Question

The temperature of the condenser in the absorption system shown in Figure 1 drops down to 34°C. Determine the concentration and enthalpy of the solution at point 4 and comment on the results obtained. Assume that all other temperatures are unchanged. (6 marks)

(5) Water vapor Dots Qe a Condenser Generator 95 °C 40 °C To L Heat exchanger t o Water vapor LiBr solution water vapor Absorber 30 °C Evaporator 10°C m = 0.65 kg/s Figure 1. A single-stage LiBr-water absorption system with a heat exchanger

Answer 1

# Solution # Mass flow rate of Libr - ma. mass flow rate of water = mw. spacific heat of Libr c = 0.59 kJ Spacific heat of water Cp - 4.18 KJ kegk Kgk total heat loss by systema total heat gain by a system hat Qc = Ag+ Qc (mat mw) C (30-1) + mwe(40-T) =(Mat Mwjc (95-7) & moq(10-T) 0.65 80.59 [95-T-30 +7] z mw x4.18 [407-10 +7] 0.65% 0.59*65 = mw x 4.18% 30 mw = 0.20 kg mar 0.65 – 0.20 see = 0.45 kg see

for Heat exchanger Heat loss by 3-2 = Heat gain by 1-2 mac (T3-T4) z mo e (T2 -T.) 0.45 (95 - Ty) = 0.65X(54-30) - Tu = 6oc Ans. Enthalpy at point is hi ma c Ty = 0.45X0.59*333 Tu= 273+60 = 333 K hu = 88 ki je Ans.