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Socket Weld Connection (heat transfer)
- Thread starter cocky2001
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btrueblood
Mechanical
The temperature of what?
The materials being welded? Depends on those materials, but the melt points are easily looked up.
Temperature of the arc? Ask a plasma scientist, as it's debatable if you can actually use those terms.
Peak temperature of something connected to the tube, but some distance away from the weld joint? Ah, that is a heat transfer problem. Will depend on the time required to complete the weld, wall thickness of tubes.
The materials being welded? Depends on those materials, but the melt points are easily looked up.
Temperature of the arc? Ask a plasma scientist, as it's debatable if you can actually use those terms.
Peak temperature of something connected to the tube, but some distance away from the weld joint? Ah, that is a heat transfer problem. Will depend on the time required to complete the weld, wall thickness of tubes.
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btrueblood
Mechanical
"at center of my product "
Argh; this does not answer my question. Are you worried about the temperature of fluid in the pipe (as in a hot tapping process, as zapster suggested)?
Or is "your product" some valve or other solid object that is being welded to?
Argh; this does not answer my question. Are you worried about the temperature of fluid in the pipe (as in a hot tapping process, as zapster suggested)?
Or is "your product" some valve or other solid object that is being welded to?
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- #6
Yes, the product is a valve. The internals can only withstands a certain temperature. So, I am looking to determine if the process of welding the pipe in the SW connection will transfer too much heat to these internals. Only concerned with the temperature of the product. So I want to know how much heat generated while welding in a piece of 3/4" pipe in a 304ss valve with a socket weld connection.
That is a little difficult to answer. I would think that the thickness of the valve, or thickness of material from the 3/4" pipe (OD) to product would need to be known. Additionally, while welding the temp is hotter than the surface of the sun.
btrueblood
Mechanical
You can bound the problem by estimating the size of the weld bead. The amount of metal melted would be some factor (say ~2x to 3x) of the size of the weld bead, and would be initially at the temperature of the molten weld pool (say 50-100F above the listed melting point for your metal). Take that amount of heat input (don't forget the heat of fusion of the metal) and apply it instantaneously to the weld area, solve numerically for temperature at the seal. Assume at first that no tube is attached.
Now do it again, but assume the weld takes some finite time to put down (do several iterations from 15 seconds to 5 minutes).
Now do it a third time with a tube stub on one side that can soak up heat.
Now do it again, but assume the weld takes some finite time to put down (do several iterations from 15 seconds to 5 minutes).
Now do it a third time with a tube stub on one side that can soak up heat.
This is a good example of the uncertainty of heat input.
The equation used by most of us involved with welding is Q=(IxEx60/V) where I is amperage, E is voltage, and V is the travel speed. What isn't accounted for is the efficiency of the welding process, i.e., how much of the energy is actual transferred into the weld puddle.
The welding process and the welding parameters influence the energy per unit time introduced into the weld. That being the case, you can select a welding process such as GTAW that has inefficient heat transfer and low welding amperage, a short arc length and high travel speeds to minimize the energy introduced into your part. A cool down period can be used to allow the heat to dissipate before doing damage to you part.
Since the material being welded is unlikely to form martensite upon cooling, you can hasten the cooling between passes with a blower or with wet rags applied to the area adjacent to the socket joint.
Best regards - Al
The equation used by most of us involved with welding is Q=(IxEx60/V) where I is amperage, E is voltage, and V is the travel speed. What isn't accounted for is the efficiency of the welding process, i.e., how much of the energy is actual transferred into the weld puddle.
The welding process and the welding parameters influence the energy per unit time introduced into the weld. That being the case, you can select a welding process such as GTAW that has inefficient heat transfer and low welding amperage, a short arc length and high travel speeds to minimize the energy introduced into your part. A cool down period can be used to allow the heat to dissipate before doing damage to you part.
Since the material being welded is unlikely to form martensite upon cooling, you can hasten the cooling between passes with a blower or with wet rags applied to the area adjacent to the socket joint.
Best regards - Al
gtaw has the right place to start. The amount of electrical energy used in the welding process is the Q formula he gave you. I believe the temperature of molten steel is about 2300 degrees F. To determine the potential temperature increase, add the heat generated, Q, divided by (the specific heat of the metals x the mass of the metal) to the starting temperature and I believe you get the theoretical maximum temperature. The problem is to determine the amount of metal getting heated as the heat flows to the point of concern.
All the heat input problems can be minimized with smaller weld pass sizes and plenty of time for cooling between passes. I do not know much about your situation so I can't tell you if cooling between passes using accelerated methods is best. In my work, we are often trying to do the opposite during welding, trying to keep all the part warm enough. Good Luck!
All the heat input problems can be minimized with smaller weld pass sizes and plenty of time for cooling between passes. I do not know much about your situation so I can't tell you if cooling between passes using accelerated methods is best. In my work, we are often trying to do the opposite during welding, trying to keep all the part warm enough. Good Luck!
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