Question

An equimolar mixture of argon and air is to be used as the working fluid in a closed-loop gas-turbine cycle. The mixture enters the turbine at 3 MPa and 1400 K and expands isentropically to a pressure of 200 kPa. Determine the work output of the turbine per unit mass of the mixture. З MPа 1400 K Ar - Air Turbine 200 kPa Select one: A. 900 kJ/kg B. 450 kJ/kg C.603 kJ/kg D. 867 kJ/kg E. 720 kJ/kg

Answer 1

0043 MP, 1400K. Solutions Given data: Equimolar mixture of Argon & Air. (Let n kmoles of each are there). P, = 3MPa, T = 1400K , P₂ = 200 kPa. Tsentropić Expansion. (2) 7 200 kPa First of all, we have to find out the properties of the mixture like C, G & r. Take the values of specific heats for Argon & Air as o (Gair = 1.005 Kolugor , GAgon = 0.5203 Koliek, Gair = 0.715 Kolkg .k, CC Arron = 0.312.2. Xo/kgk. (Molat mass) air = 18-37 Kg Jamol., Ömeled Mous)Aroma 89-948 Xy ]Kmet. How mass of each gas is given by, (n)air = No. of modes x (19clore Mass air. Mair = 5 X 28-37 kg] f Mayon = 5x39-948 Kg Now, on maiz X (G) air + mayonx (Pagon. Micture mair + mayon. nx 28.97X 1.005 t nx39.948 X 0.5203 HX98-97 + bx 39.948. → ) mixture = 0.724 kJ/kg-k.

( mitte = mairt Clair & margo + moux Cuazon. mair & Margon, = 6X28.97 X 0.78 + bx 39.948 x 0-3122 nX28.97 + nx 39.948. Cul mixture = 0.4827 KJ kg k.) ele 'mixture 0-724 0.4827. (C) mixture - 0.4055. → This here = 1.5 1 For Isenstripic expansion process, 7 = 17 - = 0.4055. TT-60-4015) → T = 567.672 K.) Nowo, work output of the turbine per whit mass is given by W = h, the = (C) mix (T-T2) = 0.724 x (1400 - 567.672). W = 602-6 ~ 603 KJ /kg / Aus.