Recent content by whlock

I read it but I am not sure how I would figure out volumn of water from that...please explain

So do i used this formula per what u said Q of mandrel = 94 lbs x 0.110 btu/lb x 650 = 6721 then use water = lbs x 1 btu/lb x (200-70) = 6721 and give me lbs of water = 51.7 lbs = 6 gallons of water? that sounds really low to me...

To be more clear, the water from the tank circulate in a closed pipe system into the mandrel and go back to the tank. The same water keep circulating in this closed system.

It is in a closed circulating system. The water come from a tank and the water flow thru the mandrel. I am looking for the min amount of water needed in this tank to cool the mandrel down.

The water inlet temp is 70 deg. Outlet unknown. I calculate the heat loss of the mandrel using Q = m x specific heat x temp change m is about 94 lbs..so 94 x .11 x -650 = -6721 BTU. The water is constantly flowing

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 ma facing: 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...