CN102565119A - Method for measuring cooling effect and thermal insulation effect of turbine blade with thermal barrier coating - Google Patents

Abstract

A method for measuring the cooling effect and the thermal insulation effect of a turbine blade with thermal barrier coating comprises the following steps: slotting the surface of a real turbine blade without a thermal barrier coating to embed a thermocouple, making a first cooling effect test under a selected test condition, then spraying thermal barrier coating on the turbine blade and making a second cooling effect test under the same condition with the first blade test. Comparison between the two tests can obtain the cooling effect of the blade with the thermal barrier coating and the thermal insulation effect of the thermal barrier coating on the real blade, so as to provide foundation for proving or improving the thermal barrier coating design of the blade. The method of the invention has the following advantages: the cooling effect of the blade with the thermal barrier coating can be measured conveniently, the comparison between the cooling effects of the blade with and without the thermal barrier coating can be realized visually, and the thermal insulation of the thermal barrier coating on the real blade can be obtained, thereby being convenient to validate and evaluate the design of the blade and the thermal barrier coating.

Description

A kind of method of measuring band thermal barrier coating turbo blade cooling effect and effect of heat insulation

Technical field

The present invention relates to coating effect of heat insulation detection method, a kind of method of measuring band thermal barrier coating turbo blade cooling effect and effect of heat insulation is provided especially.

Background technology

The turbo blade of gas-turbine unit is faced with harsh Service Environment; Common gas turbine blades serviceability temperature scope is more than 1000 ℃, and the aeromotor fuel gas temperature reaches more than 1500 ℃.For the thrust that improves engine and the thermal efficiency of fuel, must improve the temperature of combustion of fuel, the serviceability temperature of turbo blade also improves day by day; But receive the influence of high temperature alloy fusing point and mechanical property; The air capacity that is used for the high-performance enginer cooling simultaneously also is limited; In this case; Thermal barrier coating is applied in turbine blade surface, turbo blade matrix and thermal-flame are separated, thereby play the vital role of protecting blade, reduction leaf temperature, improving engine efficiency.

In the turbo blade design phase, the foundation of well designed can verified and provide to the test of carrying out the related turbine blade to the blade design; Wherein the blade of being with thermal barrier coating is carried out cooling effect is tested and the thermal barrier coating effect of heat insulation is tested for thermal barrier coating design and application very vital role is arranged.

Effect of heat insulation is one of important indicator of estimating the thermal barrier coating performance, and coating effect of heat insulation test at present mainly is to carry out on the test piece surface, can not quantitatively embody the protective effect to real blade of thermal barrier coating under engine operating condition.The turbo blade of being with thermal barrier coating is not carried out the cooling effect test at present yet, can't understand the influence of the application of thermal barrier coating specific turbo blade cooling effect.

Summary of the invention

The objective of the invention is in order to detect the cooling and the effect of heat insulation of band thermal barrier coating turbo blade exactly, the spy provides a kind of method of measuring band thermal barrier coating turbo blade cooling effect and effect of heat insulation.

The invention provides a kind of method of measuring band thermal barrier coating turbo blade cooling effect and effect of heat insulation, it is characterized in that: the method for described mensuration band thermal barrier coating turbo blade cooling effect and effect of heat insulation is:

True turbine blade surface fluting not spraying thermal barrier coating is buried thermopair underground; Carry out first round cooling effect test at the optional test state; To blade spraying thermal barrier coating, under the situation consistent, carry out second and take turns test afterwards with first round blade trystate; Contrast test can obtain blade cooling effect and the effect of heat insulation of thermal barrier coating on real blade with thermal barrier coating thus, perhaps improves the design of blade thermal barrier coating for checking foundation is provided;

First aspect provides the technological approaches that the turbo blade of band thermal barrier coating is carried out cooling effect test and effect of heat insulation test; Be included in the turbine blade surface that does not spray thermal barrier coating and choose some test points, fluting is buried the thermopair of measuring wall temperature underground, at fluting position built-up welding thermal conductivity and the approaching high temperature alloy of blade matrix, and carries out polishing; Under selected operating mode, carry out first round cooling effect test; Special; After accomplishing first round test, need carry out necessary excision and protection work to the test lead of thermopair, itself does not remove thermopair; Wait to accomplish thermal barrier coating spraying back to test lead recovery assembling, thereby guaranteed the consistance of two-wheeled test measuring point; Technical requirement according to the design point thermal barrier coating sprays thermal barrier coating at blade surface; Under the situation consistent, carry out second and take turns the cooling effect test then with the selected operating mode of first round test; Through contrasting the temperature value of identical measuring point under the identical operating mode, can intuitively find the heat-blocking action of thermal barrier coating to blade; Thereby and can on the basis of measured value, do further to analyze to obtain the variation that turbo blade has or not thermal barrier coating state cooling effect;

Second aspect proposes according to the principle of similitude turbo blade of being with thermal barrier coating to be carried out the cooling effect test, and obtains the method for thermal barrier coating at the effect of heat insulation of blade surface through test; The implication of the physical descriptor that need use when test method is described is seen table 1;

Table 1 physical descriptor and footnote and implication

When the exerciser condition can't be simulated true engine operating condition fully, need carry out the cooling effect test at a lower temperature, confirm the cooling effect calculation expression thus, extrapolate the cooling effect under the actual engine operating mode on this basis.The derivation of confirming the cooling effect calculation expression is following:

Get blade wall micro unit, negligible axial and radially heat conduction, then the monobasic thermal balance equation is:

${\left(\mathrm{hF}\right)}_g(T_g-T_{w,g})={\left(\mathrm{\λF}\right)}_w\frac{1}{\mathrm{\δ}}(T_{w,g}-T_{w,c})={\left(\mathrm{hF}\right)}_c(T_{w,c}-T_c)\·\·\·$ (1)

