Question

Cold water (Cp = 4180 J/kg · °C) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at 15°C at a rate of 0.25 kg/s and is heated to 45°C by hot water (Cp = 4190 J/kg · °C) that enters at 100°C at a rate of 3 kg/s. If the overall heat transfer coefficient is 950 W/m² · °C. determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the ε-NTU method.

Answer 1

Giren Data: Ce=4180 Thege To a : Ce = 4180 This Tc; = 150, meke 0.25 kg/see Thi= 100°c Tego = 450 , Man=3 kg/ see U= 950 m m (6 = 4190 They Poor controles Equating hear transfer save, 90-96. Me Spex C Too-Tai) = my' Conx ( Thi-Tho) 0.25% 4180x (45-15) = 3x4190x (100 - Tho) Tho=97.5°C Thi BIO a en - Heat transfer trough hear exchanger 9 = mc palTco-Tei) - My Spin C Tri-Tno) = 0.25% 4180x (45-15) = 31.35 kay

No to, effectiveness of his can be calculated as E = Tco-Tu = 45-15 = 0.353 1 Thi-To 100-15 Here, Cmin CE i Coronto Cominore z 0:25 4180 374190 MECE OFx 4480 Стую = 0.083 - Relation between att & & NTU for counter flow Hie is e = l-e-MTU (1-C) l-cie NTUCI-C) 01353 = l-e-NTU (1-0.083) 1- 0.083, e-HTUCI-0.08 3] 0.97 ēNTUx6.917 0.647 NTU=0.442 But NTV can be expressed og NTU Z UAS grox Coming 0.442 5 950x As 0.2874180 As 20.4862 m2