Thermodynamics

Chapter 5 | 235Substituting, we get0 1200 m/s22h 2 283.14 kJ/kg a1 kJ/kg2 1000 m 2 /s b 2 303.14 kJ/kgFrom Table A–17, the temperature corresponding to this enthalpy value isT 2 303 KDiscussion This result shows that the temperature of the air increases byabout 20°C as it is slowed down in the diffuser. The temperature rise of theair is mainly due to the conversion of kinetic energy to internal energy.EXAMPLE 5–5Acceleration of Steam in a NozzleSteam at 250 psia and 700°F steadily enters a nozzle whose inlet area is0.2 ft 2 . The mass flow rate of steam through the nozzle is 10 lbm/s. Steamleaves the nozzle at 200 psia with a velocity of 900 ft/s. Heat losses from thenozzle per unit mass of the steam are estimated to be 1.2 Btu/lbm. Determine(a) the inlet velocity and (b) the exit temperature of the steam.Solution Steam enters a nozzle steadily at a specified flow rate and velocity.The inlet velocity of steam and the exit temperature are to be determined.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 There are no workinteractions. 3 The potential energy change is zero, pe 0.Analysis We take the nozzle as the system (Fig. 5–26A). This is a controlvolume since mass crosses the system boundary during the process. Weobserve that there is only one inlet and one exit and thus ṁ 1 ṁ 2 ṁ.(a) The specific volume and enthalpy of steam at the nozzle inlet areThen,P 1 250 psiaf v 1 2.6883 ft 3 /lbm1Table A–6E2T 1 700°F h 1 1371.4 Btu/lbmm # 1 v 1V 1 A 1P 1 = 250 psiaT 1 = 700°FA 1 = 0.2 ft 2STEAMm = 10 lbm/sq out = 1.2 Btu/lbmFIGURE 5–26ASchematic for Example 5–5.P 2 = 200 psiaV 2 = 900 ft/s(b) Under stated assumptions and observations, the energy balance for thissteady-flow system can be expressed in the rate form asE # in E # (steady)out dE system >dt ¡0 0⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎪⎬⎪⎪⎭Rate of net energy transferby heat, work, and massm # a h 1 V 1 210 lbm>s E # in E # outV 1 134.4 ft/sRate of change in internal, kinetic,potential, etc., energies2 b Q# out m # a h 2 V 2 212.6883 ft 3 >lbm 1V 1210.2 ft 2 22 b1since W# 0, and ¢pe 02