Promptly $\frac{{\left(\mathrm{HF}\right)}_g}{{\left(\mathrm{HF}\right)}_c}=\frac{T_{w,c}-T_c}{T_g-T_{w,g}}\·\·\·$ (2)

$\frac{\mathrm{\δ}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\λF}\right)}_w}=\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}\·\·\·$ (3)

Make relative cooling effect $\mathrm{\θ}=\frac{T_g-T_{w,g}}{T_{w,g}-T_c}\·\·\·$ (4)

Then $\mathrm{\θ}=\frac{1}{\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}+\frac{T_{w,c}-T_c}{T_g-T_{w,g}}}\·\·\·$ (5)

$\mathrm{\θ}=\frac{1}{\frac{\mathrm{\δ}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\λF}\right)}_w}+\frac{{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{hF}\right)}_c}}\·\·\·$ (6)

Can find out that by (6) the cooling effect θ of blade depends on that combustion gas is to the ratio

of the coefficient of heat conductivity of the coefficient of heat transfer of the ratio

of the coefficient of heat transfer of cold air and combustion gas and blade material and the latter is depended on the coefficient of heat conductivity and the temperature range of blade material to a great extent; Study the turbo blade structure of using for us, the heat conduction item is very little to the influence of cooling effect, usually in test accuracy; For for simplicity, ignore the heat conduction item; Suppose that promptly blade is a thin wall vane, inside and outside wall temperature is even, and then thermal balance equation can be simplified to:

h _g(T _g-T _w)＝h _c(T _w-T _c)………………………………………………(7)

Promptly $\frac{h_c}{h_g}=\frac{T_g-T_w}{T_w-T_c}=\mathrm{\θ}\·\·\·\left(8\right)$

Combustion gas and cold gas are to the heat exchanging relation of wall:

Nu＝C·Re ⁿ……………………………………………………………(9)

$\mathrm{Nu}=\frac{\mathrm{hd}}{\mathrm{\λ}}\·\·\·$ (10)

Be the cold air side: $h_c={\left(C\frac{\mathrm{\λ}}{d}{\mathrm{Re}}^n\right)}_c\·\·\·$ (11)

The combustion gas side: $h_g={\left(C\frac{\mathrm{\λ}}{d}{\mathrm{Re}}^m\right)}_g\·\·\·$ (12)

(11)/(12), $\frac{h_c}{h_g}=\frac{C_c}{C_g}\·\frac{{\mathrm{\λ}}_c}{{\mathrm{\λ}}_g}\·\frac{d_c}{d_g}\·\frac{{{\mathrm{Re}}^n}_c}{{{\mathrm{Re}}^n}_g}\·\·\·\left(13\right)$

substitution equation is got

$\frac{h_c}{h_g}=\frac{C_c}{C_g}\·{\left(\frac{F_g}{F_c}\right)}^n\·\frac{{\mathrm{\λ}}_c}{{\mathrm{\λ}}_g}\·{\left(\frac{d_g}{d_c}\right)}^{1-n}{\left(\frac{{\mathrm{\μ}}_g}{{\mathrm{\μ}}_c}\right)}^n{\left(\frac{G_c}{G_g}\right)}^n\·{\mathrm{Re}}_g^{n-m}\·\·\·$ (14)

(14) physical parameter of combustion gas and cold air in

$\frac{{\mathrm{\λ}}_c}{{\mathrm{\λ}}_g}\·{\left(\frac{{\mathrm{\μ}}_g}{{\mathrm{\μ}}_c}\right)}^n={\left(\frac{T_g}{T_c}\right)}^z\·\·\·$ (15)

(14) constant term in can be merged into $\frac{C_c}{C_g}\·{\left(\frac{F_g}{F_c}\right)}^n\·{\left(\frac{d_g}{d_c}\right)}^{1-n}=C\·\·\·$ (16)

Then (14) formula can be organized into $\mathrm{\θ}=\frac{h_c}{h_g}=C\·{\left(\frac{G_c}{G_g}\right)}^n\·{\left(\frac{T_g}{T_c}\right)}^z\·{\mathrm{Re}}_g^{n-m}\·\·\·$ (17)

Perhaps $\mathrm{\θ}=C\·{\left(K_G\right)}^n\·{\left(K_T\right)}^z\·{\mathrm{Re}}_g^{n-m}\·\·\·$ (18)

Need to explain how to replace high temperature and pressure test, two kinds of trystates is similar during its theoretical foundation below with lower temperature, pressure test; Its simulated condition is: geometric similarity, kinematic similarity, kinematic similarity and heat similarity; Because testpieces is exactly the actual engine blade, guaranteed geometric similarity; Kinematic similarity and kinematic similarity regulation, the two states of being studied, all pour points have the distribution pattern of similar mobile graphic and similar power; Dynamics and thermodynamic similarity regulation have similar velocity field and temperature field in flowing;

If two states guarantee that following dimensionless parameter equates, then obtains kinematics and kinematic similarity;

$\mathrm{Cp}=\frac{P}{{\mathrm{\&rho;<figure-callout\; id="2"\; label="V"\; filenames="HDA0000038045930000011.png"\; state="\{\{state\}\}">V</figure-callout>}}^2}\&CenterDot;\&CenterDot;\&CenterDot;$ (19)

$\mathrm{Re}=\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (20)

Pressure coefficient Cp is the ratio of pressure and inertial force, and Reynolds number (Re) is the ratio of inertial force and friction force; Therefore the Reynolds number of two states and pressure coefficient equate, have guaranteed the kinematics and the kinematic similarity of two states;

The Prandtl number of two states (Pr) equates, has guaranteed heat similarity;

