Tuesday

June 28, 2016
Posted by **Danny** on Sunday, July 14, 2013 at 5:35pm.

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.

- Physics Help -
**Ms. Sue**, Sunday, July 14, 2013 at 5:44pmSince 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. - Physics Help -
**Elena**, Sunday, July 14, 2013 at 6:22pm(A)Ht=Hsζ+(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=Hsζ+ Hw Hwζ

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 =418611.120 = 929292 J = 929.292 kJ

Q₂= = mL =11.12260000 =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 - Physics Help -
**Ms. Sue**, Sunday, July 14, 2013 at 6:28pmThank you, Elena. I was wrong.

- Physics Help -
**Elena**, Sunday, July 14, 2013 at 6:35pmAnother 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% - Physics Help -
**Elena**, Sunday, July 14, 2013 at 7:16pmUsing 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 +0.7141865.2 = 2268.23 kG/kg

Heat already at water (water is at 80℃) = 4.18680= 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