Fundamental Concepts and Computations in Chemical Engineering Solution Manual

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Solutions Manual forFundamental Conceptsand Computations inChemical EngineeringVivek UtgikarA note regarding this Solutions Manual:As the problems in Chapters 1-3 are discussion problems,no formal solutions are provided for those chapters.You will find solutions for Chapters 4-9 herein.

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Chapter 44.1.Cramer’s rule and matrix inversion-multiplication offer alternative techniques to solve a system oflinear algebraic equations. Conduct a literature search to collect information about these twotechniques and the elimination and iteration techniques discussed in this chapter. Compare the varioustechniques regarding the complexity of algorithms, ease of implementation, and potential errors.No solution will be given.4.2.The Newton-Raphson technique may not converge to a solution. Inspecting equation 4.16, in whatother possible way can the technique fail?Equation 4.16 is:The technique will not yield a solution, if the absolute value of the second term on the right hand side does not tendto approach 0 with increasing number of iterations. However, the technique will also fail at any point wheref’(xn) =0. The second term becomes indeterminate at this point and no further evaluations are possible.4.3.Roots of any equation can be found using what is known as thebracketing technique. Conduct aliterature search and explain the principle behind such solution techniques.No solution will be given.4.4.The following data were obtained in an experiment where the concentration of a substance wasmonitored as a function of time. Calculate the first derivative of the concentration with respect to timefor all possible times using the forward difference formula. Can the second derivative also becalculated numerically?Time, sConcentration00100.5201.0302.0404.0505.5606.5707.0907.7The first derivative of concentration is calculated using equation 4.18. Further application of the principle yields thefollowing forward difference formula for the second derivative:1'nnnnfxxxfx2212212iiiiiCCCd CCdttttt

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Excel calculations for the first and second derivatives of concentration are shown below in columns 4 and 6,respectively.Time, sConcentrationC=Ci+1–CiC/t(C/t)(C/t)/t000.50.0500100.50.50.050.050.00520110.10.10.0130220.2-0.05-0.0054041.50.15-0.05-0.005505.510.1-0.05-0.005606.50.50.05-0.015-0.00157070.70.035907.74.5.What is the area under the concentration-time curve obtained from the data shown for problem 4.4?Use the trapezoid method. An alternative technique is to use the rectangle method. What is thedifference in the areas if the area is calculated using the rectangle method?A plot of the concentration-time data is shown below. Also shown are the trapezoids formed between any twoadjacent date points by the straight line between the two data points, the time-axis, and the two ordinates. The totalarea under the curve is found by calculating the area of each trapezium and adding all such areas. The trapezoidalrule yields an area of 377 concentration units-seconds.A simpler alternative is to draw rectangles as shown in the figure below. The area under curve in this case is 419concentration units-seconds. This is clearly an overestimate, as it assumes that the concentration in any time intervalis constant and equal to the concentration at the end of the interval. If on the other hand, it is assumed that theconcentration in any time interval is equal to the concentration at the beginning of that interval, the area obtainedwould be 335 concentration units-seconds, a clear underestimate. However, all three values will tend to converge to0123456789020406080100ConcentrationTime, sProblem 4.5Trapezoidal Rule

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a single value as the frequency of measurements increases or the time interval between measurements decreases to avery small value.0123456789020406080100ConcentrationTime, sProblem 4.5Rectangle

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Chapter 55.1Calculate the Reynolds numbers for a 1.5 in. inside diameter pipe carrying water ata flow rate of 0 to 5 gpm. Assume a temperature of 25°C.The Excel solution to the problem is shown below:The density of water is 1 g/cm3. The viscosity value is taken from the data provided in the chapter.5.2Calculate the Reynolds numbers for the following situation: (a) a 1m sizedmicrobe swimming with a speed of 30m/s; (b) a swimmer competing in anOlympic 100 m race finishing in 50 s. Make any reasonable assumptions necessaryfor the solution.The temperature is assumed to be 25oC, making the density and viscosity of water values to be 1g/cm3, and 0.009 poise. The microbe dimension is stated (1m), however, the swimmer dimensionsare not provided. It is assumed that the characteristic length dimension for the swimmer is 1 ft. TheReynolds number calculations are straightforward and are shown below.

