To elucidate the heat transfer a bit, because I have a couple long breaks today while we’re running a reaction. Taking the simplified means to describe this system, rather than messing with higher order mathematics and transient state systems, since the outcome is exactly the same.
3 things are happening from our “zero state.”
1) Heat is transferred from the combusting gases to the barrel
2) The temperature of the barrel mass rises
3) The barrel mass expels heat to the ambient air mass
1) Rate of Heat transfer to the barrel: Q=U*A*(Th-Tc)*t
Q is the total heat transferred, represented by Q=Q•*t. Q• is the rate of heat transfer, colloquially, this is the “amount of energy” transferred from the hot mass to the cold mass, while the two masses are in contact. In a firing rifle, the total heat transferred then can be calculated by multiplying Q by the contact time t. This number will be the same for any barrel diameter, as proven below.
U is the overall heat transfer coefficient, effectively how well the two materials transfer heat from one to the other. The units for U are BTU/Ft2 hr •F, which describes heat transfer per area, per amount of time, per degree of temperature difference. U-values are properties of the two materials, in this case, the propellant gases and the barrel. This is the same for any barrel diameter.
A is the area of the transfer surface. In our case, the internal surface area of the bore. This is the same for any outside-diameter of barrel.
Th and Tc are the hot and cold temperatures. In our case, the temp of the hot propellant gases and that of the relatively cold barrel mass. This is the same for any diameter of barrel at the first shot.
Since we’re in a complex, transient state system, it’s easier to just generalize here and take an average estimate for gas temp and dwell time to represent how many BTU’s are given from the hot gases to the cold barrel.
So we fire a shot, and “Q” rate of heat transfer to the barrel while the hot gas is in contact with the barrel, such Q*t (time) represents the total heat transferred. You can see above, the transfer rate is ONLY a function of the material types (U), the bore surface area (A), the dwell time (t), and the difference in temperatures (Th-Tc). At this point, barrel mass and OD don’t matter, so a single shot will transfer the same total heat to any barrel, regardless of profile.
So if we had sufficient data, we would know the heat transferred by each shot - we would know Q. For now, we simply need to know A) it is not dependent upon barrel diameter, and B) that it could be solved given sufficient data.
2) Barrel heating up: Q = M*Cp*(Tf-Ti).
Q again is the heat transferred, which we proved above is constant regardless of the barrel diameter. Effectively, it’s not an unknown, if we solved the first equation above.
M is the mass of the barrel - this is where we start having two sets of equations to compare the two different barrel diameters.
Cp is specific heat capacity, a property of the material. Same for any diameter of barrel.
Tf is the final temp, Ti is the initial temp. When you compare the heavy barrel to the smaller, Q is the same, so if M increases for one, the Tf-Ti term must decrease. Since Ti is the starting, ambient temp, then decreasing (Tf-Ti) would mean Tf has to reduce as mass increases.
As such, we have explained that larger diameter barrels take longer to heat up than a smaller diameter barrel - or more appropriately, heavier barrels take more shots to reach the same final temperature than a lower mass barrel.
3) Barrel mass expelling heat to ambient air
Again, Qcooling=UA(Th-Tc)t as it was in the first step above.
Qcooling is the heat transferred to the ambient air by the barrel mass in the given amount of time, t.
Again, U is a property of the two materials, in this case, steel and air, and it is not dependent upon Barrel diameter.
A, again, is the surface area of the barrel. This IS dependent upon the diameter of the barrel. The larger the diameter, the larger the A.
Th is the hot barrel temp, Tc is the cool air temp. We have options here - we can say we got both barrels to the same starting temp -which is Tf from above - and want to see which cools faster so then (Th-Tc) is the same for both barrels. Or we can say we took the same number of shots with both barrels, such both contain the same heat to be transferred (equal Q’s) to the air - which would mean the temp difference would not be the same, because the Tf in step 2 would be different. Either way, the same result falls out, because, as described above, Q is the amount of heat transferred, and Q• is the rate of transfer, such Q=Q•*t. But in one case, all of the variables are known to be the same, in the other, we have two unknowns and would need both equations from step 2 and 3 to solve. Possible, and simple, but harder to follow.
The simpler math case is the same starting temp. For example, we fired both until the barrel temp was uniform in both at 125F. So (Th-Tc) is fixed. So then we’ll say we’ll compare the temperature of both after 5min, so time, t, is also fixed.
Q is the heat transferred then in those 5min from each barrel. Q=UA(Th-Tc)t, and we know the heat transfer coefficient U is constant, the temperature difference between the starting temp of the barrel and the ambient air (Th-Tc) is fixed, and time t is fixed. The only difference between the inputs for the two barrels is the area A, and the difference in area will yield, proportionately, the difference in heat transferred in that time t, which is Q. Area is larger for the larger diameter barrel, such Q for the larger diameter barrel must be larger - which means it is expelling heat to the environment faster than the smaller barrel.
Such, we’ve shown mathematically a larger surface area expels heat faster than a smaller surface area. So larger diameter barrels cool faster than smaller diameter barrels.
