Question

The cider is heated from an initial temperature of 4°C to a final temperature of 65°C. The steam enters the heat exchanger as 50% quality steam and exits as water condensate at 85°C. Calculate the mass of steam required to heat 150 kg of cider. (For the cider, assume the Cp=3.651 kJ/kg°C, and latent heat=280.18 kJ/kg.)

Answer 1

TS1 = 85 C
xS1 = 0.5000

TS2 = 85 C
xS2 = 0.0 ( saturated liquid water at 85 C )

Steam Table Data at 85 C :
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T = 85 C
Psat = 57.83 kPa
hf = 355.90 kJ / kg
hfg = 2296.0 kJ / kg
hg = 2651.9 kJ / kg

Get hS1 for the steam from the quality equation :
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hS1 = hf + ( xS1 ) ( hfg )

hS1 = 355.90 + ( 0.5000 ) ( 2296.0 ) = 1503.9 kJ / kg

hS2 = hf at 85 C = 355.90 kJ / kg

Delta hS = hS2 - hS1

Delta hS = 355.9 - 1503.9 = - 1148 kJ / kg

qS = QS / mS = Delta hS = - 1148 kJ / kg

Now get QAC :
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QAC = ( mAC ) ( CpAC ) ( TAC2 - TAC1 )

QAC = ( 150 kg ) ( 3.651 kJ / kg - C ) ( 65.0 C - 4.0 C )

QAC = 33410 kJ

You have QS = - ( QAC ) :
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QS = - 33410 kJ

mS = QS / qS

mS = ( - 33410 ) / ( - 1148 kJ/kg )

mS = 29.1 kg <-------------------------