You didn't link to the relay module in question (I see you're new so no worries, please try to do it next time ;-)) but I'm assuming it's this one.
If this is the case, then I think you have mistakenly connected the 26.6 VAC supply to what is meant for a 5 VDC supply. If you have this is not good, hopefully it hasn't blown anything (very possible though)
The marking are actually VCC (not VOC) and GND. TO these you need to connect a 5VDC supply as mentioned.
Here is a schematic of one of the relays on your module:
Notice the 5V on the input side of the relay. The "switch" is from pins 3 and 4 with a fuse in series.
For the relay connections, they act like a single pole single throw switch (SPST) and need to be in series with the power source. To switch the power to the solenoids, you need one lead from the power supply (either one as it's AC) to one of the relay terminals, then one of the solenoid leads to the other terminal. Finally the other lead from the solenoid connects to the other supply lead.
Just try to think of the relay as a simple switch.
Would anybody be able to roughly quantify these extra losses to give motor commisioners a guide of what to expect?
You'd have to use a harmonics analyzer or high-resolution scope as suggested in this article to measure the apparent power with PWM. Also from that (2001) article I gather there's no industry standard for measuring efficiency of DC motors driven by PWM...
In general, you can measure losses with the calorimetric method, but that's very cumbersome for motors. I only found specifics for (AC) induction motors, which being larger and more expensive are probably worth the hassle. For those they are worried whether inverter [PWM-]generated sinewaves (which contain a lot of ripple noise) cause significant extra heating, which is similar but not quite the same problem with what you're asking.
Image from this paper, which also details the issue of having to couple a mechanical load while decoupling it thermally:
Another important difference is that the system is
equipped to a load and measure rotating machines. The
system is provided with an integral mechanical load.
Shafts may be loaded at up to 7.5 kW at speeds in the
range 1,500–4,500 rpm, and with a maximum torque of
62.5 Nm below 1,500 rpm. A Siemens controller and a
permanent magnet (PM) synchronous machine provide the
load. The load is led through the test chamber wall using
a combination of a magnetic coupling and a carbon fiber
shaft. This minimizes stray heat flow along the shaft, as
may be expected if a steel shaft were used. The 1,500-W
system is expected to have an accuracy better than 5 W.
The developed system is shown in Figure 5.
Regarding:
Are the losses bad enough to make this worthwhile ?
Given the dearth of publications on this (for PWM-driven DC motors)... I would venture a guess the answer is usually "no"... however, I found a few publications dealing with the latter, the most recent one I found being combo paper that [despite its title] also studied efficiency of DC motors in a section, which I'm quoting here, and it's indeed (like @JonRB said) mostly and issue of eddy-current losses:
PWM supply is widely used in the speed control of dc motors.
The pulsation components in the PWM supply will also
induce additional losses. The losses induced by PWM supply include
copper loss, hysteresis loss, and eddy-current loss. Among
them, eddy-current loss is the dominating loss mechanism according
to theoretical research and experiments carried out in
the past. [...] In this paper, based on the Fourier series analysis
of the PWM waveform, the eddy-current loss is calculated
under linear material assumption.
So yes, theory predicts that the losses in the motor are greater with PWM than if you supplied it with DC (of the average value of the PWM). (In those calculations they assume that PWM source voltage drops to 0 during the off period, but since the windings are indeed inductors they provide their own averaging affect).
And the paper also has some experimental data to back that up (however this part is somewhat confusing to me, see my comments after the following quote):
The experiment on a dc motor is also under no-load condition.
The dc motor is rated at 125V, 2 kW. The dc motor is tested with
pure dc power supply, as well as a full bridge dc–dc converter.
The difference between the input power of the dc motor under
PWM supply and under pure dc power supply is considered as
the loss induced by PWM converter. Fig. 14 shows the loss variation
with duty cycle at constant dc bus voltage of 100 V. The
calculated and measured iron losses are consistent and in both
measurement and calculations, the eddy-current losses reach the
maximum value at 0.5 duty ratio.
