(t 2 =5, ln(c A2 )= 0.223)] (t 2 =5, C A2 =0.80) logarithmic axis 1-21

22 Rectangular: What would you plot versus what? CHE 205 Chemical Process Principles How do you calculate the slope and intercept? What is the final equation for C A? Semilog: What would you plot versus what? Calculate the slope and intercept: What is the final equation for C A? How long will it take for C A to drop to mol A/L? Q: What could go wrong with this waste treatment process? 1-22

23 1. Hypothetical data: x y CHE 205 Chemical Process Principles 2. You believe the data has a power law relationship 3. If true, two approaches will produce linear fits: (i) Change both axis Y In both cases: X-Y on Rectangular Coordinates X ln b ln b y a x y2 ln( y1) x lnx 2 1 ln Non-linear! y lna blnx (i) Change the data 10.0 Y Y vs. X on Log Plot X ln Y vs. ln X on Rectangular Plot ln X ln Y 1-23

24 Exercise: A photometric detector sensitive to the concentration of A gives a reading R. In a calibration experiment, values of R for known values of C A are measured. A log plot of the calibration data appears linear. A line drawn through the data passes through the two points labeled below. 10 CA(mol/L) 1 (R 1 =2.7, C A1 =0.75) (R 2 =38, C A2 =4.7) Derive an expression for C A (R). Solution: 1 10 R 100 The plot shown is equivalent to a rectangular plot of vs. The form of the equation of the rectangular plot is = + ln a The form of the desired expression is therefore C A = ar b The exponent b is calculated as The intercept ln a is o a = The final expression is C A = Work through Test Yourself on p. 29. For additional practice, work through Problems 2.32 and 2.38 in the Student Workbook. Read Section 2.7e and Appendix A.1 on the Method of Least Squares. 1-24