Question

Q5 A process fluid is flowing with a velocity of 4 m/s in a steel pipe having a diameter of 187 mm and a length of 1 m and maintaining the pipe surface temperature of 96.C. The fluid is entering the pipe at a temperature of 25 C and leaving with a temperature of 76C, take Prandtl number as 12. The density and viscosity of the fluid are 1210 kg/m?, 1 x10Ns/m respectively and its thermal conductivity is 5x102 w/mk. Use the relations Nu = 0.0214(Re 0.8 Pr0.4) and Nu = and (1) Determine the heat transfer coefficient of the fluid; (ii) Determine the heat transfer rate; (iii) Analyze the effect on heat transfer rate If the density of the fluid is increased or decreased; (iv) Analyze the effect of the temperature of the steel pipe if its exposed to a surrounding with higher temperature than the pipe surface.

Answer 1

NS/ 1 given, Velocity of fluid, ve 4 mois sol: Re-evda 12108 4x 0.187 9,05,080 dia of Steel pipe, d = 0.167 m 1x103 length of pipe , l = Surface temperature, Ts = 96°C Nu 0.0214 (9,05,080) 0.8 1204] inlet fluid temperature, Ti-25c Outlet fluid temperature, To: 760 = 3368.5 Prandth number, Pi = 12 density of fluid, e = 1210 kg/m3 Nu-hl -> Overall heat transfer coefficient, K Viscosity of fluid, 4= 1x103 h = Nuk= 33685 x 5x102 = 168.42 wink thermal Conductivity, k= 5x102 Winok NU=0.0214 Creo.8 pxl.4) (11) heat transfex rate, a = ha (Is-Tm). A + Surface area A- AD1 = x 0.187% 1 Nu=hl a: 168.42x0.587(96-50.5) .. A=0.587 m² Tm- Bulk mean temperature a - 44.98.25 W Tm= T; + 10 = 50.5" (10) 16 density increases. Reynold's number increases. (Pare). there fore, Nu increases (Nudre). 7 h increases. Chanu). hence, heat transfer rate increases, (Qxh).

(cv) According to Zeroth law of thermodynamics, temperature flours from high temperature to low temperature, till it reaches equilibrium. Then, if Surface temperature is less than Surrounding temperature, temperature of Steel pipe raises till it reaches equilibrium with surroundings.