Consider the wind tunnel shown in Figures 6.7 and 6.8. Atmospheric air enters th
ID: 1842103 • Letter: C
Question
Consider the wind tunnel shown in Figures 6.7 and 6.8. Atmospheric air enters the system with a pressure and temperature of 14.7 psia and 80 F, respectively, and has negligible velocity at section 1. The test section has a cross- sectional area of 1 ft2 and operates at a Mach number of 2.5. You may assume that the diffuser reduces the velocity to approximately zero and that final exhaust is to the atmosphere with negligible velocity. The system is fully insulated and there are negligible friction losses. Find (a) The throat area of the nozzle. (b) The mass flow rate. (c) The minimum possible throat area of the diffuser (d) The total pressure entering the exhauster at startup (Figure 6.7). (e) The total pressure entering the exhauster when running (Figure 6.8). (f) The hp value required for the exhauster (based on an isentropic compression). Shock section Diruset Exhauster Nozzle Exhauster section Position in Wind Tunnel Position inWind Tunnel Figure 6.7 Supersonic tunnel at startup (with associated Mach number variation). Figure 6.8 Supersonic tunnel in running condition (with associated pressure variation)Explanation / Answer
GIVEN:- Pin = 14.7psi = 2116.8 lb/ft2, Tin = 80F, (Cross-section area of test section) Ats = 1 ft2, (Mach number of test section) Mts = 2.5,
SOLUTION:-
(a) Throat area of nozzle A2 = ?.
A2 x V2 = Ats x Vts (where V2 = M2 x 1116 ft/s and Vts = Mts x 1116 ft/s)
i.e. A2 = (1 x Mts x 1116) / (M2 x 1116) = (1 x 2.5 x 1116) / (0.9 x 1116) = 2.78 ft2
Throat area of nozzle A2 = 2.78 ft2 (ANSWER)
(b) Density of air (d) at given temperature & pressure = 0.15 lb/ft3
By formula mass flow rate M = (2.78 x 1116) x d = 2.78 x 1116 x 0.15 = 465.4 lb/s
Mass flow rate M = 465.4 lb/s (ANSWER)
(c) For diffuser, the velocity of air is negligible. (V5 = 1116 x 0.9) (Here M = 0.99 is < 1 for diffuser)
Now, Ats x Vts = A5 x V5 i.e. 1 x (2.5 x 1116) = A5 x (1116 x 0.99)
i.e. A5 = (2.5 x 1116) / (1116 x 0.99) = 2.52 ft2
Throat area of diffuser A5 = 2.52 ft2 (ANSWER)
(d) By formula --- (Pout - Pin) = d x Vout (Vout - Vin)
i.e. Pout - 2116.8 = 0.15 x (0.99 x 1116) [(0.99x1116) - (0.1x1116]
i.e. Pout = 166722.5 = 1158 psi (ANSWER)