Flushing Chemical Feed Lines - Heat of Solution for 50% w/w H2SO4
Flushing Chemical Feed Lines - Heat of Solution for 50% w/w H2SO4
(OP)
Hello -
Here is my situation: my client desires the ability to flush 50% w/w H2SO4 feed lines as to prevent chemical from sitting in the pipes.
My concern was the heat evolution to dilute the line. k value for PVC is 0.1428 W/mK.
Here is my approach:
1. System: 1" Sch. 80 PVC, 30m long, inner radius = 0.01215m, outer = 0.0167m, 50% w/w H2SO4
2. Heat of Solution, 50%w/w H2SO4 = -9.715e5 J/kg
3. Volume of Chemical: 0.01414m3 50%w/w H2SO4, comes to 9.188 kg H2SO4 PURE.
4. Dilution: calculated amount of water to dilute from 50% w/w to 1% w/w. I used c1v1=c2v2 for simplicity, This comes to 186.8 gallons of H2O to dilute to 1% w/w H2SO4. At 20 gpm H2O, the time to dilute is 560 seconds.
5. Heat Evolved: Using 9.188 kg H2SO4 pure, multiply by heat of solution, comes to 8.926E6 Joules evolved. Considering the dilution time, I divide the heat evolved by 560 seconds.
6. Temperature of the inner wall of the pipe: I assumed the no convective heat transfer. Therefore, no heat is being removed from the outer surface of the pipe. So, all heat evolved is contained in the PVC pipe, and I feel this is a conservative estimate because the actual heat evolved would be less. So, I use Fourier's law for Cylindrical geometry. Q(dot) = deltaT/ln(ratio of diameters)*2*PI*k*l
7. I solve for the inner wall temperature, assuming outer wall is constant at 298K.
I figure the temperature of the inner wall to be 400F after the dilution.
Does this seem correct? Or too high? Do I need to consider convective heat transfer?
Thanks -
Here is my situation: my client desires the ability to flush 50% w/w H2SO4 feed lines as to prevent chemical from sitting in the pipes.
My concern was the heat evolution to dilute the line. k value for PVC is 0.1428 W/mK.
Here is my approach:
1. System: 1" Sch. 80 PVC, 30m long, inner radius = 0.01215m, outer = 0.0167m, 50% w/w H2SO4
2. Heat of Solution, 50%w/w H2SO4 = -9.715e5 J/kg
3. Volume of Chemical: 0.01414m3 50%w/w H2SO4, comes to 9.188 kg H2SO4 PURE.
4. Dilution: calculated amount of water to dilute from 50% w/w to 1% w/w. I used c1v1=c2v2 for simplicity, This comes to 186.8 gallons of H2O to dilute to 1% w/w H2SO4. At 20 gpm H2O, the time to dilute is 560 seconds.
5. Heat Evolved: Using 9.188 kg H2SO4 pure, multiply by heat of solution, comes to 8.926E6 Joules evolved. Considering the dilution time, I divide the heat evolved by 560 seconds.
6. Temperature of the inner wall of the pipe: I assumed the no convective heat transfer. Therefore, no heat is being removed from the outer surface of the pipe. So, all heat evolved is contained in the PVC pipe, and I feel this is a conservative estimate because the actual heat evolved would be less. So, I use Fourier's law for Cylindrical geometry. Q(dot) = deltaT/ln(ratio of diameters)*2*PI*k*l
7. I solve for the inner wall temperature, assuming outer wall is constant at 298K.
I figure the temperature of the inner wall to be 400F after the dilution.
Does this seem correct? Or too high? Do I need to consider convective heat transfer?
Thanks -





RE: Flushing Chemical Feed Lines - Heat of Solution for 50% w/w H2SO4
Good luck,
Latexman
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RE: Flushing Chemical Feed Lines - Heat of Solution for 50% w/w H2SO4
I agree with Latexman- you're thinking about this as an equilibrium problem when it's a kinetic one. If you were doing a continuous dilution it would be another matter, but all you're doing is flushing the pipe- most of the flushing will be over before most of the dilution has had time to happen. You will have the heat capacity of the water to help mitigate the heat of dilution in the line itself.
You don't mention where the diluted mess is going though. Your receiver, whatever that may be, will receive the water/acid mixture where the two will indeed mix over time, liberating the entire heat of dilution. The solution to that problem is more water in the receiver, to reduce the mixture temperature by adding more heat capacity.