in datasheets why measure of light output of power LEDs are given in lumens (luminous flux) but standard 5mm LEDs are given in mcd (luminous intensity)?
Luminous flux measures the total visible light emitted by the LED. It's a useful spec if you're going to use the LED to illuminate a room.
Luminous intensity measure the intensity of visible light emitted in a particular direction. It's a useful spec if you want to use the LED as an indicator and need to know if it will be visible against a bright background, or how bright it will appear when viewed directly, compared to other LEDs, possibly of other colors.
Since high power LEDs are often used to illuminate other objects, while 5 mm LEDs are often viewed directly (for example, when used as panel indicators), it makes sense to specify them differently.
Which type is more efficient?
You'd have to compare the output luminous flux (or intensity, depending how you will use the device) with the input power. Since the two devices aren't specified equivalently, you might have to experiment on samples of each type to find out.
If forward voltage is the same (same color) does that mean light output of 17.5 of 5mm LEDs equal 1 power LED?
Only if the efficiencies are the same.
The power supply is 3v at 60mAh (CC/CV)
A power supply can't force both the current and voltage to specific values at the same time. It can either force a certain voltage and let the load determine the current, or force a certain current and let the load determine the voltage.
Remember, the load has its own I-V characteristic that it must obey. For example, a resistive load obeys Ohm's Law \$V=IR\$.
What a 3-V 60-mA CC/CV supply does is force 3 V, unless doing so would require more than 60 mA, in which case it just provides 60 mA at whatever voltage (lower than 3 V) it takes to make that happen.
Assuming your LEDs' forward voltage is less than 3 V, your supply will operate as a 60-mA current supply, and the scenario will play out much as you have described it.
My scenario: 3v power supply directly from a battery.
A battery is more like a constant-voltage supply (with a series resistance), so this will indeed act differently than the constant-current scenario above.
Shouldn't the other two LEDs still draw 20mAh each?
Yes, they will continue to draw whatever it is they draw at 3 V. If they're the same LEDs as in the constant-current supply scenario (where we hypothesized that the forward voltage is lower than 3 V), then they will draw much more than 20 mA because you're driving them well above their rated forward voltage.
Also, as they heat or cool, their I-V curves change. So one might be okay running on a 3 V battery when first connected, but then start drawing more current as it warms up and eventually (or very quickly) burn itself out.
Boohoo, one LED breaks. That should be all. Why not?
With a constant voltage supply, one LED failing wouldn't affect the others.
With a real battery, you have to remember the internal resistance of the battery. If one LED fails, then that will tend to increase the actual output voltage of the battery after accounting for internal resistance, and that could cause the other LEDs to fail.
Best Answer
Start below 1 mA and observe the light output of the LED. Increase current until light is bright enough for your application, 3-5 mA typical. Non-illumination grade devices won't benefit from currents above ~10 mA so you shouldn't worry about overloading. Another good indicator is the temperature of the leads, when you can feel warmth with your fingers it's time to stop.
Be careful with violet emitters, they could be UV ones and emit just a small amount of visible light. Same for clear which conduct but not emit any light, that's likely IR ones. Old IR may have opaque germanium lens - don't put them in your mouth. Other than that, LEDs are safe to experiment with and quite inexpensive and if you're out of them just salvage some form old electronics.
Take a look at this too -> http://youtu.be/GAriT4B-gkA