The base of the turbine blade maintains a constant temperature of 450°C and the tip is adiabatic is 1076.67° C.
Explanation:
The rate of heat transfer for the turbine blade is expressed.
[tex]Q=\sqrt{h P k A}\left(T_{b}-T_{\infty}\right) \tanh (m L)[/tex]
The parameters are calculated.
[tex]m=\sqrt{\frac{h P}{k A}}[/tex]
[tex]=\sqrt{\frac{\left(538 \mathrm{W} / \mathrm{m}^{2} \mathrm{K}\right)(0.11 \mathrm{m})}{(17 \mathrm{W} / \mathrm{m} \cdot \mathrm{K})\left(0.000513 \mathrm{m}^{2}\right)}}[/tex]
m = 82.38
Hence the rate of heat transfer for the turbine blade is
Q = [tex]\sqrt{(538)(0.11)(17)(0.000513)(450-1093)tanh(82.38)(0.053)}[/tex]
Q = -461.758 W
The temperature at the blade tip is calculated.
[tex]\frac{T_{x}-T_{\infty}}{T_{b}-T_{\infty}}[/tex] = [tex]\frac{1}{\cosh (m L)}[/tex]
[tex]\frac{T_{x}-1093^{\circ} \mathrm{C}}{450^{\circ} \mathrm{C}-1093^{\circ} \mathrm{C}}[/tex] = [tex]\frac{1}{\cosh (82.38 \times 0.053)}[/tex]
[tex]\frac{T_{x}-1093^{\circ} \mathrm{C}}{450^{\circ} \mathrm{C}-1093^{\circ} \mathrm{C}}[/tex] = 0.0254
[tex]T_{x}[/tex] = 1076.67° C
