Small transformers generally have fairly poor load regulation - the output voltage drops significantly between no load and full load. Assuming your power supplies are identical, the poor load regulation is an advantage because if one transformer hogs the load, its voltage will drop to bring it more in line with the others. You can get an idea of how well this will work by measuring the voltage vs. current characteristic of each of your power supplies over the full current range and then modeling each power supply as a resistance in series with a fixed voltage source. From that, you can calculate the current drawn from each when they are all in parallel driving a single load resistor.
For example, suppose you have 3 power supplies with the following characteristics:
1) No load voltage = 16 V, Voltage driving a 7.5 Ω resistor = 14 V (that's ~ 30 Watts, so be careful!)
2) No load voltage = 17 V, Voltage driving a 7.5 Ω resistor = 14.5 V
3) No load voltage = 15 V, Voltage driving a 7.5 Ω resistor = 13 V
You would model each as:
1) 16 V source, Rseries = (Rload)(Vdrop) / (Vout) = (7.5)2/14 = 15/14 = 1.07 Ω
2) 17 V source, Rseries = (7.5)2.5/14.5 = 1.29 Ω
3) 15 V source, Rseries = (7.5)2/13 = 1.15 Ω
Wired in parallel and driving a 2.5 Ω load, for a total current Itotal = I1 + I2 + I3 we get three equations:
(1) 16 V = (1.07 Ω)I1 + (2.5 Ω)(I1 + I2 + I3)
(2) 17 V = (1.29 Ω)I2 + (2.5 Ω)(I1 + I2 + I3)
(3) 15 V = (1.15 Ω)I3 + (2.5 Ω)(I1 + I2 + I3)
Converting to matrix form:
--I1---I2---I3--
3.57 2.5 2.5 | 16
2.5 3.79 2.5 | 17
2.5 2.5 3.65 | 15
Using an online equation solver, such as:
http://www.bluebit.gr/matrix-calculator/linear_equations.aspx
we solve the 3 simultaneous equations (6 in your case), and get:
I1 = 2.037
I2 = 2.465
I3 = 1.026
In this example, power supply 3 is not doing its fair share, so you could balance things by adding equalizing resistors to the other two power supplies