Hw-03_Solution - Michigan State University DEPARTMENT OF...

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v, C salt =0 v, C salt Michigan State University DEPARTMENT OF CHEMICAL ENGINEERING AND MATERIALS SCIENCE CHE 210: Modeling and Analysis of Transport Phenomena Fall, 2007 Homework Assignment #3 Solution 1. A tank contains brine consisting of 75 lb. of salt dissolved in 100 gallons of water. Pure water runs into the tank at a rate of 3 gallons per minute and the mixture runs out at the same rate. The concentration of brine in the tank is kept uniform at all times by stirring. How much salt (in lb.) would remain in the tank at the end of 1.5 hours? [ Note : you may assume that the density of the brine is constant throughout the process.] Solution : The diagram for this problem may be presented as shown on the right. We will define “salt” as the system. Since there is no input of salt into the system, the balance is accumulation= -out : () out dCV dM vC dt dt == Integrating, we get: 0 0 / 0 0 0 0 ln ln t Ct C vt V C C dC v vt C vt dt C C C e CV V =− = ∫∫ The concentration at 1.5 h is: ( ) / 0 31 . 56 0 lb lb 0.75 exp 0.0504 gal 100 gal gal vt V CC e −⋅ ⎛⎞ ==⋅ = ⎜⎟ ⎝⎠ We can use this result to calculate the mass of salt in the tank at 1.5 h: 1.5 lb 0.0504 100gal=5.04lb gal salt h m =⋅ 2. Consider a well-mixed tank initially half filled with fresh water. At time t=0 , a salt solution of concentration C o is continuously added to the tank at a volumetric flow rate of v o . Develop an expression for the concentration of salt in the tank as a function of time for the cases below: a. The outlet flow rate v out is equal to v o . b. The outlet flow rate is zero until the tank is filled after which v is equal to v o . Data : Volume of tank, V=100 L; inlet concentration of salt, C o =2 g-mol/L; inlet volumetric flow rate, v o = 10 L/min.
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2 Solution : a) The mass balance is acc=in-out . Since the volume is constant for this case, we get () o dc v cc dt V =− We can separate the variables, and directly integrate (noting that C 0 =0 in this case): [] 0 00 ln ln ln ct c oo o o dc v vt vt d t c V V V = −− = −= ∫∫ This equation can be simplified to: // ln 1 vt V vt V o o o vt c e c c e cV ⎛⎞ − = = ⎜⎟ ⎝⎠ b) This problem has two subparts – i) prior to the tank filling up (no output stream), and ii) after the tank fills up (input and output streams are equal) i) Prior to the tank filling up : the balance on the salt is accumulation=input : dCV dC dV vC V C dt dt dt =⇒ + = Note that since both the concentration and volume are changing, we cannot separate this equation. Therefore, we need to obtain a relationship for V to integrate the equation. This can be done by a balance on the brine: accumulation=input . If we assume that the density remains constant during the filling process, then we can write: 0 0 0.5 0 0.5 f Vt f V dV dV vv d V v d t V v t V dt dt ρ =⇒= = = + When we incorporate this result into the salt balance equation, we get: 0 0 0 0.5 0 f dC dV dC V C vt V v C C dt dt dt += + + = From this we get: 0 : at 0, 0 0.5 f vd t dC IC t C CC v t V == = −+ Integration gives: 0 0 0 0.5 ln ln 0.5 ln ln 0.5 t C f f f V CC v t V Cv t V + = +
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This note was uploaded on 05/21/2008 for the course CHE 210 taught by Professor Robertofoli during the Fall '07 term at Michigan State University.

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Hw-03_Solution - Michigan State University DEPARTMENT OF...

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