How to reduce 28 V to 24 V under load?
Added:

This is an intriguing problem BECAUSE it is a relatively simple one once all the facts are known BUT the more you look, the more you see that not all potentially relevant facts are yet known. I'm not going to dive down all the rabbit holes at this stage as Ray can easily add the additional information if he wishes and I'll wait until he does. But, here is what a few of the burrow entrances look like.

Ray said " ... At 28 V I'm out of specification for the fans ...".
That could mean that the voltage is too high or that they draw too much power there or ...
Which it is matters.

Ray said that the fans draw 36% more power at 28V compared to 24V. This would be true of a resistor (as P = V^2/R)_ but is less like;y to be true of a fan, but may be. Was this measured or assumed?

Ray stated that Itotal is 10.6 A at 24V and that was acceptable. How much lower is it as eg 22V and is the airflow then still acceptable.

PWM with out filtering at 85.7% duty cycle (call it 86%) and 26V in, yields effectively 24V mean out. 

PWM with no filtering imposes 28V pulses and 0-V "pulses" on the motor. This MAY cause the fans to draw pulses of current as if they had 28V applied for 86% of the time so Ifan average may be higher than if 24V DC was applied. This depends on the fan technology and other factors.

Some fans will be happy with PWM applied - although many would "like" an inverse flywheel diode. Some fans will go berserk with PWM applied and with some it may depend on PWM frequency.

If you filter PWM with and RC filter to make smoothish DC then you will lose power in the filter R as if it was a pure R dropper. The apparent R value is simply multiplied by R/Duty_cycle (as long as there are no PWM related second order effects).

If you filter with and LC filter you need an input flywheel diode for when PWM is off and you have a buck converter. This can work well but efficiency may be less improved over a straight dropper R than may be expected.

IF ... more rabbit holes ....

So:


What does " ...At 28 V I'm out of specification for the fans ..." mean?
What technology does the fan use? -> Brushed motor, Brushless DC,... ? 
If you PWM the fan at 24/28th duty cycle = 85.7% on, and 28V in, at PWM frame rates of eg 100 / 1000 / 10000 ... Hz with no smoothing will the fan controller act as if it "sees" 24V or pulsed 28V? 
What is the MEASURED fan current at eg 22 24 26 28 V?
What are the measured currents. 
Is air flow at 22 VDC at fan acceptable? 


Given  Ifan at 24V = 10.6A = "I24", Vin = 28V.

Wattage with dropper resistor = 28 x I24

Wattage with 90% efficient buck converter = 24 x I24 / 0.90 = 26.7 x I24

Wattage with 95% efficient buck converter = 24 x I24 / 0.95 = 25.3 x I24

Gains are shown in the table below. To get the power reduction of probably around 5% requires provision of a 20A+ , 30V+ rated switch a PWM source a 12A capable inductor, 10A diode, and output capacitor   OR a 100 Watt rated (ideally) resistor. The table shows the various relevant losses and gains. Whether the cost and effort is worth implementing a 90% efficient buck converter or trying even harder for a 95% buck converter is up to Ray to decide. 





Original - still relevant:

You say that 28V operation requires 36% more energy than 24V.
That would be true with a purely resistive load as \$power = V^2/R\$ so increases with voltage\$^2\$ and \$(28/24)^2 = 1.36\$. Depending on fan technology this may not be true - as fan motor back emf increases with velocity. 

However, if this is true then reducing fan voltage to 24V will reduce current and power so if you then resistively drop voltage from 28V you are still better off than using 28V on the fan directly.



If you PWM the fan directly it needs to be able to tolerate the 28V peaks and the technology used needs to be "happy" with the PWM modulation. If you smooth the PWM with an RC filter you lose about the same power as if you used a resistive dropper. ie completely smooth (impossible) is same power loss as a pure resistor and minimal smoothing = no resistive losses BUT motor characteristics will affect how the fan power changes.

Adding a series L that does not saturate and a 'flyback' diode after the PWM switch effectively yields a buck converter. 

Note that a linear regulator or voltage dropper will have an efficiency of 24V/28V = 85.7%
While you can get better than that with a buck converter you need to decide if the gain is worthwhile.  

At 90% you'd get 90/85.7 = 5.02% more run time per energy used.
At 95% you'd get 95/85.7 = 10.9% more run time per energy used.

24v x 10.6A = 254 Watts
28V x 10.6A = 296.8 Watts
The energy lost in a purely resistive dropper is 42.8 Watts = 14.42% (as you'd expect) of a total of 297W.

Using a buck converter you are unlikely to save more than perhaps another 20 Watts - halving losses to around 20 Watts.  Even with an exceptionally good converter you'd probably only an an extra 10 Watt's savings. 

It's quite likely that a resistive dropper is good enough.

If you can run the fan on say 22V you may dissipate more power in the dropper but use less energy overall. A good knowledge of fan Voltage/current/airflow characteristics and actual needs will help optimisation.