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The flow around the microbe is highly laminar, while it is highly turbulent for the slow swimmer.5.3The viscosity of 30 wt engine oil at 100°C is 0.0924 poise. What is the viscous(shear) force needed to slide an 8 cm diameter, 8 cm long piston through a cylinderon a 2 micron thick oil film with a speed of 8 m/s?The shear force is calculated using the relation:The density of 30 wt engine oil is found from the internet sources to be 0.8 g/cm3. The velocitygradient is calculated assuming a linear velocity profile between the sliding and stationary surfacesseparated by the thickness of the oil film. The shear force needed is ~74N. The results are shownbelow:rshearsheardvFAdr

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5.4For noncircular geometries, a hydraulic diameter (Dh) is used as the characteristiclength parameter for calculating the Reynolds number calculation:𝐷ℎ= 4 ∙ 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟An HVAC duct circulates 600 cfm (cubic feet per minute) of air at 85°F throughan 18 in. × 12 in. rectangular duct. What is the air velocity? What is the Reynoldsnumber if the air density and viscosity at 85°F are 1.177 kg/m3and 1.85 × 10−2mPa‧s, respectively?The solution is shown below:

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5.5It may be feasible to extract uranium from sea water (concentration 8 ppb, or partsper billion) by placing uranium absorbing plates in the ocean. The oceanic wavesresult in circulation of water through the absorbing structure. The absorbing platesare mounted so they divide a 5 ft × 5 ft square pipe into smaller square pipes. Thevolumetric flow rate through the big pipe is 1000 gal/min. What is the Reynoldsnumber if the viscosity is 1 centipoise (cP)? How small can the opening of thesmaller square be for the flow to still be in the turbulent regime? Assume that theaverage velocity remains constant.The hydraulic diameter of the big pipe is calculated from the cross sectional area and the wettedperimeter using the formula stated above. The Reynolds number calculations are straightforward,and yield a value greater than 47000 indicating that the flow is turbulent. The minimum Reynoldsnumber for turbulent flow is taken to be 4000; assuming that the velocity remains the same, thehydraulic diameter needed for this Reynolds number is found to be14.32cm, which is also thelength of the square-shaped opening.5.6Supertankers with ultra-large capacity can carry 3,166,353 barrels of crude oil (1barrel = 42 gal). If it takes two days to unload the tanker using a 24 in. diameterhose, what is the average velocity of the oil? If the oil density and viscosity are 870g/L and 0.043 poise, respectively, what is the Reynolds number? Is this flowturbulent?The calculations are shown below. The flow is highly turbulent.

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5.7What is the friction factor for the flow in problem 5.6? What is the pressure drop ifthe oil is pumped into a storage tank located 1 km away?The friction factor is calculated from the Nikuradse equation using theGoal Seekfunction.The pressure drop is calculated using equation 5.7:The pressure drop is found to be ~116 psi.1014.0log0.40Reff22fv LPd

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5.8The height of a liquid in a partially filled spherical tank can be calculated from thefollowing equation:𝑉𝑙𝑖𝑞𝑢𝑖𝑑= 𝜋3 ℎ2(1.5𝐷 − ℎ)What is the height of liquid when a 35,000 m3capacity storage tank is only 75%full? What is the area of the free liquid surface?The diameterDof the tank is calculated from the tank capacity:D= (6Vtank/)1/3= 40.58 m.The Goal Seek function is used to calculate the height of the liquid, which is found to be 27.34 m.Calculation of the free surface area requires obtaining the radiusrhof the circular free surface atheighthusing the formularh= [(D/2)2-(h-D/2)2]½. The radiusrhis found to be 19.03 m, yielding aarea of free surface to be 1138 m2.DhD/2h-D/2rh

