Effect of Flow Rate on Fan Temperature 1

[Damineh]

I am working on a fan system. The airflow rate through the fan will increase as you increase the rpm:
Q = (öND3) sin (á)/350

ö= 0.1
D=1
á=21 degrees

How would the temperature change as Q changes? Let’s say we have a few objects behind the fan which are affected by the flow rate increase. How would each component's temperature change with the increase in flow rate?

 

[DLANDISSR]

Damineh:
You need to know the fan efficiency, differential pressure across the fan and the friction loss from the fan to the items.

With these you can calculate the heat added due to fan inefficiency, compression heat and friction heat. Knowing the heat and air flow you can calculate the heat rise.

You would then need to know the velocity of the air at the component in question to calculate the film coefficient and heat transfer rate to the object from the air.

 

[Damineh]

Ok let me re-write my question. This is the same fan I have used in my above question:

I have a fan. It¡¦s running for hours (reaching steady-state temperature). As I increase the fan speed (rpm), the airflow rate is increasing. The increase in airflow rate, increases the dissipated heat of the component the fan is cooling; this in theory means that the component¡¦s steady-state temperature decreases. Now:

Dissipated heat due to convection:
Q(dissipated heat) = h[convection coefficient] * A[air-exposed area] * ( T[steady-state temp] ¡VT[ambient])

Dissipated heat due to airflow:
Q(dissipated heat) = Q[airflow] * ƒâƒnƒËdensity] * Cp * (T(outlet temperature)- T(ambient))

Equating these two equations:
Q[airflow] *ƒâƒnƒËdensity] * Cp * (T(outlet temperature)- T(ambient)) = h[convection coefficient] * A[air-exposed area] * ( T[steady-state temp] ¡VT[ambient])

Gives us Q is directly proportional to T(steady-state). Meaning as I increase the airflow rate, my temperature rises. Which we know is incorrect because as the airflow increases, the component gets cooler!!!! And I don¡¦t know why!

 

[quark]

Damineh!

I am sorry to say you are fundamentally confused. The heat gain by air at standard conditions is given by the equation Q = 1.08 x cfm x (T2-T1) where T2 is air exit temperature and T1 is air inlet temperature. For a constant Q and T1, cfm and T2 are variables. If you increase cfm, T2 will decrease and viceversa keeping the heat gain constant.

Now coming to your second term of the equation, for a fixed heat gain of air and a fixed initial object temperature, the final object temperature always remain constant (if you balanced the both equations)

Only thing you have to do is just supply sufficient air to remove the heat rejected by the object at a particular rate. More air flow increase h and cools the object faster.

Regrds,