Cooling of an electronic power transistor is assisted by attaching it to an aluminium heat sink. The electrical power to be dissipated is 0.3 watts, the exposed surface area of the heat sink is 0.001m2, the convection heat transfer coefficient is 6.0 W.m-2.K-1 and the ambient temperature is 20oC. Assuming steady state conditions and ignoring radiation effects, what will be the surface temperature of the heat sink?
How don't know how power is incorporated into this? Can someone pls explain how to do this problem, thx :)
Well, let's see if I can help you out, even though I'm more of a clown than a physicist. So, basically, you want to determine the surface temperature of the heat sink when it's helping to cool down that power transistor.
To do that, we can use a formula called Newton's Law of Cooling, which takes into account the convection heat transfer coefficient, among other things. In this case, we'll be using the formula:
Q = h * A * (Ts - Ta)
Here, Q is the amount of heat transferred (which is given as 0.3 watts), h is the convection heat transfer coefficient (6.0 W.m-2.K-1), A is the surface area of the heat sink (0.001m2), Ts is the surface temperature of the heat sink (what we're trying to find), and Ta is the ambient temperature (20oC).
To solve for Ts, we'll rearrange the equation:
Ts = (Q / (h * A)) + Ta
Now, let's plug in the values we have:
Ts = (0.3 / (6.0 * 0.001)) + 20
Ts = 0.05 + 20
Ts = 20.05oC
So, according to my calculations, the surface temperature of the heat sink should be approximately 20.05oC. Keep in mind that this is just an approximation and doesn't take into account any other factors, like radiation effects.
I hope that helps! Let me know if you have any more questions, or if you just want more clown-related jokes. 😄
To solve this problem, we can apply the principle of energy conservation. The power dissipated by the electronic power transistor is transferred to the heat sink and then dissipated into the surrounding environment through convection.
The heat transfer rate can be calculated using the formula:
Q = hc * A * (T_surface - T_ambient)
Where:
Q = Heat transfer rate (in watts)
hc = Convection heat transfer coefficient (in W.m-2.K-1)
A = Surface area of the heat sink (in m2)
T_surface = Surface temperature of the heat sink (in oC)
T_ambient = Ambient temperature (in oC)
Given:
Power to be dissipated (P) = 0.3 watts
Exposed surface area of the heat sink (A) = 0.001 m2
Convection heat transfer coefficient (hc) = 6.0 W.m-2.K-1
Ambient temperature (T_ambient) = 20 oC
We can rearrange the formula to solve for T_surface:
T_surface - T_ambient = Q / (hc * A)
Now, substituting the given values into the formula:
T_surface - 20 = 0.3 / (6.0 * 0.001)
Simplifying, we have:
T_surface - 20 = 50
T_surface = 20 + 50
T_surface = 70 oC
Therefore, the surface temperature of the heat sink will be 70 oC.
To solve this problem, we can use the principle of energy balance. The power dissipated by the electronic power transistor will be transferred to the heat sink and then to the surroundings through convection.
The energy balance equation is as follows:
Power dissipated by the transistor = Power transferred to the heat sink + Power transferred to the surroundings through convection
The power dissipated by the transistor is given as 0.3 watts.
The power transferred to the heat sink can be calculated using the equation:
Power transferred to the heat sink = (Heat transfer coefficient) x (Exposed surface area) x (Temperature difference)
Given:
- Heat transfer coefficient (h) = 6.0 W.m-2.K-1
- Exposed surface area (A) = 0.001 m2
- Temperature difference (ΔT) = (Surface temperature of the heat sink - Ambient temperature)
Now, the power transferred to the surroundings through convection is the same as the power transferred to the heat sink, as per the energy balance equation.
So, we can rewrite the energy balance equation as:
0.3 watts = (6.0 W.m-2.K-1) x (0.001 m2) x (Temperature difference) + (6.0 W.m-2.K-1) x (0.001 m2) x (Temperature difference)
Simplifying the equation:
0.3 watts = (12.0 x (Temperature difference)) x (0.001 m2)
Dividing by 0.012:
Temperature difference = 0.3 / 0.012 = 25
Now, we can calculate the surface temperature of the heat sink by adding the temperature difference to the ambient temperature:
Surface temperature of the heat sink = Ambient temperature + Temperature difference = 20°C + 25°C = 45°C
Therefore, the surface temperature of the heat sink will be 45°C.