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5.9The following data were obtained for the pressure drop for water flow through 100ft of fire hose:Hose Diameter,in.Flow Rate,gpmPressureDrop,psi1.52.02.53.012015022040045201215Plot the pressure drop as a function of Reynolds number for the flow. Makereasonable assumptions in order to perform these calculations. Calculate thefriction factors from these data and compare those with the friction factors obtainedusing the Nikuradse equation.The calculations performed are as shown below. The observed friction factor are calculated usingequation 5.7. The calculated friction factors are based on the observed Reynolds number. Nikuradseequation is used to calculate these friction factors using theGoal Seektool at each data point.

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5.10Enbridge Line 5 is a 30 in. pipeline connecting Wisconsin and Canada throughMichigan’s upper and lower peninsulas carrying ~550,000 barrel of light crude010203040502.00E+053.00E+054.00E+055.00E+05Pressure Drop, psiReynolds numberProblem 5.9010203040500.0010.0020.0030.0040.00Observed Pressure Drop, psiCalculated Pressure Drop, psiProblem 5.90.003300.003800.004300.004800.005300.003300.003800.004300.004800.00530Calculated Friction FactorObserved Friction FactorProblem 5.9

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every day. The line is split into two 20 in. pipes buried under deep water forcrossing 4.5 miles of the Straits of Mackinac. Assuming the fluid densities andviscosities of problem 5.6, calculate the Reynolds numbers for the both the 30 in.and 20 in. sections. What is the pressure drop across the Straits of Mackinac?The flow in both 20 in. and 30 in. pipes is highly turbulent, with the respective Reynolds numbersbeing ~256000 and ~341000. The pressure drop across the Straits of Mackinac is5.67atm foreach 20 in. pipeline.

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Chapter 66.1.Raw French-cut potato strips for fries are dried in a conveyer-type dryer, which operates at 60°C. The watercontent of the strips at the inlet to the dryer is 60% by mass. Dry air at 60°C is fed to the dryer at a volumetricflow rate of 1000 ft3/min. Air exiting the dryer is saturated with water vapor with the saturation moisturecontent of 3.7 g water per dry ft3of air. The feed rate of strips is 20 kg/min. What is the moisture content ofstrips exiting the dryer? What is the condensate flow rate if the air exiting the dryer is passed through acondenser for the removal of moisture?The schematic representation of the process is shown below:Symbol Definition ṁwet chips: Mass flow rate of wet chips entering the dryer = 20 kg/min ṁdry chips: Mass flow rate of dried chips exiting the dryerQdry air: Volumetric flow rate of dry air = 1000 ft3/minxw in: mass fraction of water in wet chips entering the dryerxw out: mass fraction of water in dried chips exiting the dryerCw air: concentration of water in air exiting the dryer = 3.7 g/ft3dry airMass Balance on the DryerWater Balance:Water in with wet chips + water in with dry air = water out with dried chips + water out with wet airṁwet chips·xw in+ 0 = ṁdry chips·xw out+Qdry air·Cw airStarch (Chip Dry Matter) BalanceStarch in with wet chips = Starch out with dried chipsṁwet chips· (1 -xw in) = ṁdry chips· (1 -xw out)Water Balance: 20 kg/min · 0.6 = ṁdry chips·xw out+ 1000 ft3/min · 3.7 g/ft3· 1 kg/1000 gTherefore, ṁdry chips·xw out= 8.3 kg/min… (A)Starch Balance: 20 kg/min · (1 - 0.6) = ṁdry chips· (1 -xw out)kg/minTherefore,ṁdry chips· (1 -xw out) = 8 kg/min… (B)Wet Chips20 kg/minWet Air, 3.7 g water/ft3Dry Air, 1000 ft3/minDry AirCondensateDryerCondenserCoolantDry Chips

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