Question

Hot water with a mass flow rate of 3 kg/s and a temperature of 90 degrees Celsius enters a 2-inch copper pipe with a length of 3 m. The pipe is exposed to cold air at the constant temperature of 1C and wind velocity is 5 m/s. Find the outlet temperature of the water and the rate of heat loss from the pipe.

Answer 1

Sclution :- Guiven dala mass Flow ouale = 3kgls temperature 90°C Dia of pipe Rin wind velocity 5m IS man flow rale of water = 3kgls air temperature 10 Now we will first find the external convechon heat transfer coelbicnt hout= ? .: V = velocila dyerkral flow Reynold's numbers Revo =smis De Dudley dia pipezain Res 5X 0.0508 = 0.0508m 13.37e-6 1> (18997.956 Now the appropriate alusselt alember from heat transfer data Nu cx Crem) x Cpx 0.333 ) Na = 0.193 x C18927756048 ) x .: C= 0.193 mc 0.618 C0-4340-335) Re- 18997-756 =76.82 Pr - Prand Hakember for airat i = 0-736 Here We know Nu= hout 76-82– hout X0.0508 0.02371 hout = 35.85 w/mk hcoat)? D- Outer diad pipe=2 in -0.0508m Now we are Find the internal convechve heat transter hin= ?

Reynold's number = Rez 618 3 V- velocity of Mlernal flow PA 965.3 X 0.00 2027 = 1.533mls I DV - 1 x 0.0508? 0.002029 mV :.8- dennly twocher 90°C = 2053 kg/m3 D:lin 0.0508m 1- Kinematic viscosity of water at °C – 3.2582-7 MNK ke- up 19 1.5333 0-0508 3.256e-? 239057.602 0-3) Mu= 0.623 x C Re08) X (po 0-3 Re=Q39057.602 Nu= 0.023x (239057. 601.0-5) (203) 568-6 we know Nu= hind k 568.65 hinxo :0508 0.6613 him = 7402.21 wolmul Now there are three thermal resistanced present • Interncel heat Convecton CR1) Conduchan through walls. External Heat convechon ce ) . Ri = I hout A 2. hcout) = 3585 to (nuk

A= area of heat transfer TTDL = 7 X 0.0508 X3 0.44988m? 2. D-0.0508m L= length 4 pipe 3m Thermal Resistance - 35.85 70.4788 0.05825kw Thermal resistance R2= 7402:21X0.4788 = 0.00028221W The total resistance : Rtotal ehitle = 0-05825+ 0.0002822 0.05854 (o low the total heat tranferred in steady state mycy (Tin-Tout J = Thean - Taid R total 3X4167X (90 - Tout )- 90+ Tout :. T(mean) 2 mean temp of water - Tin trout 0.08854 Tout = 89.410 finally The alm The amout of heat transfer can be Teind=inlet temp of water c- Specific heal- of water 418931 | gole Q 90489.41 Tmean - Tair Rtotal Q =1515.4426 2 0.05854

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