Question

Air at 0.8 bar and 127 Centers an adiabatic diffuser steadily at a rate of 6000 kg/ h and leaves at 1 bar. The velocity of the air stream is decreased from 230 m/st 0 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the ai rand (b) the exit area of the diffuser.

Answer 1

a) Exit temperature = 425.87 K

b) Exit area = 0.0679 metre square.

3. can use steady flow energy equation this diffuser The SFEE given as hit tg Z, &q=ha. dolo +9Zz+ Wer hyhz specific enthalpy ali out inlet and at intet Velocity of exit Elevation the inlet ZiZz and outlet above datum Qq specific heat transfer (heat transfer per kg). Wer transfer pen Work flow rate for steady - Mass E flow process. Absolute pressures at inlet Pr.P2 and excite

To, Ta Adosadoute Temperature out inlet and excit Now assume the following the adiabatic diffuser steady state process change potential energy i's zero, or negligible APE=0 7 -72=0. iii) Since adiabatic process, Since l's done, The SFEE equation for diffuser become hit ca 11 het ca Now CeTi and ha=ce Tz a'y can take. 1.005 KJ /kgk. Rair = 0.287 kJ/xgK.

Given that = 0.8x15 Pa. P = 0.8 ban T = lsic 400K. P2 = I bon los Pa. m 6ooo kg ln Gooo 3600 Kals = 1.67 kgls 230 mis ci 30 mis Now esait temperature T2. a) To find ca 2 = hat 2. old + - сета Sald C Сет. + 2 2 C2 u Сет+ 2. сета c²-c² Ti TR 2ce. 2303 – 30² 400 11 2 x 100 400 +25.87 T2. = 425.87K -

Esat Temperature 425.87K diffusor. b) Excit Area of the at inlet and = Density of air 3, S2 at intet AI, AR and of diffuser Area excit. Since Process Is Steady, inlet mass flow and frow rok same. S.A.C 32A2 (2. Az = m 82C2. Using ideal gas equation of state P RT S2 Pa los Ranx Tz 287x425. 87 32 = Tos 0.82 kg/m3 122224.69) A2 = 1.67 3202 1.67 24.6 = 0.0679 m² 0.92 x 30 Erit Area = 0.0670 m²