Posted by Danny on Sunday, July 14, 2013 at 5:35pm.
Since you've posted this question about eight times in the last few days, I think we can assume that none of our tutors can help you.
Sorry.
(A)Ht=Hs•ζ+(1-ζ)Hw,
where
Ht = total (actual) enthalpy (kJ/kg),
Hs = enthalpy of steam (kJ/kg),
Hw = enthalpy of saturated water or condensate (kJ/kg)
Using steam-pressure table
http://enpub.fulton.asu.edu/ece340/pdf/steam_tables.PDF
we can find the magnitudes for p=20 bar
(the given data in the problem is p=2250 000 Pa = 22.5 bar)
Hs =2800 kJ/kg
Hw = 908.8 kJ/kg and.
Ht=34191. 510/15 =2279.434 kJ/kg
Ht=Hs•ζ+ Hw –Hw•ζ
ζ(Hs-Hw)=Ht-Hw
ζ= (Ht-Hw)/ (Hs-Hw)=
=(2279.434-908.8)/(2800-908.8) =
=1370.634/1891.2=0.725
Answer: 72.5% (the answer may differ due to using of 20 bar instead of 22.5 bar)
(B)
Q=Q₁+Q₂+Q₃
Q₁=cmΔt =4186•11.1•20 = 929292 J = 929.292 kJ
Q₂= = mL =11.1•2260000 =25086000 J= 25086 kJ
Q₃= Ht ≈34191.510 kJ (I believe that we can use the result of part (a)
Q = 929.292+25086+ 34191.510 ≈ 60206.802 kJ
Thank you, Elena. I was wrong.
Another table http://www.chem.mtu.edu/~tbco/cm3230/steamtables.pdf
gives more precise values
Hs =2801.7 kJ/kg
Hw = 936.48 kJ/kg
As a result
ζ= (Ht-Hw)/ (Hs-Hw)=
=(2279.434-936.48)/(2801.7-936.48) =
=13.42.954/1865.22=0.7199 => ≈72%
Using Problem 5 page 234 from
http://books.google.com.ua/books?id=PQRo3XuHHvYC&pg=PA262&lpg=PA262&dq#v=onepage&q&f=false
From steam table
at 22.5 bar
Hw= 936.48 kJ/kg
He = 1865.2 kJ/kg - enthalpy of evaporation (Latent heat)
Enthalpy of 1 kg of wet steam is
H=Hw+ζ •He=
=936.48 +0.714•1865.2 = 2268.23 kG/kg
Heat already at water (water is at 80℃) = 4.186•80= 334.9 kJ/kg
Heat required per 1 kg of steam = {Enthalpy (or) heat required to raise 1 kg of steam from water at 0℃} – {Heat already present in water} = 2268.23 -334.9 = 1933.33 kJ/kg
Heat supplied per kg = 1933.33 kJ/kg.
Heat required for 11. 1 kg =11.1• 1933.33= 21459.963 kJ