Question

2. (10 points) A pharmaceutical fluid is sterilized by heating it from 25 to 75 °C as it moves at 0.2 m/s through a straight thin-walled stainless steel tube of 12.7-mm diameter that is 10 m long. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the straight tube. The pharmaceutical fluid properties may be approximated as: p = 1000 kg/mº, cp = 4 kJ/kg-K, u = 2.102 kg/s-m, k = 1 W/mK, and Pr = 10. If the fluid enters the tube with a fully developed velocity profile and a uniform temperature profile, determine: a. The mass flow rate, in kg/s. b. The heat transfer rate, in W. C. The heat flux, in W/m². d. The Reynolds number e. The hydrodynamic entry length, in m. f. The thermal entry length, in m. g. The Graetz number at the exit. h. The average Nusselt number at the exit. i. The average convection heat transfer coefficient at the exit, in W/m².K. 1. The tube surface temperature at the exit, in °C. -L-10m Tm,o-75°C U.mp=0.2 m/s Tmj-25°C D-12.7mm 195

Answer 1

Greven - Tma= 25°C , Tmio= 75°C V=0.2mis D = 12.7 mm= 0.0127 m L= 10m 8= 1000 Kg/m3 ; (p=4 kJ/kgk ; M = 2x10 3 kg/m8 K= 1 W/mK i fr=10 (p= 4000 J/K9K K L alom > Tm,o Tmi Um=0.2m18 Pocon (a) The mass flow rate :- (m) m= SAV = fx (27 D?) x v m = 1000 x 77 x (02.01273² x 0.2 m= 0.02533 Kg 18 (le) Heat tor. rate :-(2) À = Heat required to change the temperature of fluid ä - im Cp (Tmio - Tm, :) Q = 0.02533 (4000) ( 75-25) CS Scanned with CamScanner Q 5066 W

(c) Heat flux (9): - Apply Energy balance → Total heat to. by coil = Heat gained by fluid 9.A q. (ADL) = 5066 q 5666 12697.306 W/m 2 TX 0.0127 10 q=12,697.306 W/m2 (d) Reynolds Number PVD Re= M Re= 1000x0 2x0.0127 2x103 1270 Re=1270 (e) Hydrodynamic Entrance length :-(Le) Since IRe < 2300 So flow is laminan. For Laminan flow Lesh=0.05 Re D Le, h= 0.05X1270 X 0.0127 Le, h= 0.80 645m CS Scanned with CamScanner Leh=0.80645 m

(f) Thermal Entry length (Le,t) For laminan flow of Let ~ 0.05 Rep. Por Let = 0.05 Red. Por xD L2,4 = 0.05x1270x se to X 0.0127 ['G Pr=10 Given Let 8.0 645 m I Grautz Number at the exit :- 2 Re Por Gz = 0.0127 10 X1270X10 = 16.129 Gz = 16.129 h) The average Nusselt Number at the esit: - Since flow is fully developed and laminan so for uniform Heat flun f Nu= constant = 4.36 Nu=4.36 (2) Average convective Heat for. coefficient at the exit - (hl hD Nu = = 4.36 CS Scanned with CamScanner K

tip - 4.36 4.36 x h = 4.36XK D 0.0127 h = 343.3 W/m²K (2) The average surface temperature at the exit :- q=h (Ts, e-To) 12697-306 = 343.3 (Tse-75) Tse = 111-98°C CS Scanned with CamScanner