CO2 suppression skids come with a lockout valve on the main CO2 discharge line but not on the pneumatic horn. I understand the pneumatic horn is fed by CO2 to warn those in the enclosure CO2 is present prior to releasing the main isolation valve discharging CO2 (time delayed valve).
NFPA 12...
I think the confusing part is that this skid has a set flow rate and a set heater...can't be changed. So when the oil passes through one time is will not reach the deltaT desired the first time, may take 2-3 loops. But I think I can use a lot of this information posted on here, thanks for the help.
Understood, but we know how much power we have right now.
Wait, I think I'm getting this now, lol. If I know the current power of the heater I can calculate what the delta T will be at the flow rate of the heater circulation line. This in turn can be applied to the total mass of the tank (this...
If we are wanting to raise the temperature of the tank from X degrees to Y degrees, wouldn't that difference in temperatures be the same for both equations?
Our desired delta is the the difference between X and Y, then what is the other delta T represent?
IRstuff. Your equation cancels delta T and specific heat and just leave total mass over mass flow rate. This may be why I am getting such a low number. Any thoughts?
After re-working the problem in Mathcad I ended up getting a much lower heating time using IRstuff's method versus my original method...this doesn't quite make sense to me since IRstuff's method is more accurate and takes into account the flow rate of the circulation loop.
Or am I just...
I ended up with 15 hours using the equation above and 9 hours using the original method I posted, so the 15 hours makes more sense since this is a circulation loop.
Thanks IRstuff for the input!
Understood. The inlet and outlet of the circulating loop are 1-2 feet from each other in the top of the tank, but are at different elevations in the tank. Due to mixing in the oil I was trying to stay away from doing an unsteady state calculation since the exact time is not urgent, but rather an...
Yes I understand, but due to the reasoning behind heating the oil, an exact time is not crucial. Would it be wise to take this time and double it, or even triple it? Just to add some margin
This is what I did..
Converted 4 kW to BTU/hr. Multiplied this by the specific heat of the oil to get X.
Divided X by the total weight of the oil to get Y.
Divided Y by the temperature difference I am trying to achieve to get Z.
This left me with Z in units of 1/hr
Just took the inverse of...
I'll search around. If I know the specific heat of the oil and the kW rating of the heater, it should be a pretty straight forward equation.
Thanks for pointing me in the right direction.
If I know the kW rating of the heater outside the tank (through which the oil is passing), would this simplify the analysis. I want to treat the tank as a insulated body.
An approximation is satisfactory because an exact time to reach a specific temperature is unecessary.
We are filtering these rectangular lube oil tanks and want to use the skids to heat up the oil during cold startup.
I pulled out the old Heat Transfer book but I think I'm over thinking this. If we are taking cold oil out from the bottom of the tank, heating it (specific temp.), and discharging...
