3 replies to this topic

I am trying to calculate the temperature drop of the slurry across a vacuum belt filter.

The filter is feed with 225 tph (liquid volume) slurry at 65 Deg and the filter vacuum is operating at about 20 kpaA. The vacuum pump is rated for 3250 m3/hr.

I have commenced the calculation by trying to do a mass and energy balance across the filter, however I would really appreciate a worked example to ensure my assumptions are correct. Can someone please help !!!

[sgkim]

Phildb

Heat Balance: Heat lost from slurry = Heat of vaporization ......-W*Cp*dT = Λ*dW ........(1)
Mass Balance: Mass lost from slurry = Mass exhausted by vacuum..dW = (M*P*v/RT)*dt..(2)
where,
W=Mass Rate, Cp=Sp. Heat, T=Absolute temp. of the slurry=Abs. temp. of vapor, Λ = Latent Heat of the liquid; t=time duration under vacuum. P=pressure, v=actual volume rate, M=mol. mass of vapor, R=Universal gas constant.

Solve (1)and (2) simultaneously or simplify eq(1) as -Wo Cp dT = Λ*dW --(3) then solve equation (2) and (3) to get T a function of t, where Wo=initial mass of the slurry.

Firstly Thanks for your input.

If I understand your response correctly, solving (3) and (2) simultaneously the solution becomes

dT = ^*(M*P*v/RT)/-Wo*Cp *dt

Whereby:

^ = Latent heat of the liquid at the vacuum condition (kJ/kg)
M = mol mass of the vapour (mol/hr)
P = Pressure (kPaA)
T = Temp (oC)
v = Actaul Volume rate from the vacuum pump (Am3/hr)
R = 8.314 J/K/mol
T = temp of the vapour being drawn by the vacuum pump (oC)

If so I am correct in assuming that to calculate the mol mass of vapour drawn by the vacuum an assumtion of the vapour conditions will need to be made ?

Thanks again

Phildb

Heat Balance: Heat lost from slurry = Heat of vaporization ......-W*Cp*dT = Λ*dW ........(1)
Mass Balance: Mass lost from slurry = Mass exhausted by vacuum..dW = (M*P*v/RT)*dt..(2)
where,
W=Mass Rate, Cp=Sp. Heat, T=Absolute temp. of the slurry=Abs. temp. of vapor, Λ = Latent Heat of the liquid; t=time duration under vacuum. P=pressure, v=actual volume rate, M=mol. mass of vapor, R=Universal gas constant.

Solve (1)and (2) simultaneously or simplify eq(1) as -Wo Cp dT = Λ*dW --(3) then solve equation (2) and (3) to get T a function of t, where Wo=initial mass of the slurry.

After placing the last post I did the calcualtion assuming the vapour partial pressure being drawn by the vacuum pump will be the vapour pressure of water at the temp of the solution entering the filter.

I.e. at 51.7 Deg this will 13.4 kPaA

Then as the vacuum pressure is 50 KpaA the mol % of the water vapour is 26.9%

Calcuating the molar mass of water being evaporated is then relativly straight forward, and then finally conducting the overall heat mass balance by using:

dT = ^*(M*P*v/RT)/-Wo*Cp *dt

Can be done easily.

My question is is the assumption I used to calcuate the vapour pressrue in the gas stream to the vacuum pump correct ?

thanks in advance