Theoretical Otto and Brayton cycles are reversible and have efficiency less than the Carnot efficiency, so appear to violate 2nd law. Help

The reversible Otto and Brayton cycles don’t operate between two fixed temperature reservoirs as does the Carnot cycle. That makes them less efficient.

Regarding your statement:

"The efficiency of all reversible engines operating between the same two temperatures is the same."

The key point is the reversible cycle has to operate between two fixed temperature reservoirs. The Otto and Brayton cycles do not operate between two fixed temperature reservoirs. They operate between a series of reservoirs between the maximum and minimum temperatures.

The Carnot efficiency is

$$eta =1-frac{T_L}{T_H}$$

Where $T_H$ is a single high temperature reservoir where all heat is taken from and $T_L$ is a single low temperature reservoir where all heat is rejected.

For the Otto and Brayton cycles $T_H$ is replaced by $T_{Hm}$ where $T_{Hm}$ is the mean temperature at which heat is taken in. That value is less than $T_H$.

Then $T_L$ is replaced by $T_{Lm}$ where $T_{Lm}$ is the mean temperature at which heat is rejected. That value is always greater than $T_L$ of the Carnot cycle.

So if you plug in the mean high and low temperatures of the Otto and Brayton cycles into the Carnot efficiency equation you will find that the efficiency is always less than the Carnot efficiency.

I have provided TS diagrams of the idealized reversible Otto and Brayton cycles gas cycles with that of the Carnot Cycle, for both heat engines and reverse heat engines, in case they may be of help to you.

The issue seems to be that reversing the inefficient engine creates a really efficient heat pump. So I know that's where the error lies, but not exactly what the error is. Can anyone clarify?

In fact, a less efficient heat engine does make a more "efficient" heat pump, and vice versa. This may see counter intuitive until you look at it closely. Consider the following:

First, the “efficiency” of the heat pump is actually called its Coefficient of Performance (COP) and equals the heat delivered to the high temperature environment divided by the work required. In your example, it would be 2500/250 = 10. If your reverse your heat pump cycle to make it a reversible heat engine cycle the efficiency of the heat engine would be 0.1. This means the relationship between the COP of the heat pump (HP) and the efficiency of the heat engine (HE) is

$$COP_{HP}=frac{1}{eta _{HE}}$$

Which gives the interesting, but important, result that the lower the efficiency of the heat engine the greater the COP of the heat engine when operating in reverse as a heat pump, and vice versa. As I said, this may seem counter intuitive until you look at it closely.

You know the efficiency of a reversible heat engine cycle is greater the larger the difference in temperature between the high and low temperature reservoirs. The greater that difference in temperature, the more output work you get for the same heat input.

On the other hand, for the heat pump the greater the difference in temperature the more work is required to move the same amount of heat from the low temperature environment to the high temperature environment. It’s kind of the equivalent of pulling a weight up a hill. The higher the hill, the more work required to move the same weight up the hill. Or, to put it another way, moving heat from a low temperature to a high temperature is not a natural thing, just like things don’t naturally roll up hill. It takes work. And the greater the temperature difference the harder it is (the more work required) to move the same amount of heat.

This is one of the reasons that heat pumps are used primarily in temperate climates where the difference between the indoor and outdoor temperature is not too great.

Hope this helps.