Question

Hot water with the mass flow rate of 3 kg/s and temperature of 90 C enters a 2 inch copper pipe with the 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 water and the rate of heat loss from the pipe.

Answer 1

so 0.0508m Mass flow hate of water, h = 3kg / Sec. Inlet temprature of water, Ti = 90°c Diometer of pipe, D = 2 in Length of pipe, to zm Temprature of aur, Too Velocity of air, to a 5m/sec Now, Mean film temprature, Tja Tit Too 2 got12 45.5°C 2 989.6 Kg/m3 - 5.8068x10-7 m²/sec 0.63899 W/m.k 3.6570 So Thermophysical properties of water at 45.50 Density, p = Kinematic Vescosity Thermal Conductivity, K = Prondtl dember, fra to, la velocity of water, der = Mass flow rate Density X 989.6% (0.05083 Two 104964 m/sec X Area

Vo.D Reynoldy Number, (Re) Rea (1.4964) (0.0508) 5.8068X107) Rez 130910.51 > 4000 So flow of water is TURBULENT so for turbulent flow Neusselt alumber, Neu= ho D = 0.023 (Re Jo 8 (fr) 13 x (Bu) Y2 0.0508 5505.14 hoz K X 0.023x (Re) 0.8 he 0.63899 x 0.023 x (130910.51908 (36570) X W/m²K So, Imere heat tromsfor Coefficientis 5505.14 w/m2 K By water is Q = mx 6 (Ti-To) 3 x 4.0665x103) (90° 12199.5 (90 - TO Heat loss w

9150 Date. AIR Now, FOR Thermophysical properties of air at joc. is Density, Co- 1.289 Kg/m2 Kinematic Viscosity, da Thermal Conductivity, 1.341810-S m²/sec K = 0.02415 w/mik 0.7195 Prondtl Number Da 1.341810-S 74000 So flow of Now, Reynolds alumbert, (Re) - V.A (Re) - 5x0.0508 (Re) = 18941.08 AiR is TURBULENT Now, for turbulent flow Nusselt Number, Nu= hoo 0.023 x(fe)".0 x (Polo no = KX 0.023 x (Rejo.8 x (Por) 0.33 = 0.02415 X 0.023x (18941.08)98 (0.7195)0:33 ho W/m²K Duter Heat trenifer Coefficient 25:315 wparik 10.33 ho 0.0508 25.913

iso Date. fo, Overall meat trmsfer Coefficient is tl 1 + (5505.14) (25.913) 25.791 W/m²K Son Heat transfer between aid and water U A On न withere, One 2 Arithmatic Meon temprature Difference Om² (Ti-To) + (To-Too) Om a 90-1)+(To-1) Om x (44 +0.5 To) 2 2 1212 so, Q UA, Om 2 25.791 x Tx(0.0508)X 3 (44+0.57.) 120341 (44+0.5 To 2 Q

Heat loss by water 12199.5 (go-To) Heat tronsfor between 12.341(44+0.57) air and water 988.53 (go-To) 44+ 0.5 To 88967.7 988.53 To = 44+ 0.5 To 88923.7 = 989.03 To To = 89.91°C Qutlet temperature of water Ử 89.91°C Loss also, Heat loss from the pipe is eragird 12199.5 (90.-T.) 12199.5 ( 90-89.91) 1097.955 wote walts so, Heat loss by the pipe is 1097955 watts Loss