Mixing temperature

[Lukezo]

Hi,

I have all data on mixing vapor, starch and water (temperatures, pressure) and when I make a balance:

m_vapor_in*h_vapor_in+m_water*cp_water*T_water=(m_water_in+m_vapor_in)*cp_water*T_mix_measured+m_starch*
cp_starch_new*T_mix_measured+E_endo)

I get a result, that expression on left is 1,9x bigger than on the right side. I took into account that cp,new (starch) is temperature dependent.

Regarding new mixing temperature (95˙C) I understand that I get some heat loss, but not almost by factor 2? I neglected reactor mass. Mass of vapor is determined by measuring mixture level before and after measurment. (V=3,14*D^2/4*(h2-h1)...m_water=V*ro_water=m_vapor_in

thanks

 

[goutam_freelance]

First consider a control volume and consider enthalpies. Use c_p for calculating enthalpies.

Now do a first law heat balance around the CV

 

[Lukezo]

I did exactly the same calculation, but the balance doesn't match. What comes in (left side) is by factor 1,7x bigger. So I presume that we have huge heat loss.

 

[goutam_freelance]

Your left side expression does not have starch enthalpy as input. So probably starch is at lower temperature and enthalpy of which should be minus in left side.

 

[IRstuff]

I neglected reactor mass.

Did you do the math on ignoring the reactor mass? Unless the reactor is pre-heated to the final temperature, you WILL had a heat loss.

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