Question

A turbine 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.

Answer 1

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