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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 coecient) (Weisman 1 , Schrock and Grossman 2 ) 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 R co htc (3) T ciT co = q 2 k c ln R co R ci (4) T foT ci = q 2 R g htc g = q ( R ci + R fo ) htc g (5) T maxT fo = q 4 k f (6)...
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 Spring '11
 Schubring

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