For a forced convection cooling of an axial flow turbine blade, the following data are given:
Gas temperature T_{\text{g}} = 1500 K
Coolant temperature at inlet T_{\text{cr}} = 320 K
Coolant mass flow rate \dot{m}_{\text{C}} = 10 kg/s
Gas flow rate \dot{m}_{\text{g}} = 33 kg/s
Blade height L = 0.1m
Gas heat transfer coefficient h_{\text{g}} = 1500 W/m² K
Wetted perimeter of blade profile S_{\text{g}} = 0.15 m
Parameter (h_{\text{g}}S_{\text{g}}/h_{\text{c}}S_{\text{c}}) = 2
It is required to
(a) Calculate and plot the temperature of the blade material and coolant along the blade height.
(b) Calculate the parameter \phi = T_{\text{G}}-T_{\text{M}}/T_{\text{G}}-T_{\text{C}} from Figure 14.39, where T_{\text{M}} is the maximum blade temperature at tip as calculated above, and T_{\text{C}} is the coolant temperature at the compressor outlet; here it is 695 K.
(c) What is its value for forced convection?
(d) Next use the same bleed mass flow rate ratio \dot{m}_{\text{c}}/\dot{m}_{\text{g}} to calculate the maximum blade temperature T_{\text{m}} at the blade tip for other cooling techniques shown in Figure 14.43.
TABLE 14.5
Variations of the Temperatures of Coolant, Blade, and Gas Along the Blade Height
l | l/L | T_{\text{C}} | T_{\text{b}} | T_{\text{g}} |
0.0 | 0.0 | 350 | 950 | 1200 |
0.01 | 0.1 | 350.36 | 950.1 | 1200 |
0.02 | 0.2 | 350.716 | 950.21 | 1200 |
0.03 | 0.3 | 351.074 | 950.316 | 1200 |
0.04 | 0.4 | 351.432 | 950.421 | 1200 |
0.05 | 0.5 | 351.789 | 950.526 | 1200 |
0.06 | 0.6 | 352.146 | 950.631 | 1200 |
0.07 | 0.7 | 352.504 | 950.736 | 1200 |
0.08 | 0.8 | 352.86 | 950.841 | 1200 |
0.09 | 0.9 | 353.218 | 950.946 | 1200 |
0.1 | 1 | 353.574 | 951.05 | 1200 |
1. The blade temperature is expressed by the relation (14.90), from which
T_{\text{b}}=T_{\text{cr}}+(T_{\text{g}}-T_{\text{cr}})\left[1-\frac{e^{-Kl/L}}{1+h_{\text{g}}S_{\text{g}}/h_{\text{c}}S_{\text{c}}} \right] =320+1180\left[1-\frac{e^{-Kl/L}}{3} \right] \\ \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{1500 \times 0.15}{1.0 \times 1005(1+2)} =0.0746 \\ \therefore T_{\text{b}}=320+1180\left[1-\frac{e^{-0.0746 \ l}}{3} \right] \quad \quad \quad (1)
The coolant temperature is to be calculated from the relation
T_{\text{C}}=T_{\text{g}}-(T_{\text{g}}-T_{\text{b}})[1+(h_{\text{s}}S_{\text{g}}/h_{\text{c}}S_{\text{c}})] \\ T_{\text{C}}=1500-3(1500-T_{\text{b}}) \quad \quad \rightarrow (2)
From Equations 1 and 2, T_{\text{b}} \text{ and } T_{\text{g}} are calculated at different blade heights
The blade and coolant temperature distribution along the blade height are plotted in Figure 14.43.
Variation of blade and coolant temperatures along the blade height.
2. Calculate the parameter (φ) and (T_{\text{M}})
\text{Since} \ \phi = \frac{T_{\text{G}}-T_{\text{M}}}{T_{\text{G}}-T_{\text{C}}} \\ T_{\text{M}}=T_{\text{G}}-\phi(T_{\text{G}}-T_{\text{C}})
Since \dot{m}_{\text{c}}/\dot{m}_{\text{g}} = 3%, from Figure 14.39, the value of Φ is determined and the maximum blade temperature for different cooling techniques is calculated and listed in Table 14.6.
As seen from the table, transpiration method gives the minimum value of the maximum temperature, which assures that it is the best cooling technique.
I(m) | T_{\text{b}}(K) | T_{\text{C}}(K) | |
0 | 1106.6 | 319.8 | Root |
0.025 | 1107.4 | 322.2 | |
0.05 | 1108.1 | 324.3 | Mean |
0.075 | 1108.8 | 326.4 | |
0.1 | 1109.9 | 329.7 | Tip |
TABLE 14.6 Variation of the Maximum Blade Temperature Based on Different Cooling Techniques |
|||
Cooling Technique | φ | \mathbf{T_{m}(K) } | |
1 | Simple radial flow | 0.395 | 1182 |
2 | L.E Impingement/no film | 0.466 | 1124.9 |
3 | Multipass/trip strips | 0.536 | 1068.5 |
4 | Film/cross flow impingement | 0.611 | 1008 |
5 | Transpiration | 0.69 | 944.6 |