Input: what is put in, taken in, or operated on by any process or system.
Output: the amount of something produced by a person, machine, or industry.
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
Foot protection could be required with Sharps, Slippery areas, Hazardous liquids,and Falling objects.
Hope this helps!!
Answer:
A)
D = 158.42 kmol/h
B = 191.578 kmol/h
B) Rmin = 1.3095
Explanation:
<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>
F = D + B ----- ( 1 )
<em>Given data :</em>
F = 350 kmol/j
Xf = 0.45 mole
yD ( distillate comp ) = 0.97
yB ( bottom comp ) = 0.02
back to equation 1
350(0.45) = 0.97 D + 0.02 B ----- ( 2 )
where; B = F - D
Equation 2 becomes
350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3
solving equation 3
D = 158.42 kmol/h
resolving equation 2
B = 191.578 kmol/h
<u>B) Determine the minimum reflux ratio Rmin</u>
The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve
first we calculate the value of the enriching line
Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )
q - line ; y = ( 9 / 9-1 ) x - xf/9-1
therefore ; x = 0.45
Finally Rmin
= (( 0.97 / (Rm + 1 )) = 0.42
0.42 ( Rm + 1 ) = 0.97
∴ Rmin = 1.3095
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