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.
1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam.
ANS = %
2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry.
ANS = kJ
Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank you.
5 answers
(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
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%
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
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
0 answers