Question

heat exchanger , NTU method

2.6. Assume that in a condenser, there will be no subcooling, and condensate leaves the condenser at saturation temperature, T, Show that variation of the coolant temperature along the con- denser is given by 2.7. In a boiler (evaporator), the temperature of hot gases decreases from Ty to Th2, while boiling occurs at a constant temperature T. Obtain an expression, as in Problem 2.6, for the variation of hot- fluid temperature with the surface area. Show that Equation 2.46 is also applicable for C, > C, that is, C* = C/C, Obtain the expression for exchanger heat transfer effectiveness, e, for parallel flow given by Equation 2.47 2.8. 29.

Answer 1

when one fluid is under phase change then we can draw any type of configuration ie counter or parallel , both will give some results. Te both will ICI Now Qcondensate = na (AT) since, AT of Condensate is zero 80 (c) condensate - x - Chot = ca (say) cminimum (menimum - 80 Meat capacity Ratio = Cmaximum Heat Capacity Ratio = o=c Now Effectiveness = Actual Heat Tromsfer Maximum Heat transfer Тc— 1с1 me Cel Tc-TC1) mcce (Th- Tci) Tha Tou

Μο 'Go (ΤΟ UA (min) Her min = Cc Cc- cold fluid heat UA Capgerty NTU = Now as inė (1-с) NTU I e Es - (1-CONTU 1- Се @ Не съо, because Crex c Cheo :(1-0NTU N -NTU TO & E = Iné (1-0) MTU - Ox{ e(1-0)NTOS - I exp(-NTO) 80) TC-TI The Tal Tr-Tcy - t- expo (-UA) In- Tell

Strilanly cc=0 Thi t c= 0 The Effectiveness = min ahl Thi Tha) na Cah (The Tc) le Thi - The The To Chr specific heat of hot fluid = 0,80 C=0 UA NT- - (en) Cmin. i-expl- - exp(- NTU) a E = (for c=0) Th. The Th - The Thi - Tc