At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.
Answer:
Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..
Answer: Solid state welding
Explanation: Solid state welding is the welding procedure which is based on the temperatures and pressure but without any liquid or vapor as aid for welding.This process is carried mainly cohesive forces and considering forces as less important. Brazing,soldering and adhesive is the process in which material are joint which the help of solid welding agent.Thus solid state welding is the correct option.
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
(a) The stress on the steel wire is 19,000 Psi
(b) The strain on the steel wire is 0.00063
(c) The modulus of elasticity of the steel is 30,000,000 Psi
Explanation:
Given;
length of steel wire, L = 100 ft
cross-sectional area, A = 0.0144 in²
applied force, F = 270 lb
extension of the wire, e = 0.75 in
<u>Part (A)</u> The stress on the steel wire;
δ = F/A
= 270 / 0.0144
δ = 18750 lb/in² = 19,000 Psi
<u>Part (B)</u> The strain on the steel wire;
σ = e/ L
L = 100 ft = 1200 in
σ = 0.75 / 1200
σ = 0.00063
<u>Part (C)</u> The modulus of elasticity of the steel
E = δ/σ
= 19,000 / 0.00063
E = 30,000,000 Psi
