A) chilled water from evaporator
Answer:
it depends on the but i would recommend check in the front next to the turbo intake.
Answer:
Explanation:
From the information given:
The total load is distributed across both the rod and tube:
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
Replace 
into equation (1)
Finally, to determine the normal stress in aluminum rod:
Thus, the normal stress = 23.523 MPa in compression.
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:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ
