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Transcript
Design of a Tubular Heat Exchanger Design of a Tubular Heat Exchanger We will use the following assumptions: 1. Heat transfer is under steady-state conditions. 2. The overall heat-transfer coefficient is constant throughout the length of pipe. 3. There is no axial conduction of heat in the metal pipe. 4. The heat exchanger is well insulated. Change in heat energy in a fluid stream, if its temperature changes from T1 to T2, is expressed as: = mass flow rate of a fluid (kg/s), where m cp = specific heat of a fluid (kJ/kgC), temperature change of a fluid is from some inlet temperature T1 to an exit temperature T2. Hot fluid, H, enters the heat exchanger at location (1) and it flows through the inner pipe, exiting at location (2). Its temperature decreases from TH, inlet to TH,exit . Fluid C, is a cold fluid that enters the annular space between the outer and inner pipes of the tubular heat exchanger at location (1) and exits at location (2). Its temperature increases from TC,inlet to TC,exit. Conducting an energy balance, the rate of heat transfer between the fluids is: q m H c pH (TH ,inlet TH ,exit ) m C c pC (TC ,exit TC ,inlet ) where cpH is the specific heat of the hot fluid (kJ/kgºC), cpC is the specific heat of the cold fluid (kJ/kgºC), m H is the mass flow rate of the hot fluid (kg/s), m C is the mass flow rate of the cold fluid (kg/s). The preceding equation has limited application 1 Fluid c Insulation Fluid h Fluid h Fluid c 2 1 T Th, inlet dTh dq Th, exit Tc, exit dTc Tc, inlet A Consider a thin slice of the heat exchanger, the rate of heat transfer, dq, from fluid H to fluid C may be expressed as: where T is the temperature difference between fluid H and fluid C. The temperature difference, T, between the two fluids H and C is For a small differential ring element as shown in Figure, using energy balance for the hot stream H and, for cold stream C 2 dq m C c pC dTC In Eq , dTH is a negative quantity, therefore we added a negative sign to obtain positive value for dq. Solving for dTH and dTC, we obtain dTH dq m H c pH and dTC dq m C c pC Then subtracting 1 1 dTH dTC d (TH TC ) dq m c H pH mC c pC or 1 d (TH TC ) 1 U (TH TC ) m H c pH m C c pC dA Integrating ln 1 1 UA TC ,inlet ) m H c pH m C c pC (TH ,exit TC ,exit ) (TH ,inlet Noting that we get ln 1 T2 1 UA T1 m H c pH m C c pC Substituting T T T T T ln 2 UA H ,inlet H ,exit C ,exit C ,inlet q q T1 Rearranging terms 3 T UA ln 2 (TH ,inlet TC ,inlet ) (TH ,exit TC ,exit ) q T1 Rearranging terms, q UA( Tlm ) where Tlm T2 T1 T ln 2 T1 Tlm is called the log mean temperature difference. A Counter-Current Heat Exchanger Fluid c Insulation Fluid h Fluid h Fluid c 2 1 T Th, inlet Tc, exit dq Th, exit Tc, inlet A 4 Class Problem Calculate the log mean temperature difference for a heat exchanger with the following data, the temperature of hot stream decreases from 150C to 50C, while the temperature of cold stream increases from 40C to 80C. Consider another case when temperature of hot stream decreases from 150C to 50C and cold stream temperature increases from 40C to 45C. 5