Question

The condenser of a large steam power plant is a heat exchanger in which steam is condensed to liquid water. Assume the condenser to be a shell-and-tube heat exchanger consisting of a single shell and 30,000 tubes, each executing two passes (see figure below). The tubes are of thin wall construction with D=25mm, and steam condenses on their outer surface with an associated convection coefficient ho=11,000W/ m K The heat transfer rate that must be effected by the heat exchanger is q-2x10'w, and this is accomplished by passing cooling water through the tubes at a rate of 3x10*kg/s (the flow rate per tube is therefore 1kg/s). The water enters at 20°C, while the steam condenses at 50°C. What is the temperature of the cooling water emerging from the condenser? What is the required tube length L per pass (NTU method will be used)? We can neglect the thermal resistance of tube material and fouling, as well as the heat transfer between the heat exchanger and surroundings. Properties and data: Cp=4179J/kg.K k=0.613W/m.K Pr-5.83 Water density=997kg/m u=855x10 "N.S/m2 Nu = 0.023Re, SP 0.4 Condenser: Ch 00 NTU(for C-Cmin Cmax=0)=-In(1-5) Steam 50°C 4 - 2 x 10°W Tube data The -50°C Water = 3 x 104 kg's ( N = 30,000 D=0.025 m h = 11,000 W/m2.K L = length/pass 1 kg's T = 20°C

Answer 1

- For tomp of cooling water Exit using Relation a A7 CAT Ans coellini 'Ster q= mcpot Exh tomp is Tz Ti= 200 q= 24109 w given in = mass flow fate of cooling water = 34104 Kgls Ep = 41799 lege 2709 – 3810444199(T3-20) 1 T2 = 35.95°c Exit tomp of cooling water Heat Tronstable coefficient? Steam: Stream side he=11000 W/m²k - walen 0 Now we have to calculate heat fransfer coefficient on water side ho. Renold No Re= fup = 997 xvx25/1000 855 810-6 in = fxAXV = 1=997XX (25) ²xV . [V=2.043 m/s velocity of cooling water Re=59566-62 Nu = 0.0234 (Rejor por jou Nu = 0.0 23x (59586-62.00 (5.83)04 I NU = 307.5711

using Relation Nu hwla Les D'amotes = 0.025m k 307.57 = hw x 0.0 25 0613 chw = 7542.55 W/m2K led overall heat Gons for coeficient = u het - Neglecting conductive Resistance of Pipe s chw ū= Itu a nd Tai = Ao loncelui Nol Tu= 4474047 W/m²K 11000 2542:55 Efectivoness = min oT = 35.95-20 Emin (Thin Tein) 5o - 20 TE=0.5316 using Relation NTU= - In (IFE) NTU = - ln (1-0.5316) NTU = 0.758% NTU = UA Emin No of Pers Oox L x 2 0.7586 = 4474.47X (30,000)x(x 0.XL) X 2 - 38104 x 4197 L = 4.529 m) = Length of Tube Per Pass Ang
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