I don't know what you mean by "contemplates". I'll assume you mean "implements" or "operates in" CC or CV mode.
You stated that your DC-DC converter has a maximum output current of 2.7A. This implies that whatever output voltage the converter produces, you cannot ever expect more than 2.7A current from that output, at 0.8V, 10V, 20V or any voltage.
You also told us that the maximum output voltage that this converter can produce is 20V. This means that, in CC mode, whatever current you draw from the output, from 0A to 2.7A, the output voltage can never exceed 20V.
Given these constraints, the only thing you can know about power is the maximum available:
$$ P_{OUT(MAX)}=I_{OUT(MAX)} \times V_{OUT(MAX)}= 2.7A \times 20V = 54W $$
Without knowing characteristics of the load, you cannot possibly infer anything about the actual power that the load will draw. You can only make these statements:
The output voltage of the DC-DC converter will never exceed 20V.
The output current of the DC-DC converter will never exceed 2.7A.
This DC-DC converter can produce up to 54W of power, but not more.
In CV mode, it is the DC-DC converter load that decides voltage, and the load that determines the resulting current. In CC mode, it is the DC-DC converter that sets current, and load's own properties that determines what voltage would be required to pass that particular current. In both cases, to predict actual power, you need to know the characteristics of the load itself. You would require a formula or graph that yields the relationship between current through and voltage across that particular load.
In the case of your notebook, you set the converter to CV mode, at 19.55V. By experimentation you were able to determine that this device draws 2.7A when 19.55V is applied across it. With this information you can infer that the output power of the DC-DC converter, under those very specific conditions, is \$19.55V \times 2.7A = 53W\$, but this says nothing about what power would be drawn if you connected an LED or battery charger or banana across the output. To know that, you would have connect one, and measure the current and voltage.
The take-away here is that power draw, whether in CC or CV mode, isn't predictable unless you know the current vs. voltage characteristics of the load itself. You can't assume that 6V across the notebook would result in 8.8A of current flowing, for two reasons:
You don't know anything about the notebook's behaviour at 6V. You only know what it does at 19.55V.
This DC-DC converter can't produce more than 2.7A, you said so yourself.
As an example, let's use loads where we do know the exact characteristics. A resistor, and an LED. First the resistor. A resistor doesn't care what voltage we apply across it, and we have a very clear relationship between voltage \$V\$ across the resistor and the resulting current \$I\$ through it, a relationship called Ohm's law:
$$ V = I \times R $$
$$ I=\frac{V}{R} $$
I will use the DC-DC converter in CV mode, set the voltage to 12V, and attach a resistive load of 6Ω:
simulate this circuit – Schematic created using CircuitLab
The voltmeter shows that the supply is indeed applying 12V across the resistor load. We aren't in control of current; that's decided by the load which obeys Ohm's law:
$$ I = \frac{V}{R} = \frac{12V}{6\Omega} = 2A $$
That's shown on ammeter AM1. Now that we know both current \$I\$ through the load, and voltage \$V\$ across it, we can calculate power:
$$ P = IV = 2A \times 12V = 24W $$
Importantly, we know the relationship between current through and voltage across a resistor. It's Ohm's law, \$I=\frac{V}{R}\$. Therefore, even without measuring current we could have predicted this result:
$$ P = IV = \frac{V}{R}V = \frac{V^2}{R} = \frac{{(12V)}^2}{6\Omega} = 24W $$
Now I will reduce the load's resistance to 3Ω. According to Ohm's law, if we could successfully apply 12V across the load, the resulting current would be:
$$ I = \frac{V}{R} = \frac{12V}{3\Omega} = 4A $$
However, we know that the maximum current available is only 2.7A, and there's no way it can apply the full 12V across the load. When this 2.7A limit is reached, the DC-DC converter enters CC mode, and outputs whatever voltage is necessary to conform with this upper current limit. Again, this can be predicted with Ohm's law, but this time we know the current will be \$I=2.7A\$, and we wish to find the corresponding voltage \$V\$:
$$ V = IR = 2.7A \times 3\Omega = 8.1V $$
When the DC-DC converter is operating in constant current mode, it becomes a current source, instead of a voltage source, and in that role it adjusts \$V\$ to whatever voltage is necessary to produce 2.7A. Let's see if our prediction of 8.1V is correct, in a simulation:
simulate this circuit
The power being delivered to R is:
$$ P = IV = 2.7A \times 8.1V = 22W $$
This has nothing to do with the 54W maximum we calculated before, and nothing to do with the 24W that the 6Ω resistor drew. Power is therefore a function of both the load's own current/voltage relationship, and the limitations of the power supply.
Using an LED for a load:
simulate this circuit
We probably won't have a nice algebraic relationship to use, as we did with a resistor. Perhaps you have a graph of current versus voltage, like this:
When powering LEDs, generally we let the diode decide its voltage, and try to control current ourselves, instead of voltage. This will require the power supply (your DC-DC converter) to be operating in CC mode, with current limit set to some desired value, say 500mA. You would also want the supply to output up to its maximum of 20V. This does not mean it will output \$V=20V\$, it means that it will output whatever voltage is necessary to achieve \$I=500mA\$, up to a maximum of 20V.
From the graph above, the green markers represent the point corresponding to the voltage that will appear across the LED when 500mA is passing through it, which is 3.6V. That agrees nicely with the schematic above.
The power being received from the supply is now:
$$ P = IV = 0.5A \times 3.6V = 1.8W $$
Again I want to point out that this power value could have been predicted before you even connected the LED load, by referring to the graph, but you need the graph ahead of time. In other words, you need to know the current/voltage relationship for the load before you can make any predictions about power. Furthermore, that power is not related at all to calculations made for other loads, it's dependent entirely on this particular load.
Do you know the notebook computer's relationship between current and voltage? No, except for one data point: 19.55V → 2.7A. Worse still, that's at a particular moment in time, where its internal battery might or might not be charging, and the computer was performing a specific set of tasks. Of course this will change over time, so it doesn't even make sense to assume that your measurement of 2.7A at 19.55V will be the same 5 minutes later.
There are only three things you know for sure:
The output voltage of the DC-DC converter will never exceed 20V.
The output current of the DC-DC converter will never exceed 2.7A.
This DC-DC converter can produce up to 54W of power, but not more.
This information is not sufficient to make any predictions about power, except one; it can't be over 54W.
