- #1

sci0x

- 83

- 5

- Homework Statement
- At end of fermentation, beer is cooled to 2 degrees C. Height of beer in vessel is 20m, top pressure of 0.1 bar g of co2 is applied. Calculate the equilibrium conc of co2 in the beer in g/l assuming that a reasonable estimate of its average conc can be obtained by considerimg the prevailing pressure at the mid-point of beer in the vessel

Data:

Henrys constant at 2 degrees C = 84.1 x 106Pa (mole fraction) -1

Beer density = 1008kg m-3

Acc due to grav = 9.81 m s-2

Mol weight CO2 = 44

Mol mass beer = 18

Atm pressure = 1.013 bar 1 bar = 105 Pa

- Relevant Equations
- P1 = P0 + qgh

Absolute pressure = Gauge press + Atmos press

Atmos press = 105 Pa

Co2 press = 0.1 bar g = 10000Pa

Abs press = 10,105 Pa

Hydrostatic pressure = absolute press + (density)(grav)(height at midpoint)

= 10,105 + (1008)(9.81)(10)

10,8989.8 Pa

Calc CO2 conc by henrys law

P=KHC

C=P/KH

= 108989.8/84.1x10^6Pa

= 1.29 x 10^-3

Ans x 44 = 0.056 g/l

Can i get some help please:

Have i calculated absolute pressure correctly.

Atmos pressure CO2 should be used in calc of absolute pressure from gauge pressure

Hydrostatic pressure is added for the pressure at the mid-point, calculated by multiplying density x acceleration due to grav x liquid height (10m)

Calc of Co2 conc is by henry's law

Mole fraction is converted to g/L

Answer should be 6.1 g/L

Atmos press = 105 Pa

Co2 press = 0.1 bar g = 10000Pa

Abs press = 10,105 Pa

Hydrostatic pressure = absolute press + (density)(grav)(height at midpoint)

= 10,105 + (1008)(9.81)(10)

10,8989.8 Pa

Calc CO2 conc by henrys law

P=KHC

C=P/KH

= 108989.8/84.1x10^6Pa

= 1.29 x 10^-3

Ans x 44 = 0.056 g/l

Can i get some help please:

Have i calculated absolute pressure correctly.

Atmos pressure CO2 should be used in calc of absolute pressure from gauge pressure

Hydrostatic pressure is added for the pressure at the mid-point, calculated by multiplying density x acceleration due to grav x liquid height (10m)

Calc of Co2 conc is by henry's law

Mole fraction is converted to g/L

Answer should be 6.1 g/L