$\mathrm{Pr}=\frac{\mathrm{Cp}\&CenterDot;\mathrm{\&mu;}}{\mathrm{\&lambda;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (21)

Prandtl number is a physical parameter, and in the cooling effect process of the test, the Prandtl number of combustion gas and cooling air changes very little, can think constant; Therefore two kinds of trystates are guaranteeing that it is similar just can to satisfy two states under the situation that pressure coefficient, this two other dimensionless group of Reynolds number equate; In fact in the cooling effect test, these two dimensionless groups comprise fuel gas temperature, pressure and flow, cold air temperature, pressure and flow; Because cold air is discharged in the combustion gas and goes, so cold air pressure depends on cold air temperature, flow and combustion gas state; In fact have only 5 from variable element;

Keep the combustion gas Reynolds number and the pressure coefficient of two states to equate, can get by equation (20):

${\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (22)

Replace the ρ V in the equation (22) with G/Fg, put in order

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (23)

Use

Replace the ρ V in the equation (19) ², put in order

${\left(\frac{P}{G\sqrt{T}}\right)}_{g,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (24)

Merging (23) (24) gets

$P_{g,L}=P_{g,H}{\left(\frac{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_L}{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_H}\right)}_g\&CenterDot;\&CenterDot;\&CenterDot;$ (26)

Show that (26) when the fuel gas temperature of the fuel gas temperature of known or the selected condition of high temperature and pressure and low-temperature condition, the gaseous-pressure of the needed low-temperature condition of simulation test can be confirmed; Fuel gas temperature, pressure confirm that flow is confirmed thereupon;

For cold gas system, make two states similar, can handle by the method for similar gas burning system; The Reynolds number and the pressure coefficient that guarantee the cold gas system of two states equate, obtain following equation:

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ …(27)

()/() obtains the ratio of combustion gas to the cold air parameter

${\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_L={\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_H\&CenterDot;\&CenterDot;\&CenterDot;$ (28)

Keep the cold air of these two states equal, i.e. (G to the throughput ratio of combustion gas _c/ G _g) _L=(G _c/ G _g) _H, can obtain the relational expression of two state viscosity:

μ _c，L＝μ _c，H(μ _g，L/μ _g，H)……………………………………………………(29)

If known or when having selected the cold air temperature of fuel gas temperature and the condition of high temperature of the condition of high temperature and low-temperature condition, can obtain the cold air coefficient of viscosity of required low-temperature condition by equation (29), and then obtain cold air temperature;

Needed cold air pressure also can write out following equation:

${\left(\frac{P}{G\sqrt{T}}\right)}_{c,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (30)

In case after cold air temperature and flow and combustion gas state were confirmed, cold air pressure promptly was determined value; The cold air that guarantees the low-temperature condition and the condition of high temperature during test is equal to the throughput ratio of combustion gas, required cold air flow in the time of can obtaining low-temperature test; For two kinds of trystates being studied, state parameter carries out modelling as stated above and calculates, and can realize the similar of two states, and the cooling effect of two similar states equates also, promptly

$A={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ (31)

The expression-form that sometimes also is organized into relative cooling effect also is feasible;

$\mathrm{\&theta;}={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ (32)

Obtain local cooling effect A by low-temperature condition _nWith average cooling effect

Can obtain condition of high temperature parameter substitution (31) the corresponding topical wall temperature and the mean wall temperature of the condition of high temperature; Promptly

T _w，H，n＝T _g，H-(T _g-T _c) _H·A _n……………………………………………(32)

$\stackrel{\&OverBar;}{T_{w,H}}=T_{g,H}-{(T_g-T_c)}_H\&CenterDot;\stackrel{\&OverBar;}{A}\&CenterDot;\&CenterDot;\&CenterDot;$ (33)

Utilize formula (32) (33) to be converted into the wall temperature value under the condition of high temperature wall surface temperature that records under the low-temperature condition (comprising the gentle mean wall temperature of partial wall); Relatively under the identical test state; Not with thermal barrier coating with the band thermal barrier coating the wall temperature value, its difference is heat insulation temperature value; So just can simulate the effect of heat insulation of coating under engine condition more intuitively.

Advantage of the present invention:

Can conveniently record the cooling effect of the blade of band thermal barrier coating state; And can contrast the not cooling effect of the blade of band coating state intuitively; And obtain the effect of heat insulation of thermal barrier coating on real blade, thereby easily the design conditions of blade and thermal barrier coating is verified and estimated.

Description of drawings

Below in conjunction with accompanying drawing and embodiment the present invention is done further detailed explanation:

Fig. 1 is typical thermal barrier coating structural drawing;

Fig. 2 is that the tangential thermopair of blade is arranged synoptic diagram, and blade only illustrates the part;

Fig. 3 is that blade footpath thermoelectric couple is arranged synoptic diagram, and blade only illustrates the part;

Fig. 4 buries the thermopair partial schematic diagram for the blade fluting;

1 is the high-temperature alloy blades body portion among Fig. 1, and 2 is metal bonding coating, and 3 is ceramic topcoats;

4 is the position of thermopair layout points among Fig. 2;

5 is the position of blade footpath thermoelectric couple layout points among Fig. 3;

6 is thermopair among Fig. 4, and 7 use high temperature alloy for built-up welding.

