Question

A liquid (NOT WATER) is passed through a tube with a square cross section of 0.01m x 0.01m at a Reynolds number of 2,000. The liquid has a Prandtl number of 1, a thermal conductivity of 0.2 W/mK, and a heat capacity of 4,000 J/kg*K. Simultaneously, air is blown at a speed of 20 m/s and a temperature of 25°C over the outside of the tube such that the flow is perpendicular to one of the square sides. The liquid enters the tube at 200°C and exits at 50°C. Presume the tube wall is extremely thin. a. For what length of tube can the developing entrance length be ignored and the internal flow approximated as fully developed throughout ? b. If the tube is 20m long, what is the heat transfer between the two fluids ? c. If the tube is 20m long, what is the mass flow rate of the liquid through the tube ? Air: k = 0.03 W/mK, p = 1 kg/m?, v = 2e-5 méls, u = 2e-5 Pas, a =2.9e-5 m²/s, B= 30-3K-!, Pr = 0.7, Cp = 1,000 J/kgK,

Answer 1

K= 0.2 W/mk fage-1 liquid for liquid Air २००८ Re = 2000 Pr=1 orol ✓ Air À liquid 50% 256 Quoten F0101 expaused - 4000 J/kgle To determine the entrance length, Reynolds number loe 2000 As per given Hence it will be laminar flow So, the entrance length required to establish fully developed laminar flow is given by By using this relations Le 0007 Re - D where, D= (equivalent diameter) 4A {na & A=cs Area P p = Perimeter - = 4x 0.01X0.01 480.0] ooolm

Le Page=2 0.87Re D Le oool = 0.0FX 2000 Lez 1.4m Ary # By = Heat transfer first have to find out heat transfer coefficient for liquid relationship of laminar flow for constant using surface temperature of pipe Nu = 3.66 hip = 3.66 Klin 73-2 W/m²K hy- 3.66X0-2 oool # Heat transfer coefficient for air which is flowing through one of the plane surface of tube ine abcd first have to check the nature of How whether it is laminar or turbulant, for that have to find out Reynolds number (Re SVE IX 20X0.01 air 2x; 10000 < 28105 'Hence flow is laminar By lying relationship 0.664 CRebosch - u (Relair 0.33 Nu =

Page hol Kair = 0.664 x (ResouS6)0323 05 0.33 hoxoool 0:03 50-664x (1000) (0.7) ho 177.08 Wymak overall heat transfer co-efficient 1 1 1 hi t U no I 1 + 73-2 177.08 U = 51.79 W Work 7 34103 Pair 2 Tair Tmean Tinlet & Tout (where Tink] 2 3X103 (linele a 1x2 298+ Tout Tout lair = 25+273 = 298K 368.66k or 95.66 G Hence outlet temperature of aff is more than the outlet temperature of liquid, hence i. will use counter flow heat exchanger relations,

Page-li 2ooc (0,=104.39) 95-66 liquid soc Air O₂=25°c 25 LMTDC Logarithmic mean temperature difference) 0,- 104.34-25 In/oi In ( In/104034 O2 25 LMID =55.52% Heat transfer when length of the tube is zom g YA (LMTD) 51-79 x 20X0.01 X55.52 A = contandt A Surface area for a = 575.07W And Theat transfor ine. abcd Mass flow rate of the liquid for length zom. ca inlet (200-50) 575.07 =mliq X 40008 Imlig = 9.588104 Kys ©
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