A fin 30 cm long & 10mm diameters throughout is made of steel alloy of thermal conductivity 43 W/m-deg. the fin attached to a plane heated wall at 200 deg.C temp. extends into surroundings at 25 deg.C.& Unit surface conductance of 120W/m^2-deg. Workout the heat flow rate from the fin to surroundings.Presume that the tip of the fin is insulated & thermal radiation effects are negligible.

To find the heat flow rate from the fin to the surroundings, we can use the concept of thermal conduction through a fin.

The formula for heat flow rate through a fin is given by:

Q = (k * A * (Tb - Ts))/L

Where:
Q = Heat flow rate (in watts)
k = Thermal conductivity of the material (in W/m-deg)
A = Cross-sectional area of the fin (in m^2)
Tb = Base temperature (in deg C)
Ts = Surrounding temperature (in deg C)
L = Length of the fin (in m)

In this case, we are given:
k = 43 W/m-deg
A = (π * (d/2)^2), where d is the diameter of the fin
Tb = 200 deg C
Ts = 25 deg C
L = 30 cm = 0.3 m

First, let's calculate the cross-sectional area of the fin:

d = 10 mm = 0.01 m
A = π * (0.01/2)^2 = 0.00007854 m^2

Now, we can substitute the values into the formula:

Q = (43 * 0.00007854 * (200 - 25))/0.3
Q ≈ 0.615 W

So, the heat flow rate from the fin to the surroundings is approximately 0.615 watts.

To find the heat flow rate from the fin to the surroundings, we can use the formula for heat conduction through a fin:

Q = (k * A * ΔT) / (L1 + L2)

where:
Q is the heat flow rate
k is the thermal conductivity of the fin material
A is the cross-sectional area of the fin
ΔT is the temperature difference between the fin and the surroundings
L1 and L2 are the thermal resistances at the base and tip of the fin, respectively.

Let's calculate each of these values step-by-step.

1. Cross-sectional area of the fin (A):
The cross-sectional area of a fin can be calculated using the formula for the area of a cylinder:

A = π * (r^2)

where:
r is the radius of the fin, which is half the diameter.

Given that the diameter of the fin is 10 mm, we can calculate the radius:

r = 10 mm / 2 = 5 mm = 0.005 m

Therefore, the cross-sectional area of the fin is:

A = π * (0.005^2) = 0.00007854 m^2

2. Temperature difference (ΔT):
The temperature difference between the fin and the surroundings is:

ΔT = T1 - T2

where:
T1 is the temperature of the heated wall (200°C) and
T2 is the temperature of the surroundings (25°C).

Converting the temperatures to Kelvin:

T1 = 200°C + 273.15 = 473.15 K
T2 = 25°C + 273.15 = 298.15 K

Thus, the temperature difference is:

ΔT = 473.15 K - 298.15 K = 175 K

3. Thermal resistances (L1 and L2):
Since the tip of the fin is insulated and thermal radiation effects are negligible, we can assume that the thermal resistance at the tip (L2) is infinity. Therefore, the total thermal resistance of the fin is equal to the thermal resistance at the base (L1).

The thermal resistance of a fin can be calculated using the formula:

L = 1 / (h * A)

where:
h is the unit surface conductance provided (120 W/m^2-deg)
A is the cross-sectional area of the fin.

Therefore, the thermal resistance is:

L1 = 1 / (120 W/m^2-deg * 0.00007854 m^2) = 10874.78 K/W (Kelvin per Watt)

4. Heat flow rate (Q):
Finally, we can calculate the heat flow rate using the formula mentioned earlier:

Q = (k * A * ΔT) / (L1 + L2)

Since L2 is infinity, we can consider L2 = 0. Therefore:

Q = (k * A * ΔT) / L1

Given that the thermal conductivity of the fin material is 43 W/m-deg, we can calculate the heat flow rate:

Q = (43 W/m-deg * 0.00007854 m^2 * 175 K) / 10874.78 K/W

Calculating the above expression will give us the heat flow rate from the fin to the surroundings.