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
Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.
<h3>What is the mass of casein in wet casein?</h3>
The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg
Mass of water 250 kg
The mass of casein is constant while the moisture content can be changed.
At 12% moisture content;
750 kg = 88%%
100 % = 100 ×750/88 = 852.27 kg
Therefore, the mass of dried casein produced os 852.3 kg.
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Answer:
E= 15 GPa.
Explanation:
Given that
Length ,L = 0.5 m
Tensile stress ,σ = 10.2 MPa
Elongation ,ΔL = 0.34 mm
lets take young modulus = E
We know that strain ε given as



We know that

Therefore the young's modulus will be 15 GPa.
Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now

If consider that density of air is not changing from inlet to exit then by using continuity equation

So 
m/s
Now from first law for open system

Here Q=0 and w=0

When air is treating as ideal gas

Noe by putting the values



So the exit temperature is 293.74 K.