Solutions Manual for Fundamental Concepts and Computations in Chemical Engineering Vivek Utgikar A 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. Chapter 4 4.1. Cramer’s rule and matrix inversion -multiplication offer alternative techniques to solve a system of linear algebraic equations. Conduct a literature search to collect information about these two techniques and the elimination and iteration techniques discussed in this chapter. Compare the various techniques 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 what other 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 tend to approach 0 with increasing number of iterations. However, the technique will also fail at any point where f’ ( x n ) = 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 the bracketing technique . Conduct a literature 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 was monitored as a function of time. Calculate the first derivative of the concentration with respect to time for all possible times using the forward difference formula. Can the second derivative also be calculated numerically? Tim e, s Concentration 0 0 10 0.5 20 1.0 30 2.0 40 4.0 50 5.5 60 6.5 70 7.0 90 7.7 The first derivative of concentration is calculated using equation 4.18. Further application of the principle yields the following forward difference formula for the second derivative:     1 ' n n n n f x x x f x      2 2 1 2 2 1 2 i i i i i C C C d C C dt t t t t                   Excel calculations for the first and second derivatives of concentration are shown below in columns 4 and 6, respectively. Time, s Concentration  C = C i+1 – C i   C/  t   (  C/  t )   (  C/  t )/  t  0 0 0.5 0.05 0 0 10 0.5 0.5 0.05 0.05 0.005 20 1 1 0.1 0.1 0.01 30 2 2 0.2 - 0.05 - 0.005 40 4 1.5 0.15 - 0.05 - 0.005 50 5.5 1 0.1 - 0.05 - 0.005 60 6.5 0.5 0.05 - 0.015 - 0.0015 70 7 0.7 0.035 90 7.7 4.5. 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 the difference 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 two adjacent date points by the straight line between the two data points, the time-axis, and the two ordinates. The total area under the curve is found by calculating the area of each trapezium and adding all such areas. The trapezoidal rule 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 419 concentration units-seconds. This is clearly an overestimate, as it assumes that the concentration in any time interval is constant and equal to the concentration at the end of the interval. If on the other hand, it is assumed that the concentration in any time interval is equal to the concentration at the beginning of that interval, the area obtained would be 335 concentration units-seconds, a clear underestimate. However, all three values will tend to converge to 0 1 2 3 4 5 6 7 8 9 0 20 40 60 80 100 Concentration Time, s Problem 4.5 Trapezoidal Rule a single value as the frequency of measurements increases or the time interval between measurements decreases to a very small value. 0 1 2 3 4 5 6 7 8 9 0 20 40 60 80 100 Concentration Time, s Problem 4.5 Rectangle Chapter 5 5.1 Calculate the Reynolds numbers for a 1.5 in. inside diameter pipe carrying water at a 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/cm 3 . The viscosity value is taken from the data provided in the chapter. 5.2 Calculate the Reynolds numbers for the following situation: (a) a 1  m sized microbe swimming with a speed of 30  m/s; (b) a swimmer competing in an Olympic 100 m race finishing in 50 s. Make any reasonable assumptions necessary for the solution. The temperature is assumed to be 25 o C, making the density and viscosity of water values to be 1 g/cm 3 , and 0.009 poise. The microbe dimension is stated (1  m), however, the swimmer dimensions are not provided. It is assumed that the characteristic length dimension for the swimmer is 1 ft. The Reynolds number calculations are straightforward and are shown below. The flow around the microbe is highly laminar, while it is highly turbulent for the slow swimmer. 5.3 The 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 cylinder on 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/cm 3 . The velocity gradient is calculated assuming a linear velocity profile between the sliding and stationary surfaces separated by the thickness of the oil film. The shear force needed is ~ 74N . The results are shown below: r shear shear dv F A dr          5.4 For noncircular geometries, a hydraulic diameter ( D h ) is used as the characteristic length parameter for calculating the Reynolds number calculation: 𝐷 ℎ = 4 ∙ 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 An HVAC duct circulates 600 cfm (cubic feet per minute) of air at 85°F through an 18 in. × 12 in. rectangular duct. What is the air velocity? What is the Reynolds number if the air density and viscosity at 85°F are 1.177 kg/m 3 and 1.85 × 10 − 2 mPa ‧ s, respectively? The solution is shown below: 5.5 It may be feasible to extract uranium from sea water (concentration 8 ppb, or parts per billion) by placing uranium absorbing plates in the ocean. The oceanic waves result in circulation of water through the absorbing structure. The absorbing plates are mounted so they divide a 5 ft × 5 ft square pipe into smaller square pipes. The volumetric flow rate through the big pipe is 1000 gal/min. What is the Reynolds number if the viscosity is 1 centipoise (cP)? How small can the opening of the smaller square be for the flow to still be in the turbulent regime? Assume that the average velocity remains constant. The hydraulic diameter of the big pipe is calculated from the cross sectional area and the wetted perimeter 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 Reynolds number for turbulent flow is taken to be 4000; assuming that the velocity remains the same, the hydraulic diameter needed for this Reynolds number is found to be 14.32 cm, which is also the length of the square-shaped opening. 