Question

(0.03m) At section the static pr-4temp. and Mach Number are (277K, 690 kpalabs) and 0.6) At section @ the terp is. 260k Draw on Tos diagrams for the flow between state 3) Air flow through an insulated duct with constant area and , showing both static and stagnation states. calculate, Mrope and friction force exerted by the air on the wall. Ans.( M=0.844, P2=475 kepa. F=2650N)

Answer 1

Pll Are 9 of Flowd A=0.03 mm (p= 1005 legk Y = 1.4 T2 = 260k Pstatic = P = Gookpan Istatic = Ti = 277k Mach Number MY0.8 duct beacure flow through Insulated from SHEEE hz + V2² hit v² (1) 2 at section i hisha = enthlpy Via V₂ = velocity at section o s Mia Vi = MI VERT VYRTI 1-4X 0.287X103 x 277 nia - 200.168 m/s 2 V From eq - w? - V₂2 ha. ha V? - V22 Cp (T2-T ) = Vg² - v,2 = 240(To - T2 ) 200.1082 = 2x1:005 ( 277-260) x103 272.46 m/s VERT Mach number at carit section 6 M₂ = 272.446 1.4 x 277 x 260 tha: 1/4

h8.0 =rul stagation temperature at section @ = (Tola (Tola = Tz [1+ you m?] 1-40! x0.84] - 260 [ It 2 (Tol = 296.95k 690 Pi RT 0.287 X 277 density at section section or Si = 8.679 lg 1 m3 lubs si tonp out be quere mass flow rate through Pa Vq Az ū x'd=&d [ og A, 7 A2=0.03 m] Vz S2 = 8.697 200.169 272.46 S2 = 6.376 kg/m3 - P2 static pressure at section P2 6. 376 x 0.297 X 260 P2 R Tz "P2 = 475.81 kpq 2/4

stagation presega at section - Puz Pon U oslav P2 [ ut erat ing] [winancomy] Poz - 1.5873 P2 Por 755.25 kpq - Friction forme exerted by Arron quet Consider a control volume -PIA P₂A-F = m (V₂_W) m= mass flow rate = SAV m: 9.6740.03 X 200.168 m= 52.117 kg ls. 690x10240.03 – 475.81*103 X0.03 --F = 52.117 ( 272.46-206-465) F = 2650.0 N Friction Forme. Force. 3/4

= Toq = Toz T My = 0:6 2cp 277k somatic 0.84 abok 1 s, sz S beaure flow is irrevesible due to Friction S₂-5 = change in coln ( 72 ) - Rin (1/2) entropy 1.005 xin (– 0.287 in ( in ( 475 47 ) So-s, = 0.17 ko lagi к 4/4