Embodiment

Embodiment 1

Present embodiment provides a kind of method of measuring band thermal barrier coating turbo blade cooling effect and effect of heat insulation, and it is characterized in that: the method for described mensuration band thermal barrier coating turbo blade cooling effect and effect of heat insulation is:

True turbine blade surface fluting not spraying thermal barrier coating is buried thermopair underground; Carry out first round cooling effect test at the optional test state; To blade spraying thermal barrier coating, under the situation consistent, carry out second and take turns test afterwards with first round blade trystate; Contrast test can obtain blade cooling effect and the effect of heat insulation of thermal barrier coating on real blade with thermal barrier coating thus, perhaps improves the design of blade thermal barrier coating for checking foundation is provided;

First aspect provides the technological approaches that the turbo blade of band thermal barrier coating is carried out cooling effect test and effect of heat insulation test; Be included in the turbine blade surface that does not spray thermal barrier coating and choose some test points, fluting is buried the thermopair of measuring wall temperature underground, at fluting position built-up welding thermal conductivity and the approaching high temperature alloy of blade matrix, and carries out polishing; Under selected operating mode, carry out first round cooling effect test; Special; After accomplishing first round test, need carry out necessary excision and protection work to the test lead of thermopair, itself does not remove thermopair; Wait to accomplish thermal barrier coating spraying back to test lead recovery assembling, thereby guaranteed the consistance of two-wheeled test measuring point; Technical requirement according to the design point thermal barrier coating sprays thermal barrier coating at blade surface; Under the situation consistent, carry out second and take turns the cooling effect test then with the selected operating mode of first round test; Through contrasting the temperature value of identical measuring point under the identical operating mode, can intuitively find the heat-blocking action of thermal barrier coating to blade; Thereby and can on the basis of measured value, do further to analyze to obtain the variation that turbo blade has or not thermal barrier coating state cooling effect;

Second aspect proposes according to the principle of similitude turbo blade of being with thermal barrier coating to be carried out the cooling effect test, and obtains the method for thermal barrier coating at the effect of heat insulation of blade surface through test; The implication of the physical descriptor that need use when test method is described is seen table 1.

When the exerciser condition can't be simulated true engine operating condition fully, need carry out the cooling effect test at a lower temperature, confirm the cooling effect calculation expression thus, extrapolate the cooling effect under the actual engine operating mode on this basis.The derivation of confirming the cooling effect calculation expression is following:

Get blade wall micro unit, negligible axial and radially heat conduction, then the monobasic thermal balance equation is:

${\left(\mathrm{hF}\right)}_g(T_g-T_{w,g})={\left(\mathrm{\&lambda;F}\right)}_w\frac{1}{\mathrm{\&delta;}}(T_{w,g}-T_{w,c})={\left(\mathrm{hF}\right)}_c(T_{w,c}-T_c)\&CenterDot;\&CenterDot;\&CenterDot;$ (1)

Promptly $\frac{{\left(\mathrm{HF}\right)}_g}{{\left(\mathrm{HF}\right)}_c}=\frac{T_{w,c}-T_c}{T_g-T_{w,g}}\&CenterDot;\&CenterDot;\&CenterDot;$ (2)

$\frac{\mathrm{\&delta;}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\&lambda;F}\right)}_w}=\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}\&CenterDot;\&CenterDot;\&CenterDot;$ (3)

Make relative cooling effect $\mathrm{\&theta;}=\frac{T_g-T_{w,g}}{T_{w,g}-T_c}\&CenterDot;\&CenterDot;\&CenterDot;$ (4)

Then $\mathrm{\&theta;}=\frac{1}{\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}+\frac{T_{w,c}-T_c}{T_g-T_{w,g}}}\&CenterDot;\&CenterDot;\&CenterDot;$ (5)

$\mathrm{\&theta;}=\frac{1}{\frac{\mathrm{\&delta;}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\&lambda;F}\right)}_w}+\frac{{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{hF}\right)}_c}}\&CenterDot;\&CenterDot;\&CenterDot;$ (6)

Can find out that by (6) the cooling effect θ of blade depends on that combustion gas is to the ratio

of the coefficient of heat conductivity of the coefficient of heat transfer of the ratio

of the coefficient of heat transfer of cold air and combustion gas and blade material and the latter is depended on the coefficient of heat conductivity and the temperature range of blade material to a great extent; Study the turbo blade structure of using for us, the heat conduction item is very little to the influence of cooling effect, usually in test accuracy; For for simplicity, ignore the heat conduction item; Suppose that promptly blade is a thin wall vane, inside and outside wall temperature is even, and then thermal balance equation can be simplified to:

h _g(T _g-T _w)＝h _c(T _w-T _c)………………………………………………(7)

Promptly $\frac{h_c}{h_g}=\frac{T_g-T_w}{T_w-T_c}=\mathrm{\&theta;}\&CenterDot;\&CenterDot;\&CenterDot;\left(8\right)$

Combustion gas and cold gas are to the heat exchanging relation of wall:

Nu＝C·Re ⁿ……………………………………………………………(9)

$\mathrm{Nu}=\frac{\mathrm{hd}}{\mathrm{\&lambda;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (10)

Be the cold air side: $h_c={\left(C\frac{\mathrm{\&lambda;}}{d}{\mathrm{Re}}^n\right)}_c\&CenterDot;\&CenterDot;\&CenterDot;$ (11)

The combustion gas side: $h_g={\left(C\frac{\mathrm{\&lambda;}}{d}{\mathrm{Re}}^m\right)}_g\&CenterDot;\&CenterDot;\&CenterDot;$ (12)

(11)/(12), $\frac{h_c}{h_g}=\frac{C_c}{C_g}\&CenterDot;\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;\frac{d_c}{d_g}\&CenterDot;\frac{{{\mathrm{Re}}^n}_c}{{{\mathrm{Re}}^n}_g}\&CenterDot;\&CenterDot;\&CenterDot;\left(13\right)$

substitution equation is got

$\frac{h_c}{h_g}=\frac{C_c}{C_g}\&CenterDot;{\left(\frac{F_g}{F_c}\right)}^n\&CenterDot;\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;{\left(\frac{d_g}{d_c}\right)}^{1-n}{\left(\frac{{\mathrm{\&mu;}}_g}{{\mathrm{\&mu;}}_c}\right)}^n{\left(\frac{G_c}{G_g}\right)}^n\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (14)

