I am not really a heat transfer guy so I hope someone here can help me out here. I would like to know the formula to get this.
Here is what I have:
I have a mandrel (basically a piece of cylindrical H-11 steel). It has OD of 3.5 inches and 37” in length. It has a core hole of about 7/8” Dia and 37 “ length. Water pumped into the core hole at 10 GPM thru a 3/4” dia pipe. The H-11 steel has .110 BTU/lb specific heat capacity. Water enter at 70 deg and not sure the exist temperature.
I want to be able to lower the temp of the mandrel from 850 to 200 deg F. How much water do I need?
My take on the problem is like this.
Q = m* Cp * T
Mandrel about 94 lbs
Q = 94 x 0.11 x 650 = 6721 BTU
Water has 1 BTU/lb capcity.
so 6721 BTU will give me 806 gallon of water since 1 lb = 8.33 gallon. Is that correct?