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Unformatted text preview: o on left there is an initial velocity, specific volume, enthalpy o at center flow is deflected by a stopper with increasing area as you move up o on the right is a final velocity with a new specific volume, and new enthalpy o If v 1 = v 2 if incompressible Diameter is equivalent at either side V 1 = V 2 Δke Assume Qdot Assume Wdot Assume Δpe ~ 0 • 0 = mdot(Δh) Δh = 0 o h 2 = h 1 o state 2 = (h 2 – h f )/h fg Chapter 5 #195E o 15psia, 20ºF, 10 ft 3 /s o 100psia, Wdot = -45 hp o Find mdot, T 2 o Assume adiabtatic No mention of heat o Need saturation tables and superheat tables o mdot = ρAV = ρV o –wdot = mdot(Δh) 45hp(conversion) = mdot(h 2 – h 1 ) solve for h 2...
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This note was uploaded on 04/07/2008 for the course MECH 213 taught by Professor Knisley during the Fall '08 term at Bucknell.
- Fall '08