(14) physical parameter of combustion gas and cold air in

$\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;{\left(\frac{{\mathrm{\&mu;}}_g}{{\mathrm{\&mu;}}_c}\right)}^n={\left(\frac{T_g}{T_c}\right)}^z\&CenterDot;\&CenterDot;\&CenterDot;$ (15)

(14) constant term in can be merged into $\frac{C_c}{C_g}\&CenterDot;{\left(\frac{F_g}{F_c}\right)}^n\&CenterDot;{\left(\frac{d_g}{d_c}\right)}^{1-n}=C\&CenterDot;\&CenterDot;\&CenterDot;$ (16)

Then (14) formula can be organized into $\mathrm{\&theta;}=\frac{h_c}{h_g}=C\&CenterDot;{\left(\frac{G_c}{G_g}\right)}^n\&CenterDot;{\left(\frac{T_g}{T_c}\right)}^z\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (17)

Perhaps $\mathrm{\&theta;}=C\&CenterDot;{\left(K_G\right)}^n\&CenterDot;{\left(K_T\right)}^z\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (18)

Need to explain how to replace high temperature and pressure test, two kinds of trystates is similar during its theoretical foundation below with lower temperature, pressure test; Its simulated condition is: geometric similarity, kinematic similarity, kinematic similarity and heat similarity; Because testpieces is exactly the actual engine blade, guaranteed geometric similarity; Kinematic similarity and kinematic similarity regulation, the two states of being studied, all pour points have the distribution pattern of similar mobile graphic and similar power; Dynamics and thermodynamic similarity regulation have similar velocity field and temperature field in flowing;

If two states guarantee that following dimensionless parameter equates, then obtains kinematics and kinematic similarity;

$\mathrm{Cp}=\frac{P}{{\mathrm{\&rho;<figure-callout\; id="2"\; label="V"\; filenames="HDA0000038045930000011.png"\; state="\{\{state\}\}">V</figure-callout>}}^2}\&CenterDot;\&CenterDot;\&CenterDot;$ (19)

$\mathrm{Re}=\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (20)

Pressure coefficient Cp is the ratio of pressure and inertial force, and Reynolds number (Re) is the ratio of inertial force and friction force; Therefore the Reynolds number of two states and pressure coefficient equate, have guaranteed the kinematics and the kinematic similarity of two states;

The Prandtl number of two states (Pr) equates, has guaranteed heat similarity;

$\mathrm{Pr}=\frac{\mathrm{Cp}\&CenterDot;\mathrm{\&mu;}}{\mathrm{\&lambda;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (21)

Prandtl number is a physical parameter, and in the cooling effect process of the test, the Prandtl number of combustion gas and cooling air changes very little, can think constant; Therefore two kinds of trystates are guaranteeing that it is similar just can to satisfy two states under the situation that pressure coefficient, this two other dimensionless group of Reynolds number equate; In fact in the cooling effect test, these two dimensionless groups comprise fuel gas temperature, pressure and flow, cold air temperature, pressure and flow; Because cold air is discharged in the combustion gas and goes, so cold air pressure depends on cold air temperature, flow and combustion gas state; In fact have only 5 from variable element;

Keep the combustion gas Reynolds number and the pressure coefficient of two states to equate, can get by equation (20):

${\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (22)

Replace the ρ V in the equation (22) with G/Fg, put in order

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (23)

Use

Replace the ρ V in the equation (19) ², put in order

${\left(\frac{P}{G\sqrt{T}}\right)}_{g,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (24)

Merging (23) (24) gets

$P_{g,L}=P_{g,H}{\left(\frac{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_L}{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_H}\right)}_g\&CenterDot;\&CenterDot;\&CenterDot;$ (26)

Show that (26) when the fuel gas temperature of the fuel gas temperature of known or the selected condition of high temperature and pressure and low-temperature condition, the gaseous-pressure of the needed low-temperature condition of simulation test can be confirmed; Fuel gas temperature, pressure confirm that flow is confirmed thereupon;

For cold gas system, make two states similar, can handle by the method for similar gas burning system; The Reynolds number and the pressure coefficient that guarantee the cold gas system of two states equate, obtain following equation:

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ …(27)

()/() obtains the ratio of combustion gas to the cold air parameter

${\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_L={\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_H\&CenterDot;\&CenterDot;\&CenterDot;$ (28)

Keep the cold air of these two states equal, i.e. (G to the throughput ratio of combustion gas _c/ G _g) _L=(G _c/ G _g) _H, can obtain the relational expression of two state viscosity:

μ _c，L＝μ _c，H(μ _g，L/μ _g，H)……………………………………………………(29)

If known or when having selected the cold air temperature of fuel gas temperature and the condition of high temperature of the condition of high temperature and low-temperature condition, can obtain the cold air coefficient of viscosity of required low-temperature condition by equation (29), and then obtain cold air temperature;

Needed cold air pressure also can write out following equation:

${\left(\frac{P}{G\sqrt{T}}\right)}_{c,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (30)

In case after cold air temperature and flow and combustion gas state were confirmed, cold air pressure promptly was determined value; The cold air that guarantees the low-temperature condition and the condition of high temperature during test is equal to the throughput ratio of combustion gas, required cold air flow in the time of can obtaining low-temperature test; For two kinds of trystates being studied, state parameter carries out modelling as stated above and calculates, and can realize the similar of two states, and the cooling effect of two similar states equates also, promptly

$A={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ (31)

