Currently there are 3 baseboard heaters (750, 1000 and 1500) 3250 watts total running on 2 tied 15 amp breakers. I want to change all three heaters. What is the maximum number of combined watts I can install?
Resistive heating loads are required to be calculated as continuous loads so the circuit can only be loaded to 80% of its rating.
Therefore a 15 amp circuit at 120 volts is good for 12 amps which yields 1440 watts of power. More likely the heaters are 240 volts which yields 2880 watts per circuit for a totoal of 5760 watts for both circuits. (240 volt circuits will require a 2 pole breaker for each circuit occupying 2 spaces in your panel for each one.)
However, you would have to put two on one circuit and one on the other so you have to make sure the circuit with two heaters does not exceed 2880 watts at 240 volts and then put the largest one on a circuit by itself.
Here is part of the pertinant article from the 2017 National Electrical Code:
424.3 Branch Circuits.
(A) Branch-Circuit Requirements. Individual branch circuits shall be permitted to supply any volt-ampere or wattage rating of fixed electric space-heating equipment for which they are rated. Branch circuits supplying two or more outlets for fixed electric space-heating equipment shall be rated not over 30 amperes. In other than a dwelling unit, fixed infrared heating equipment shall be permitted to be supplied from branch circuits rated not over 50 amperes.
(B) Branch-Circuit Sizing. Fixed electric space-heating equipment and motors shall be considered continuous load.
Good luck with your project!
The "quadplex" breaker you linked, the "innermost two" are a 2-pole breaker and common trip. You are all set. (otherwise, adding a handle-tie to two breakers does not guarantee common trip**; only common maintenance shut-off. That won't do for 240V loads, but is OK for separate circuits or MWBC***. I'll address that part for others with a similar problem.)
See what ArchonOSX says about factoring 80% (or inverse of 125%) for continuous loads.
Two 120V circuits
In this case you have two 15A, 120V breakers. Each of the two circuits can drive 12A of heaters, or 1440W. Since the breakers are handle-tied, they can be configured as an MWBC, sharing neutral.
You must balance your heaters so neither circuit exceeds 1440W. For instance, a 2000W heater is out of the question.
One 240V circuit
You must change your breaker to a 2-pole type, which for most panels is cheap. Amperage does not double; voltage does instead. With one 240V, 15A breaker, you can drive 12A of heater -- giving 2880 watts. A 2000W heater is fine in this configuration.
** Common trip means if one side overloads, both sides trip.
*** There would be no reason to tie separate circuits, unless there is. MWBCs share neutral, so must be on opposite poles and must have common maintenance shut-off for safety, both assured by using a listed handle-tie.
15A breakers at 240V = 3,600W max
You mention that the existing heaters are not enough and you want to put out more heat. That suggest the load will be used continuously, so you need to derate the circuit to 80%.
That means your current circuit can safely support only 3600 * 0.8 = 2,880W.
Why? When you're using a toaster or a hair dryer, you use it for a short amount of time, so you can use 100% of the circuit capacity. But if you use a load for a long time, the heat builds up in the wires, calling for a safety derating of the circuit. The goal is to protect the wires in the walls, and prevent them from starting a fire.
You are already beyond the capacity of your circuit. Your current setup is already unsafe. Please don't add more to it.
I suggest you need rewire your circuit to a 30A 240V circuit, using 10-3 wires.