Design Problem Superheater for a Polymer Solution 1. Task 1. Task 2. Task 3. Data. Note:

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1 Design Problem Superheater for a Polymer Solution 1 Ethylene-propylene rubber (EP) is polymerized in a solvent. The product of the reaction is a 6% (by weight) solution of EP in perchloroethylene. The polymer is recovered as "crumbs" from a drum dryer. Production capacity is limited by the capacity of the dryer. It is believed that concentrating the feed to the dryer will provide the sufficient increase in capacity. Your problem is to specify the design of a superheater that would heat the solution sufficiently so that upon flashing to atmospheric pressure the solution is concentrated to at least 12%. Task 1 Determine all the physical properties to use in this problem heat capacities, densities, chemical activity as a function of concentration Task 2 Write a computer program or spreadsheet to calculate the prescribed pressure and temperature required at the end of the superheater Task 3 Write a computer program or spreadsheet to calculate the size of the heat exchanger. Data The production of rubber is 3lbs./hr or.379 kg/hr The feed temperature is 35 C Note: This is an open-ended problem with insufficient information given for you to solve the problem. You have to find the data and make a reasonable set of assumptions about the fluid to be used to heat the rubber. 1 NOTE This is an open ended, somewhat poorly defined, problem. ChE 333 1

2 Heat Transfer Analysis in Pipe Flow Consider the problem of flow in a long pipe of circular cross-section. The inside diameter of the pipe is D and is maintained at a constant temperature T o. The fluid flow through the pipe at a flow rate, Q. The goal is to describe the average temperature as a function of distance in the pipe. MODEL Energy balance T ρc p u z z = k 1 r Momentum balance r T = p z + u 1 r Initial and Boundary conditions at z = T = Ti for all r at r = T = u z = at r = T = T ; u z = + 2 T z 2 r u z ChE 333 2

3 Adimensionalization and Scaling The convective heat transfer equation T ρc p u z z = k 1 r r T can be scaled using a set of reference parameters + 2 T z 2 The equation is θ = T T T ; ζ = z L ; η = r ; v = u z v ρc p v v T L θ ζ = k T 2 1 η η η θ η + 2 L 2 2 θ ζ 2 which after some multiplication becomes v θ ζ = kl ρc p v 2 1 η η η θ η + 2 L 2 2 θ ζ 2 It follows that if /L is amall then only the first term in the Laplacian is important and the equation can be written ignoring axial heat conduction. ChE 333 3

4 The dimensional form of the equation is: T ρc p u z z = k 1 r r T along with the boundary and initial conditions T = T1 at z = T = T at r = T = at r = We could solve for the temperature profile in detail, but it might be better if we seek a solution for the average temperature by integrating over the crossection at position z. ρ T C p u z z r dr = k If the velocity field is independent of axial position, we can write r T dr ρc p z u z Tr dr = k r T = k T Examine an average temperature <T>, the mixing cup temperature, the mean temprature of the fluid that leaves cross section at z = z T = u z T 2πr dr u z 2πr dr = u z T 2πr dr Q ChE 333 4

5 This last equation can be re-written as u ztr dr = Q 2π T The integrated energy equation is : ρc p d dz Q 2π T = kt The material balance teaches us that So that we can write: ρq = w = constant d T dz = k2πt ecall that we can define a heat transfer coefficient by an expression such as : k T =h[ T T ] and the equation for the mixing cup temperature is : d T dz = πdh[ T T ] ChE 333 5

6 This can be prepared for integration 2 : d( T T) T T = πdh dz The relation is integrated readily if h is not a function of z 3 T T = exp πdhz T = exp 4St z D The definition of the Stanton Number is : where Pe = e Pr St = h ρc p U = Nu epr = Nu Pe T 2 T = exp 4St L T D πdlh = exp 2 From here on,,we drop the brackets on T<>T> for convenience and the experinced players benefit!!! 3 We can still use the same relation for St = h/ρc p U where h= 1 L L hdz ChE 333 6

7 Other Ways of Defining and using Heat Transfer Coefficients Q H = ha( T) Questions What is Q H and what is T We know that an energy balance contains: We can rewrite as QH = +wcp(t1 T2) T = + ( T 2) πdlh After integrating we find that πdlh = ln T 2 T T It follows that I can write Q H = ha( T) if the temperature difference is T T = T 2 ln T 2 T T = T 2 ln T T 2 T ( ) ( T 2 T ) = T ln T T 2 T = T ln ChE 333 7