In order to study the effect of duty ratio on PWM loss under
the same rotation speed, the output dc voltage of PWM converter
is kept constant and the dc bus voltage is adjusted according
to the duty ratio. Fig. 15 shows the relationship of PWM loss
versus duty ratio at constant speed of 396 rpm. It can be seen
that from both measured and calculated losses, the extra PWM
iron losses decrease with the increase of duty ratio. Therefore, in
order to minimize eddy-current loss in dc machines, it is beneficial
to maintain a large duty ratio by varying the dc link voltage
to achieve the desired speed.
The discrepancy shown in Fig. 15 is believed to be attributed
by the exclusion of effects of PWM supply on hysteresis loss
and copper loss in the calculations, which are included in the
measurements.
It's not really clear to me if this "full bridge dc–dc converter" they used has its own output filter or not. Anyway those extra 15W losses on a 2kW motor is however under 1% (ok at 1/4 power so half duty it would be 3%)... so I suspect that's why you probably don't hear much about it. But it's not clear to me if that is with a pure PWM drive or if their DC-DC converter has its own filter, in which case the ripple sent to the motor would be a lot less... so the losses would perhaps be more significant without it. (Normally a full bridge dc–dc converter would have an LC output filter, but perhaps for the purpose of their experiment it doesn't have that so that it's the same signal shape [i.e. dropping to 0 when off] as in their theory section? I couldn't figure from the rest of the paper what this dc-dc converter was exactly.)
Actually one reason I couldn't find more materials is terminology. A lot of the motor people refer to PWM drive as "chopper controlled" instead. With that in mind, more can be found:
What PWM freq would be best from this motor utilisation standpoint?
As high frequency as possible it seems. That was among their conclusions & recommendations. However, back in the day they could only test up to 400Hz... so presumably more can be said; alas the much more recent paper (quoted previously) that used a dc-dc converter didn't mention its frequency.
This info on frequency choice is repeated in a 2004 book that explains it by "reduced harmonic content of the current at higher frequencies"", and which cites precisely that 1982 DOE/NASA report in support. Apparently no more recent research had been done. I'm hoping that with the trendy electric vehicles research nowadays, more would exist on this, but insofar I wasn't able to locate much more.
Best Answer
The SSRs you have linked use triacs to control the output. A triac is a semiconductor switch. When the triac is triggered on its gate it turns low resistance between its two anodes. Triacs have the odd characteristic that they will remain on after the gate signal is removed until the current falls below a very low hold-on current. This renders them almost useless in DC circuits but quite useful in AC circuits where the current falls to zero at every zero cross of the AC supply.
Figure 1. Common triac packages.
Figure 2. Triac symbol.
For on-off control the triac will be switched to give a load waveform as shown in Figure 3.
Figure 3. On-off AC time control.
Typically these circuits use zero-cross circuits to switch the load on at zero-cross to minimise electromagnetic interference. The triac itself switches off at the end of the next half-cycle. This approach works well for loads such as heaters which respond slowly to power.
For circuits which respond more quickly to pulses of power (such as lamps or motors) the on-off control gives too much flicker or jerk. In these cases phase angle control is used to vary the on-time of the AC supply to the load.
Figure 4. Phase-angle control.
Phase angle control requires instant-on (non zero-cross) SSRs but the control circuit needs to monitor the mains and give the pulses at the appropriate time relative to mains zero-cross.
You can't use PWM to control an SSR. The first pulse will turn it on and it will stay on until the next zero-cross.
What is the reason?
What relay? Heating of what?
You can't use PWM to control an SSR. You can only control to the nearest mains half-cycle.
You can't use PWM to control an SSR.
You can't use PWM to control an SSR.
Finally, you can't use PWM to control an SSR.
Update
Figure 5. The result of switching a triac with a PWM signal.
At the start of this answer I emphasised the statement, "Triacs have the odd characteristic that they will remain on after the gate signal is removed until the current falls below a very low hold-on current." This is key to the problem. In Figure 5 we can see that the triac output is off until the first PWM pulse is received. It then turns on and stays on until the next zero-cross regardless of the PWM switching. The result with continuous PWM will be that the triac turns on with the next PWM pulse after each zero-cross and stays on. Note the last half-cycle in the AC waveform. The triac is still on even though the PWM has stopped.