5.6 Supertankers with ultra-large capacity can carry 3,166,353 barrels of crude oil (1 barrel = 42 gal). If it takes two days to unload the tanker using a 24 in. diameter hose, what is the average velocity of the oil? If the oil density and viscosity are 870 g/L and 0.043 poise, respectively, what is the Reynolds number? Is this flow turbulent? The calculations are shown below. The flow is highly turbulent. 5.7 What is the friction factor for the flow in problem 5.6? What is the pressure drop if the oil is pumped into a storage tank located 1 km away? The friction factor is calculated from the Nikuradse equation using the Goal Seek function. The pressure drop is calculated using equation 5.7: The pressure drop is found to be ~ 116 psi .   10 1 4.0log 0.40 Re f f   2 2 fv L P d    5.8 The height of a liquid in a partially filled spherical tank can be calculated from the following equation: 𝑉 𝑙𝑖𝑞𝑢𝑖𝑑 = 𝜋 3 ℎ 2 (1.5𝐷 − ℎ) What is the height of liquid when a 35,000 m 3 capacity storage tank is only 75% full? What is the area of the free liquid surface? The diameter D of the tank is calculated from the tank capacity: D = (6 V tank /  ) 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 radius r h of the circular free surface at height h using the formula r h = [( D /2 ) 2 -( h - D /2) 2 ] ½ . The radius r h is found to be 19.03 m, yielding a area of free surface to be 1138 m 2 . D h D/2 h - D/2 r h 5.9 The following data were obtained for the pressure drop for water flow through 100 ft of fire hose: Hose Dia meter, in . Flow Rate , gpm Pressure D rop, psi 1.5 2.0 2.5 3.0 120 150 220 400 45 20 12 15 Plot the pressure drop as a function of Reynolds number for the flow. Make reasonable assumptions in order to perform these calculations. Calculate the friction factors from these data and compare those with the friction factors obtained using the Nikuradse equation. The calculations performed are as shown below. The observed friction factor are calculated using equation 5.7. The calculated friction factors are based on the observed Reynolds number. Nikuradse equation is used to calculate these friction factors using the Goal Seek tool at each data point. 5.10 Enbridge Line 5 is a 30 in. pipeline connecting Wisconsin and Canada through Michigan’s upper and lower peninsulas carrying ~550,000 barrel of light crude 0 10 20 30 40 50 2.00E+05 3.00E+05 4.00E+05 5.00E+05 Pressure Drop, psi Reynolds number Problem 5.9 0 10 20 30 40 50 0.00 10.00 20.00 30.00 40.00 Observed Pressure Drop, psi Calculated Pressure Drop, psi Problem 5.9 0.00330 0.00380 0.00430 0.00480 0.00530 0.00330 0.00380 0.00430 0.00480 0.00530 Calculated Friction Factor Observed Friction Factor Problem 5.9 every day. The line is split into two 20 in. pipes buried under deep water for crossing 4.5 miles of the Straits of Mackinac. Assuming the fluid densities and viscosities 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 numbers being ~256000 and ~341000. The pressure drop across the Straits of Mackinac is 5.67 atm for each 20 in. pipeline. Chapter 6 6.1. Raw French-cut potato strips for fries are dried in a conveyer-type dryer, which operates at 60°C. The water content 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 volumetric flow rate of 1000 ft 3 /min. Air exiting the dryer is saturated with water vapor with the saturation moisture content of 3.7 g water per dry ft 3 of air. The feed rate of strips is 20 kg/min. What is the moisture content of strips exiting the dryer? What is the condensate flow rate if the air exiting the dryer is passed through a condenser for the removal of moisture? The schematic representation of the process is shown below: Symbol Definition ṁ wet chips : Mass flow rate of wet chips entering the dryer = 20 kg/min ṁ dry chips : Mass flow rate of dried chips exiting the dryer Q dry air : Volumetric flow rate of dry air = 1000 ft 3 /min x w in : mass fraction of water in wet chips entering the dryer x w out : mass fraction of water in dried chips exiting the dryer C w air : concentration of water in air exiting the dryer = 3.7 g/ft 3 dry air Mass Balance on the Dryer Water Balance: Water in with wet chips + water in with dry air = water out with dried chips + water out with wet air ṁ wet chips · x w in + 0 = ṁ dry chips · x w out + Q dry air · C w air Starch (Chip Dry Matter) Balance Starch in with wet chips = Starch out with dried chips ṁ wet chips · (1 - x w in ) = ṁ dry chips · (1 - x w out ) Water Balance: 20 kg/min · 0.6 = ṁ dry chips · x w out + 1000 ft 3 /min · 3.7 g/ft 3 · 1 kg/1000 g Therefore, ṁ dry chips · x w out = 8.3 kg/min … (A) Starch Balance: 20 kg/min · (1 - 0.6) = ṁ dry chips · (1 - x w out ) kg/min Therefore, ṁ dry chips · (1 - x w out ) = 8 kg/min … (B) Wet Chips 20 kg/min Wet Air, 3.7 g water/ft 3 Dry Air, 1000 ft 3 /min Dry Air Condensate Dryer Condenser Coolant Dry Chips