Biochemical Engineering & Thermodynamics (BT-507) PDF

Biochemical Engineering & Thermodynamics (BT-507) Objectives  Heat Transfer  Need for Heat Transfer  Methods for Heat Transfer  Heat Transfer Equipment  Design Equation for Heat transfer system Heat Transfer:  Heat transfer is an important unit operation in chem...

Biochemical Engineering & Thermodynamics (BT-507) Objectives  Heat Transfer  Need for Heat Transfer  Methods for Heat Transfer  Heat Transfer Equipment  Design Equation for Heat transfer system Heat Transfer:  Heat transfer is an important unit operation in chemical and bioprocess plants.  Heat transfer not necessarily means heating of fluid, it is the heat exchange between a hot & cold streams.  From the energy balance, we know the heating or cooling requirement for a reactor.  When the rate of heat transfer is known, we can calculate the surface area & other conditions for that purpose.  There are few possibilities: A. A counter current heat exchanger, in which both fluids flow in opposite directions without phase change B. A co-current heat exchanger, in which both fluids flow in the same directions without phase change C. A fluid heater by condensing vapor, such as steam, or a vapor condenser cooled by a fluid, such as water D. A fluid cooler by a boiling liquid. Need of Heat Transfer  Two applications of heat transfer are common in bioreactor operation. 1. Batch sterilization of liquid medium.  In this process, the fermenter vessel containing medium is heated using steam and held at the sterilization temperature for a period of time. cooling water is then used to bring the temperature back to normal operating conditions. 2. Heat transfer for temperature control during reactor operation.  Metabolic activity of cells generates a substantial amount of heat in fermenters; this heat must be removed to avoid rise in temperature. Most fermentations take place in the range 30-37 Forms of heat Exchange in Bioreactor For bioreactors with cooling coils Need of Heat Transfer  In any bioprocess industry, we need to transfer heat for different operations (like cooling, heating, vaporizing, or condensing) to or from various fluid streams in various equipment like condensers, water heaters, re-boilers, air heating or cooling devices etc., where heat exchanges between the two fluids.  In a chemical process industry, the heat exchanger is frequently used for such applications.  A heat exchanger is a device where two fluids streams come into thermal contact in order to transfer the heat from hot fluid to cold fluid stream. Cell growth obeys law of conservation of matter  Mechanism of Heat Transfer  There are mainly three methods for the transfer of heat A. Conductive heat transfer B. Convective heat transfer C. Radiative heat transfer Conduction  In most heat transfer equipment, heat is exchanged between fluid separated by a solid wall.  Heat transfer through wall occurs by conduction.  Fourier’s law is used to find the rate of conduction, Fourier’s law According  to this law “the rate of heat flux is directly proportional to the temperature gradient” Mathematically: Q/A dT/dy Q/A dT/dy *where k= thermal conductivity of solid Steady state conduction  At steady state, there is no accumulation or depletion of heat within wall. Therefore rate of heat flow Q must be same at each point in the wall. The Fourier's equation changes to  Q = kAT/dy let dy= B  Q = kAT/B B/kA= Rw  Q = T/Rw  Here Rw = to heat transfer Thermal Resistance in series (composite wall)  If the wall consist of several layers of different material  Consider a three layer wall as shown in fig.  Each layer would offer resistance to heat R1,R2,R3  Overall resistance is the sum of individual resistance  Rw =R1+R2+R3  The overall temperature(T) drop is the sum temperature drop across each layer T= T1+ T2 + T3 Design Equation for Heat transfer The  design equation is used by the engineers to design heat exchangers for a specific purpose, with required heat duty(Q) Q = UA Tm  The heating surface area is calculated as  A = Q/U Tm  Where Tm= log mean temperature difference Tm  Why do we need LMTD?? Home Assignment Design Equation for Heat transfer   The two fluids involved need to be identified  The heat capacity of each fluid is needed  The required initial and final temperatures for one of the fluids are needed  The design value of the initial temperature for the other fluid is needed  An initial estimate for the value of the Overall Heat Transfer Coefficient, U, is needed. LMTD  A liquid stream is cooled from 70 to 32 in a double-pipe heat exchanger. Fluid flowing countercurrently with this stream is heated from 20 to 44 Calculate the log-mean temperature difference. Problem Q. Estimate the heat exchanger area needed to cool 55,000 lb/hr of a light oil (specific heat = 0.74 Btu/lb.°F) from 190°F to 140°F using cooling water that is available at 50°F. The cooling water can be allowed to heat to 90°F. An initial estimate of the Overall Heat Transfer Coefficient is 120 Btu/hr.ft².°F. Also estimate the required mass flow rate of cooling water. Problem  A shell and tube heat exchanger is to be used for the light oil cooling described in last example. How many tubes of 3 inch diameter and 10 ft length should be used? Thank You   Biochemical Engineering & Thermodynamics (BT-507) Objectives  Individual Heat Transfer coefficient  Overall Heat Transfer coefficient  Fouling factor  Heat Transfer Equipment  Problems on Heat transfer system Problem A  liquid stream is cooled from 70 to 32 in a double-pipe heat exchanger. Fluid flowing counter-currently with this stream is heated from 20 to 44 Calculate the log-mean temperature difference. Q. Estimate the heat exchanger area needed to cool 55,000 lb/hr of a light oil (specific heat = 0.74 Btu/lb.°F) from 190°F to 140°F using cooling water that is available at 50°F. The cooling water can be allowed to heat to 90°F. An initial estimate of the Overall Heat Transfer Coefficient is 120 Btu/hr.ft².°F. Also estimate the required mass flow rate of cooling water. Convection & Thermal Boundary layer  Convective heat transfer in bulk fluid is rapid & is directly linked to mixing & turbulence.  The figure shows the heat transfer at any point on the pipe wall of heat exchanger.  Hot & cold fluid flow on either side of wall.  We assume that both fluid are in turbulent Flow.  When liquid contact a solid, a fluid boundary Layer develops as a result of viscous drag.  A thin layer near the wall is observed which is Called Stagnant or “thermal boundary layer”  In turbulent part of fluid, rapidly moving eddies Transfer heat quickly. Convection & Thermal Boundary layer  The velocity of liquid near the wall is very low. The temperature in this layer almost constant  Most of the resistance to heat transfer is contained in this film because heat transfer in this layer is mainly by conduction rather than convection because of low velocity of fluid.  As we move away from the wall, the temperature reaches to the bulk fluid temperature when the velocity reaches the bulk fluid velocity. Individual heat transfer coefficient  In the heat exchange through a solid wall, three major resistances are in series; the hot-fluid film resistance at the wall, resistance due to the wall itself, and the cold-fluid film resistance  Equation for the rate of conduction through the wall is: Q= UA  Rate of heat transfer through each thermal boundary layer is given by Q= A  Unit of heat transfer coefficient is W.m -2K-1  Resistance to heat transfer on either side of pipe is given by:  Rh=1/hhA & Rc=1/hcA Overall Heat Transfer Coefficient  For the heat transfer rate calculation, we need the value of , which is the difference between temperatures at wall (hot & cold side). For the sake accuracy & easiness we consider the bulk temperature on both sides of wall. This problem is overcome by introducing overall heat transfer coefficient ’ for the total heat-flow process through both fluids and the wall.  is defined by the equation: Q = UA T  The overall resistance then becomes RT =1/UA Total Resistance is the sum of all resistance RT = Rh + Rw +RC Overall Heat Transfer Coefficient  = When fluids are separated by a flat wall then surface area of for heat transfer through each boundary layer & wall is same, cancelling A both sides of the equation we have: = Thank You 