- #1
sci0x
- 83
- 5
- Homework Statement
- A brewery requires wort to undergo 5% by mass evaporation in one hour. The wort is boiled at atmospheric. Using the data below, calculate flow of steam required.
Saturated steam pressure = 3 bar gauge
Condensate pressure in wort boiler = 3 bar gauge
Enthalpy of steam, at 3 bar gauge = 2739 kg kj-1
Enthalpy of water at 3 bar g = 605 kj kg-1
Initial wort volume at 100 deg c = 500 hl
Enthalpy of vapourisation of water at 100 deg C and atmospheric pressure = 2257 kj kg-1
Density of water at 100 deg C = 958 kg m-3
- Relevant Equations
- Q = U A dT
I need help solving this please. Its a past exam paper.
requires wort to undergo 5% by mass evaporation in one hour
does this mean starting vol is 500hl and after one hour vol should be 475hl
Wort volume = 500 hl = 50,000 L / hr
1 L = 10^-3 m^3
50,000 L = 50 m^3 /hr
= 0.833 m^3 / min
= 0.01388 m^3 / sec
Final Enthalpy
= (0.01388)(2739)
= 38.0173 kJ
Initial Enthalpy
= (0.01388)(605)
= 8.3974 kJ
Final - Initial = 29.6199 kJ
Am i on the right track here?
requires wort to undergo 5% by mass evaporation in one hour
does this mean starting vol is 500hl and after one hour vol should be 475hl
Wort volume = 500 hl = 50,000 L / hr
1 L = 10^-3 m^3
50,000 L = 50 m^3 /hr
= 0.833 m^3 / min
= 0.01388 m^3 / sec
Final Enthalpy
= (0.01388)(2739)
= 38.0173 kJ
Initial Enthalpy
= (0.01388)(605)
= 8.3974 kJ
Final - Initial = 29.6199 kJ
Am i on the right track here?