So doing a few *rough calculations I attempted to find out if the heat-sink I used for the design is big enough for the thermal load of the 20 Watt LED. 

So first I found a simplified equation to work off of. 

V= (Q*Rv)/Delta T 

V = Volume needed

Q = Heat source power in Watts 

Rv = Volumetric thermal resistance (amount of air flow)

Delta T = Tjunction - maximum ambient temperature

Ok so we are solving for V, For Q or Watts we have a 20Watt LED (calculating for a power efficiency of around 50%)

Rv I chose 200 which is about the middle of gentle to low noise air flow

Delta T came out to 40 °C with Tjuction at 100 °C and a max ambient air temp of 60 °C

Entering the above into the calculation:

V=(20*200)/40 solve for V and we get 100 cm³

So 100 cm³ is the size of heat-sink we need to dissipate the heat load of the LED. Now it's best to oversize the heat-sink so that we have a bit of play with airflow and to extend the life of the LED. 

We then need to calculate the volume of the heat-sink I chose. 

Volume is a bit more simple we use the following:

a*b*c

Yes I know it is simplified equation to properly account for the shape of a heat sink...

We use 6.3cm * 8.2cm * 7.0cm = 361.6  cm³ 

So with the fan on we have a over sized heat-sink by 261.6  cm³

Now what if the fan dies or is removed? We then use 700 for Rv (close to or natural convection)

V=(20*700)/40  solve for V and we get 350 cm³ needed for a heat-sink to handle the heat load

This tells us if the fan dies the heat-sink should be enough to handle the load of the LED at full brightness. A better calculation would also include the the junction between the LED. 

We can also calculate the luminous efficacy  of the LED, by taking Lumens (2741) divided by Watts (20) we get 137%. Though the specs seem a bit misleading on Watts, I may have to revisit the calculations. 

Apollo