a motor cycle cylinder consists of 10 fins ; each 15 cm outside diameter and inside diameter 7.5 cms. calculate rate of heat dissipation from cylinder fins when
(i) motorcycle is stationary
(ii) motorcycle is running at 60 km/hr.

atmospheric air is at 20 degree celsius and avg fin temip is 480 celsius. the relevant thermo-physical properties at avg tem of 250 celsius are
density = 0.674 kg/m^3 SPECIFIC HEAT = 1038 J/Kg K
thermal conductivity k = 0.427

prandtl no. Pr = 0.677

kinematic viscosity v = 40.61 X 10^(-6) meter square/s

B coefficient of cubical expanson = 1.912 X 10^(-3) per degree kelvin

Approx value of heat transfer coefficient may be evaluated by idealing the fins as single horizontal flate plate of same area . use significant length l = 0.9 d

Nu = 0.54 ( Gr Pr)^ (0.25) for laminar flow
Nu = 0.036 (Re)^(0.8) X (Pr)^(0.33) for turbulent flow

where Nu nussel no. Gr grashof no. Pr prandtl no. Re REYNOLDS NO.



plz solve the problem urgently. i have exam tomorrow

To calculate the rate of heat dissipation from the motorcycle cylinder fins, we need to analyze two scenarios: when the motorcycle is stationary and when it is running at 60 km/hr. We'll use the provided thermo-physical properties and formulas to find the heat transfer coefficients and ultimately the heat dissipation rates.

Understanding the Cylinder Fins

The motorcycle cylinder has 10 fins, each with an outside diameter of 15 cm and an inside diameter of 7.5 cm. The average temperature of the fins is 480 °C, while the atmospheric air temperature is 20 °C. The relevant properties at an average temperature of 250 °C are given, which we will use in our calculations.

Step 1: Calculate the Area of One Fin

First, we need to find the surface area of one fin. The fins can be approximated as flat plates. The outer diameter (D) is 0.15 m and the inner diameter (d) is 0.075 m. The length of the fin (L) can be assumed based on the average fin height, which we will take as 0.15 m for this calculation.

• Surface area of one fin, A = π * D * L
• A = π * 0.15 m * 0.15 m = 0.0707 m²

Step 2: Calculate the Total Surface Area of All Fins

Since there are 10 fins, the total surface area (A_total) is:

• A_total = 10 * A = 10 * 0.0707 m² = 0.707 m²

Step 3: Calculate the Grashof Number (Gr)

The Grashof number is calculated using the formula:

• Gr = (g * β * (T_f - T_a) * L^3) / ν²

Where:

• g = 9.81 m/s² (acceleration due to gravity)
• β = 1.912 x 10^(-3) K^(-1)
• T_f = 480 °C = 480 + 273.15 K
• T_a = 20 °C = 20 + 273.15 K
• L = 0.15 m
• ν = 40.61 x 10^(-6) m²/s

Substituting the values:

• Gr = (9.81 * 1.912 x 10^(-3) * (480 - 20) * (0.15)³) / (40.61 x 10^(-6))²
• Gr ≈ 1.45 x 10^6

Step 4: Calculate the Nusselt Number (Nu) for Stationary Condition

For laminar flow, we use:

• Nu = 0.54 * (Gr * Pr)^(0.25)

Calculating Prandtl number (Pr) using the given value:

• Pr = 0.677

Now substituting the values:

• Nu = 0.54 * (1.45 x 10^6 * 0.677)^(0.25)
• Nu ≈ 0.54 * (9.81 x 10^5)^(0.25) ≈ 0.54 * 31.5 ≈ 17.00

Step 5: Calculate the Heat Transfer Coefficient (h)

The heat transfer coefficient can be calculated as:

• h = (Nu * k) / L

Substituting the values:

• h = (17.00 * 0.427) / 0.15 ≈ 48.7 W/m²K

Step 6: Calculate the Rate of Heat Dissipation (Q) for Stationary Condition

The rate of heat dissipation can be calculated using:

• Q = h * A_total * (T_f - T_a)

Substituting the values:

• Q = 48.7 * 0.707 * (480 - 20) ≈ 48.7 * 0.707 * 460 ≈ 15,000 W

Moving Motorcycle at 60 km/hr

Now, we need to calculate the heat dissipation when the motorcycle is moving. First, we need to calculate the Reynolds number (Re).

Step 1: Calculate the Reynolds Number (Re)

Reynolds number is given by:

• Re = (ρ * U * L) / μ

Where:

• ρ = 0.674 kg/m³
• U = velocity = 60 km/hr = 60 * (1000/3600) m/s ≈ 16.67 m/s
• μ = ρ * ν = 0.674 * 40.61 x 10^(-6) ≈ 2.73 x 10^(-5) kg/m.s

Substituting the values:

• Re = (0.674 * 16.67 * 0.15) / (2.73 x 10^(-5)) ≈ 1.58 x 10^5

Step 2: Calculate Nusselt Number (Nu) for Turbulent Flow

For turbulent flow, we use:

• Nu = 0.036 * (Re)^(0.8) * (Pr)^(0.33)

Substituting the values:

• Nu = 0.036 * (1.58 x 10^5)^(0.8) * (0.677)^(0.33) ≈ 0.036 * 1.04 x