The expression-form that sometimes also is organized into relative cooling effect also is feasible;

$\mathrm{\&theta;}={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ (32)

Obtain local cooling effect A by low-temperature condition _nWith average cooling effect

Can obtain condition of high temperature parameter substitution (31) the corresponding topical wall temperature and the mean wall temperature of the condition of high temperature; Promptly

T _w，H，n＝T _g，H-(T _g-T _c)H·A _n……………………………………………(32)

$\stackrel{\&OverBar;}{T_{w,H}}=T_{g,H}-{(T_g-T_c)}_H\&CenterDot;\stackrel{\&OverBar;}{A}\&CenterDot;\&CenterDot;\&CenterDot;$ (33)

Utilize formula (32) (33) to be converted into the wall temperature value under the condition of high temperature wall surface temperature that records under the low-temperature condition (comprising the gentle mean wall temperature of partial wall); Relatively under the identical test state; Not with thermal barrier coating with the band thermal barrier coating the wall temperature value, its difference is heat insulation temperature value; So just can simulate the effect of heat insulation of coating under engine condition more intuitively.

Need carry out suitable test repacking to blade before the test, generally be employed near the tangential layout in the edge of blade characteristic cross-section than multi-measuring point, as shown in Figure 2, this is in order to obtain the Temperature Distribution of blade characteristic cross-section.Also should arrange several measuring points on demand along blade radial, thereby obtain the blade radial Temperature Distribution, as shown in Figure 3.As shown in Figure 4, the test repacking need be opened the groove of the certain width and the degree of depth at the selected measuring point place of blade, buries the thermopair that is used for surveying wall temperature underground.After having buried thermopair underground, should be at some temperature conductivitys of fluting place built-up welding and the approaching metal of blade matrix material, such benefit is that thermopair is fixing on blade, avoids damaging in test.After built-up welding is accomplished, need be to the surfacing part fairing of polishing, purpose is to keep blade level and smooth, can be because of burying the thermoelectric aeroperformance that influence combustion gas occasionally underground.

Testpieces just can carry out first round cooling effect test through after burying galvanic couple and other test repacking work underground.Should be noted that testpieces repacking should not carry out when blade surface sprays as yet, and should under the optional test state, carry out first round cooling effect test after the repacking; The benefit of doing like this is conspicuous; Gathered the blade surface wall temperature data of no thermal barrier coating under this state, just can with carry out second subsequently and take turns the data contrast of the band coating that test gathers, front and back two-wheeled data all are identical point positions; The identical test state, thereby have stronger comparability.In the cooling effect process of the test,, also need measure the state parameter of combustion gas and cold air except the wall surface temperature data of blade.In the two-wheeled process of the test of front and back, the state parameter of adjustment combustion gas and cold air should be consistent, has guaranteed identical trystate thus.

After accomplishing first round cooling effect test, the test lead of thermopair is done to extract protection handle in the joint, purpose is to avoid test lead in thermal barrier coating spraying and heat treatment process, to damage.After the cleaning of blade surface before spraying, just can spray certain thickness thermal barrier coating at blade surface according to the state of the art that designs.After spraying is accomplished, test lead is carried out functional rehabilitation in the joint, can carry out second according to the method for first round test and state and take turns cooling effect and test.

In the test of two-wheeled cooling effect, the thermopair that is embedded on the turbo blade collects the two groups of wall temperature measurement data T in front and back _{W, x, 1}And T _{W, x, 2}

According to the wall surface temperature of being surveyed, calculate the local cooling effect A of cooling effect front and back two-wheeled test according to equation (31) _{1, x}And A _{2, x}, and average cooling effect With

And calculate the blade wall temperature under the condition of high temperature: comprise the blade wall temperature T that is not with thermal barrier coating according to equation (32) (33) _{W, H, x, 1}, mean wall temperature The blade wall temperature T of band thermal barrier coating _{W, H, x, 2}, mean wall temperature

Can obtain the effect of heat insulation Δ T of thermal barrier coating thus.

The effect of heat insulation Δ T of blade partial points _x=T _{W, H, x, 1}-T _{W, H, x, 2}

Average effect of heat insulation $\mathrm{\&Delta;}\stackrel{\&OverBar;}{T}=\stackrel{\&OverBar;}{T_{w,H,1}}-\stackrel{\&OverBar;}{T_{w,H,2}}$

This example has been introduced the turborotor cooling effect and the effect of heat insulation test of band thermal barrier coating; But should be noted that; In the time can't being rotated the test simulation of state, also can carry out the cooling effect and the effect of heat insulation test of turbine rotor blade according to the method described above.Test unit also is not included in this example, yet can be according to this routine described method, and test reaches the physical quantity of regulating in test and controlling as required, comes the appropriate design test unit.

Claims (2)

1. measure the method for being with thermal barrier coating turbo blade cooling effect and effect of heat insulation for one kind, it is characterized in that: the method for described mensuration band thermal barrier coating turbo blade cooling effect and effect of heat insulation is:

True turbine blade surface fluting not spraying thermal barrier coating is buried thermopair underground; Carry out first round cooling effect test at the optional test state; To blade spraying thermal barrier coating, under the situation consistent, carry out second and take turns test afterwards with first round blade trystate; Contrast test can obtain blade cooling effect and the effect of heat insulation of thermal barrier coating on real blade with thermal barrier coating thus, perhaps improves the design of blade thermal barrier coating for checking foundation is provided;

