This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: the fact that q is usually given. Must consider adjustments to htc due to rod bundle geometry. Example Problem I P = 14.608 MPa I Geometry: P = 12 mm, D =10 mm, R fo = 4.6 mm, R ci = 4.7 mm I htc g = 5000 W m2 K1 I Equation for htc (heat transfer coeﬃcient) (Weisman 1 φ , Schrock and Grossman 2 φ ) I k c = 20 W m1 K1 and k f = 3 W m1 K1 I q = 25 kW/m I ˙ m = 0 . 15 kg/s (per channel) I T 1 = 320 ◦ C , T 2 = T sat = 340 x 2 = 0 . 2 Equations for Solution T coT m = q 2 π R co htc (3) T ciT co = q 2 π k c ln ± R co R ci ² (4) T foT ci = q 2 π R g htc g = q π ( R ci + R fo ) htc g (5) T maxT fo = q 4 π k f (6)...
View
Full Document
 Spring '11
 Schubring
 Convection, Heat, convective heat transfer, HTC, Rco htc

Click to edit the document details