Question

Problem 4.018 SI Air enters a horizontal, constant-diameter heating duct operating at steady state at 290 K, 1 bar, with a volumetric flow rate of 0.25 m3/s, and exits at 325 K, 0.95 bar. The flow area is 0.06 m2 Assuming the ideal gas model with k 1.4 for the air, determine: (a) the mass flow rate, in kg/s, (b) the velocity at the inlet and exit, each in m/s, and (c) the rate of heat transfer to the air, in kw. Part A Determine the mass flow rate, in kg/s. kg/s the tolerance is +/-2% Click if you would like to Show Work for this question: Attempts: 0 of 15 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

290K, 1 bar theta = 0.25 m^3/sec gas follows K = 1.4 1) Characteristic gas Eqn PV = mRt P V = RT V = (RT/P) = (287 times 290/1 times 105) (Therefore R = 287 J/Kg K) V = 0.8323 m^3/Kg rightarrow Specific volume of Air Mass Flow rate (m) = (theta/V) = (0.25/0.8323) m = 0.300 kg/sec Mass Flow Rate 2) Also Mass Flow Rate (m) = Rho_1 A V_2_1 = rho_2 A Ve_2 Density at 1 rho_1 = (i/V_1) = 1.2014 Kg/m^3 Velocity at 1 V_e1 = (m/rho_1 A) = (0.300/1.2014 times 0.06) V_e_1 = 4.1615 m/sec Velocity at point 1 Density at 2 rho_2 = (rho_2/R T_2) = (0.95 times 10^5/287 times 325) = 1.0184 rho_2 = 1.0184 Kg/m^3 Velocity at 2 V_rho_2 = (m/rho_2a) = (0.300/1.0184 times 0.06) = 4.9092 V_e_2 = 4.90924/sec velocity of point 2 3) S.F.E.E, h_1 + V_1^2/2 + theta = h_2 + V_2^2/2 + W (As no work) Heat added