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.

Please help. its very urgentW

3 a fin 30 cm long and 10 cm diameter throughout is made of steel alloy of thermal conductivity 43 w/m-k. the fin attached to a plane heated wall at 200 oc temp. extends into surroundings at 25 oc and heat transfer coefficient of 120 w/m2-k. find fin efficiency and fin effectiveness. assume that the tip of the fin is insulated and thermal radiation effect is negligible. a 1.11

Well, well, well, it seems like we've got ourselves into some hot and cold thermodynamics business! Buckle up, my friend, we're about to calculate some heat flow.

First things first, let's find the surface area of the fin. Since the diameter is 10mm (or 0.01m) and the length is 30cm (or 0.3m), the surface area can be calculated as:

Surface Area = π * D * L
= π * 0.01m * 0.3m
= 0.00942 m²

Now that we have the surface area, let's calculate the conduction heat transfer from the fin using Fourier's law of heat conduction:

Q_conduction = (k * A * ΔT) / L

where:
k is the thermal conductivity (43 W/m-deg)
A is the surface area (0.00942 m²)
ΔT is the temperature difference between the fin and surroundings (200°C - 25°C = 175°C)
L is the length of the fin (0.3m)

Plugging in the numbers:

Q_conduction = (43 W/m-deg * 0.00942 m² * 175°C) / 0.3m
= 440.677 W

So, the heat flow rate from the fin to the surroundings due to conduction is approximately 440.677 Watts. That's quite a hot fin, if you ask me!

To calculate the heat flow rate from the fin to the surroundings, we can use the fin equation:

Q = (k * A * ΔT) / (L + (h * A))

Where:
Q = Heat flow rate
k = Thermal conductivity of the fin material (43 W/m-deg)
A = Cross-sectional area of the fin
ΔT = Temperature difference between the fin base and the surroundings (200 deg.C - 25 deg.C)
L = Length of the fin (30 cm = 0.3 m)
h = Surface conductance (120 W/m^2-deg)

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

Area of a circle = π * (radius)^2
Radius = diameter / 2

Given that the diameter is 10 mm, which is equivalent to 0.01 m, we have:
Radius = 0.01 m / 2 = 0.005 m

Cross-sectional area = π * (0.005 m)^2

Next, we calculate the temperature difference:

ΔT = 200 deg.C - 25 deg.C = 175 deg.C

Now, we can substitute the values into the equation:

Q = (43 W/m-deg * π * (0.005 m)^2 * 175 deg.C) / (0.3 m + (120 W/m^2-deg * π * (0.005 m)^2))

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

To calculate the heat flow rate from the fin to the surroundings, we can use the concept of conduction. The heat flow rate can be determined using the formula:

Q = (k * A * (Th - 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)
Th = Temperature at the base of the fin (in deg C)
Ts = Temperature at the tip of the fin (in deg C)
L = Length of the fin (in meters)

Let's calculate step by step:

1) Calculate the cross-sectional area of the fin:
The cross-sectional area of a cylinder is given by the formula: A = π * r^2
Given the diameter of the fin is 10mm, the radius can be calculated as r = 10mm / 2 = 5mm = 0.005m
Therefore, A = π * (0.005m)^2 = 0.0000785 m^2

2) Calculate the temperature difference between the base and the tip of the fin:
Given the base temperature (Th) is 200 deg C and the tip temperature (Ts) is 25 deg C.
Therefore, (Th - Ts) = 200 deg C - 25 deg C = 175 deg C

3) Calculate the heat flow rate:
Given the thermal conductivity (k) is 43 W/m-deg and the length of the fin (L) is 30cm = 0.3m.
Therefore, Q = (43 W/m-deg * 0.0000785 m^2 * 175 deg C) / 0.3m
= 3.91125 W

Therefore, the heat flow rate from the fin to the surroundings is approximately 3.91125 Watts.