Aturbine blade 6.3 cm long with cross sectional area 4.6x10-4 m2 and perimeter 0.12 m is made of stainless steel (k=18 w/mk). the temperature of the root is 482oc. the blade is exposed to a hot gas flow at 871oc with a heat transfer coefficient 454 w/m2k. determine the temperature of the blade tip and the rate of heat transfer at the root of the blade. tip could be assumed to be insulated.

There are two equations to be used for this problem. The firs one is the general formula for heat transfer:

Q = kAΔT
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area
ΔT is the temperature difference

Substituting the values:
Q = (454 W/m²·K)(4.6x10⁻⁴ m²)(871°C - 482°C)
Q = 81.24 W

Thus, the rate of heat transfer is 81.24 W.

We use the Q for the second equation. The radial heat transfer would be:

Q = 2πkL(T₁ - T₂)/ ln (r₂/r₁)
where 
L is the length of the turbine
r₂ is the distance of tip blade to the center
r₁ is the distance of root blade to the center
T₂ is temperature at the tip blade
T₁ is temperature at the root blade

Perimeter = 2πr₂
0.12 m = 2πr₂
r₂ = 0.019 m

Cross-sectional area = πr₁²
4.6x10⁻⁴ m² = πr₁²
r₁ = 0.012 m

Substituting to the equation:

81.24 W = 2π(454 W/m²·K)(6.3 cm * 1m/100 cm)(482°C - T₂)/ln (0.019 m/0.012 m)

Solving for T₂,
T₂ = 481.79°C

Examples: mental health problems cardiovascular disease obesity and other eating disordersmenstrual problems skin and hair problems