Question

A flexible air supply hose is connected to a compressed air tank as shown in the figure. The hose is 5 m long and has an internal diameter of 10 mm. At its free end the hose discharges to the atmosphere. The air leaves the hose at sonic velocity (M = 1), with p = 100 kPa and T = 273 K. The friction coefficient C (the same as f in the Anderson text books) within the hose is 0.0227 (a) Find the Mach number, temperature and pressure at the entrance to the hose. (10 marks] (b) Find the flow rate of air entering the hose from the tank. [3 marks] (C) Find the location in the hose where the Mach number is 0.5. [3 marks) (d) What is the change in stagnation (total) temperature from the entrance to the exit of the hose? [1 mark] (e) State any assumptions used to solve this problem [1 mark] (0) Comment on the validity of your solution method. [2 marks] Compressed Air Tank M= 1 p = 100 kPa T = 273 K Flexible Hose Diameter = 10 mm Length = 5 m Hose Entrance Compressed air tank and hose Notes: 1. If using tables, it is acceptable to use the nearest value. 2. For air R = 287 J/kg.K and k = 1.4. 3. Please use the problem-solving protocol, state all assumptions, and show all working.

Answer 1

Salution Givendata Mach number at M = 1 Length of duct is, =sm temperature and flow is considered pressure to To calculate se mach number at the entrance to the hose be hammo How . . Using the gas table 4th = 4X0.022715 - 0.01 from the table for 4th Mach number is Min 20.12 At inlet m=0.12 P At enit M = 1 Pe a looklu Te = 273k ) =9. 1156 De 1.1966 De - Now = 9.1156 1.1156 Pia 9. 1156xle = = 9.11568100 911-56kPa

= 1.1966 - Te = 1.1966 7 = 1.1968x Te = 1.1966* 273 Ti = 326.671 Hence at Pressure is the entrance 911.56 kPa to the and hose mach number is 0.12. Temperature is 326.61k. (6) Calculating the flow rate of air entering the duct erlin - Sitici s aparto normas = pi x A MiVERTE S 90.36X163 Pe g T (loxio 3m) yo. 12 2875 kgkX326.6711C RTi 1.4% 287 X 326.67 = 0.0332 g/ Olio SA Mach number 0.5 (c ) At 4 th = 1.6691 buch () at inlet M = 0.12 46 h = 45.5 - Calculating Resultant length the ab and an 44 h = 45-5-).0691 - 44.4309

= 44.4309 xd qof = 44.4309 x (LOX 16-) 4 x 0.022+ 12> 4.893 m 7 Hence at mech number ons location in the hose is. 4.893 m. (d) Temperature is is small change enhance to the remain constant because there in total temperature from the exist of the hose- - AT = (Ti-Tec •621-273) TAT. 33:6211 AT change in temperature. e Assumption used in problem is a (a) Area of duct remain constant (b) = 1.1 R=287 I kyok (c) Viscous fore are used to find the main Number (d) flow is considered to be adrastic How f) Solution remain valid. ation remain valideret -