Question 14.7: An axial flow turbine is to be cooled by forced convection. ......

The above data can be rewritten as

T_{\text{g}}=1200 \text{ K}, \quad T_{\text{cr}}=350 \text{ K}, \quad \dot{m}_{\text{g}}=50 \text{ kg/s}, \quad \dot{m}_{\text{c}}=1 \text{ kg/s}, \\ h_{\text{g}}=1200 \text{ W/m}^2 \text{/K}, \quad h_{\text{c}}=1000 \text{ W/m}^2 \text{/K} , \quad S_{\text{g}}=0.12 \text{ m}, \quad S_{\text{c}}=0.06 \text{ m} \\ L=0.1 \text{ m}

From Equation 14.85a

T_{\text{g}}-T_{\text{b}}=(T_{\text{g}}-T_{\text{br}})e^{-Kl/L} \quad \quad \quad (14.85\text{a}) \\ \frac{K}{L} =\frac{h_{\text{g}}S_{\text{g}}}{\dot{m}_{\text{c}}C_{\text{PC}}[1+(h_{\text{g}}S_{\text{g}}/h_{\text{c}}S_{\text{c}})]} =\frac{1200 \times 0.12}{1 \times 1005\left[1+\frac{1200 \times 0.12}{1000 \times 0.06} \right] } =0.04214

(i) The temperature of the coolant (T_{\text{c}}) is obtained from Equation 14.88, namely,

\frac{T_{\text{g}}-T_{\text{c}}}{T_{\text{g}}-T_{\text{cr}}} =e^{-Kl/L} \\ \frac{1200-T_{\text{c}}}{1200-350} =e^{-0.042141}

Thus, an expression for the coolant temperature along the blade height is expressed as

T_{\text{C}}=1200-850e^{-0.04214 l}

(ii) The Temperature of the blade (T_{\text{b}}) is obtained from Equation 14.90, or

\frac{T_{\text{b}}-T_{\text{cr}}}{T_{\text{g}}-T_{\text{cr}}} =1-\frac{e^{(-K/L)}l}{[1+(h_{\text{g}}S_{\text{g}}/h_{\text{c}}S_{\text{c}} )]} \\ \frac{T_{\text{b}}-350}{1200-350} =1-\frac{e^{-0.04214l}}{\left[1+\frac{1200 \times 0.12}{1000 \times 0.06} \right] }

Thus, the blade temperature at any point along its height is listed in Table 14.5.