First aspect provides the technological approaches that the turbo blade of band thermal barrier coating is carried out cooling effect test and effect of heat insulation test; Be included in the turbine blade surface that does not spray thermal barrier coating and choose some test points, fluting is buried the thermopair of measuring wall temperature underground, at fluting position built-up welding thermal conductivity and the approaching high temperature alloy of blade matrix, and carries out polishing; Under selected operating mode, carry out first round cooling effect test; Special; After accomplishing first round test, need carry out necessary excision and protection work to the test lead of thermopair, itself does not remove thermopair; Wait to accomplish thermal barrier coating spraying back to test lead recovery assembling, thereby guaranteed the consistance of two-wheeled test measuring point; Technical requirement according to the design point thermal barrier coating sprays thermal barrier coating at blade surface; Under the situation consistent, carry out second and take turns the cooling effect test then with the selected operating mode of first round test; Through contrasting the temperature value of identical measuring point under the identical operating mode, intuitively find the heat-blocking action of thermal barrier coating to blade, thereby and can on the basis of measured value, do further to analyze to obtain the variation that turbo blade has or not thermal barrier coating state cooling effect;

Second aspect proposes according to the principle of similitude turbo blade of being with thermal barrier coating to be carried out the cooling effect test, and obtains the method for thermal barrier coating at the effect of heat insulation of blade surface through test;

When the exerciser condition can't be simulated true engine operating condition fully, need carry out the cooling effect test at a lower temperature, confirm the cooling effect calculation expression thus, extrapolate the cooling effect under the actual engine operating mode on this basis.

2. according to the method for the described mensuration band of claim 1 thermal barrier coating turbo blade cooling effect and effect of heat insulation, it is characterized in that: the derivation of described definite cooling effect calculation expression is following:

Get blade wall micro unit, negligible axial and radially heat conduction, then the monobasic thermal balance equation is:

${\left(\mathrm{hF}\right)}_g(T_g-T_{w,g})={\left(\mathrm{\&lambda;F}\right)}_w\frac{1}{\mathrm{\&delta;}}(T_{w,g}-T_{w,c})={\left(\mathrm{hF}\right)}_c(T_{w,c}-T_c)\&CenterDot;\&CenterDot;\&CenterDot;$ (1)

Promptly $\frac{{\left(\mathrm{HF}\right)}_g}{{\left(\mathrm{HF}\right)}_c}=\frac{T_{w,c}-T_c}{T_g-T_{w,g}}\&CenterDot;\&CenterDot;\&CenterDot;$ (2)

$\frac{\mathrm{\&delta;}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\&lambda;F}\right)}_w}=\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}\&CenterDot;\&CenterDot;\&CenterDot;$ (3)

Make relative cooling effect $\mathrm{\&theta;}=\frac{T_g-T_{w,g}}{T_{w,g}-T_c}\&CenterDot;\&CenterDot;\&CenterDot;$

Then $\mathrm{\&theta;}=\frac{1}{\frac{T_{w,g}-T_{w,c}}{T_g-T_{w,g}}+\frac{T_{w,c}-T_c}{T_g-T_{w,g}}}\&CenterDot;\&CenterDot;\&CenterDot;$ (5)

$\mathrm{\&theta;}=\frac{1}{\frac{\mathrm{\&delta;}{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{\&lambda;F}\right)}_w}+\frac{{\left(\mathrm{hF}\right)}_g}{{\left(\mathrm{hF}\right)}_c}}\&CenterDot;\&CenterDot;\&CenterDot;$ (6)

Can find out that by (6) the cooling effect θ of blade depends on that combustion gas is to the ratio

of the coefficient of heat conductivity of the coefficient of heat transfer of the ratio of the coefficient of heat transfer of cold air and combustion gas and blade material and the latter is depended on the coefficient of heat conductivity and the temperature range of blade material to a great extent; Suppose that blade is a thin wall vane, inside and outside wall temperature is even, and then thermal balance equation can be simplified to:

h _g(T _g-T _w)＝h _c(T _w-T _c)………………………………………………(7)

Promptly $\frac{h_c}{h_g}=\frac{T_g-T_w}{T_w-T_c}=$

(8)

Combustion gas and cold gas are to the heat exchanging relation of wall:

Nu＝C·Re ⁿ……………………………………………………………(9)

$\mathrm{Nu}=\frac{\mathrm{hd}}{\mathrm{\&lambda;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (10)

Be the cold air side: $h_c={\left(C\frac{\mathrm{\&lambda;}}{d}{\mathrm{Re}}^n\right)}_c\&CenterDot;\&CenterDot;\&CenterDot;$ (11)

The combustion gas side: $h_g={\left(C\frac{\mathrm{\&lambda;}}{d}{\mathrm{Re}}^m\right)}_g\&CenterDot;\&CenterDot;\&CenterDot;$ (12)

(11)/(12), $\frac{h_c}{h_g}=\frac{C_c}{C_g}\&CenterDot;\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;\frac{d_c}{d_g}\&CenterDot;\frac{{{\mathrm{Re}}^n}_c}{{{\mathrm{Re}}^n}_g}$ (13)

substitution equation is got

$\frac{h_c}{h_g}=\frac{C_c}{C_g}\&CenterDot;{\left(\frac{F_g}{F_c}\right)}^n\&CenterDot;\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;{\left(\frac{d_g}{d_c}\right)}^{1-n}{\left(\frac{{\mathrm{\&mu;}}_g}{{\mathrm{\&mu;}}_c}\right)}^n{\left(\frac{G_c}{G_g}\right)}^n\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (14)

(14) physical parameter of combustion gas and cold air in

$\frac{{\mathrm{\&lambda;}}_c}{{\mathrm{\&lambda;}}_g}\&CenterDot;{\left(\frac{{\mathrm{\&mu;}}_g}{{\mathrm{\&mu;}}_c}\right)}^n={\left(\frac{T_g}{T_c}\right)}^z\&CenterDot;\&CenterDot;\&CenterDot;$ (15)

(14) constant term in can be merged into $\frac{C_c}{C_g}\&CenterDot;{\left(\frac{F_g}{F_c}\right)}^n\&CenterDot;{\left(\frac{d_g}{d_c}\right)}^{1-n}=C\&CenterDot;\&CenterDot;\&CenterDot;$ (16)

Then (14) formula can be organized into $\mathrm{\&theta;}=\frac{h_c}{h_g}=C\&CenterDot;{\left(\frac{G_c}{G_g}\right)}^n\&CenterDot;{\left(\frac{T_g}{T_c}\right)}^z\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (17)

Perhaps $\mathrm{\&theta;}=C\&CenterDot;{\left(K_G\right)}^n\&CenterDot;{\left(K_T\right)}^z\&CenterDot;{\mathrm{Re}}_g^{n-m}\&CenterDot;\&CenterDot;\&CenterDot;$ (18)

Need to explain how to replace high temperature and pressure test, two kinds of trystates is similar during its theoretical foundation with lower temperature, pressure test; Its simulated condition is: geometric similarity, kinematic similarity, kinematic similarity and heat similarity; Because testpieces is exactly the actual engine blade, guaranteed geometric similarity; Kinematic similarity and kinematic similarity regulation, the two states of being studied, all pour points have the distribution pattern of similar mobile graphic and similar power; Dynamics and thermodynamic similarity regulation have similar velocity field and temperature field in flowing;

If two states guarantee that following dimensionless parameter equates, then obtains kinematics and kinematic similarity;

$\mathrm{Cp}=\frac{P}{{\mathrm{\&rho;V}}^2}\&CenterDot;\&CenterDot;\&CenterDot;$ (19)

$\mathrm{Re}=\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (20)

Pressure coefficient Cp is the ratio of pressure and inertial force, and Reynolds number (Re) is the ratio of inertial force and friction force; Therefore the Reynolds number of two states and pressure coefficient equate, have guaranteed the kinematics and the kinematic similarity of two states;

The Prandtl number of two states (Pr) equates, has guaranteed heat similarity;

$\mathrm{Pr}=\frac{\mathrm{Cp}\&CenterDot;\mathrm{\&mu;}}{\mathrm{\&lambda;}}\&CenterDot;\&CenterDot;\&CenterDot;$ (21)

Prandtl number is a physical parameter, and in the cooling effect process of the test, the Prandtl number of combustion gas and cooling air changes very little, can think constant; Therefore two kinds of trystates are guaranteeing that it is similar just can to satisfy two states under the situation that pressure coefficient, this two other dimensionless group of Reynolds number equate; In fact in the cooling effect test, these two dimensionless groups comprise fuel gas temperature, pressure and flow, cold air temperature, pressure and flow; Because cold air is discharged in the combustion gas and goes, so cold air pressure depends on cold air temperature, flow and combustion gas state; In fact have only 5 from variable element;

Keep the combustion gas Reynolds number and the pressure coefficient of two states to equate, can get by equation (20):

${\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{\mathrm{\&rho;Vd}}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (22)

Replace the ρ V in the equation (22) with G/Fg, put in order

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (23)

Use

Replace the ρ V in the equation (19) ², put in order

${\left(\frac{P}{G\sqrt{T}}\right)}_{g,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{g,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (24)

Merging (23) (24) gets

$P_{g,L}=P_{g,H}{\left(\frac{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_L}{{\left(\mathrm{\&mu;}\sqrt{T}\right)}_H}\right)}_g\&CenterDot;\&CenterDot;\&CenterDot;$ (26)

Show that (26) when the fuel gas temperature of the fuel gas temperature of known or the selected condition of high temperature and pressure and low-temperature condition, the gaseous-pressure of the needed low-temperature condition of simulation test can be confirmed; Fuel gas temperature, pressure confirm that flow is confirmed thereupon;

For cold gas system, make two states similar, can handle by the method for similar gas burning system; The Reynolds number and the pressure coefficient that guarantee the cold gas system of two states equate, obtain following equation:

${\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,L}={\left(\frac{G}{\mathrm{\&mu;}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (27)

()/() obtains the ratio of combustion gas to the cold air parameter

${\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_L={\left(\frac{{(G/\mathrm{\&mu;})}_g}{{(G/\mathrm{\&mu;})}_c}\right)}_H\&CenterDot;\&CenterDot;\&CenterDot;$ (28)

Keep the cold air of these two states equal, i.e. (G to the throughput ratio of combustion gas _c/ G _g) _L=(G _c/ G _g) _H, can obtain the relational expression of two state viscosity:

μ _c，L＝μ _c，H(μ _g，L/μ _g，H)…………………………………………………(29)

If known or when having selected the cold air temperature of fuel gas temperature and the condition of high temperature of the condition of high temperature and low-temperature condition, can obtain the cold air coefficient of viscosity of required low-temperature condition by equation (29), and then obtain cold air temperature;

Needed cold air pressure also can write out following equation:

${\left(\frac{P}{G\sqrt{T}}\right)}_{c,L}={\left(\frac{P}{G\sqrt{T}}\right)}_{c,H}\&CenterDot;\&CenterDot;\&CenterDot;$ (30)

In case after cold air temperature and flow and combustion gas state were confirmed, cold air pressure promptly was determined value; The cold air that guarantees the low-temperature condition and the condition of high temperature during test is equal to the throughput ratio of combustion gas, required cold air flow in the time of can obtaining low-temperature test; For two kinds of trystates being studied, state parameter carries out modelling as stated above and calculates, and can realize the similar of two states, and the cooling effect of two similar states equates also, promptly

$A={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_g-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ …(31)

The expression-form that sometimes also is organized into relative cooling effect also is feasible;

$\mathrm{\&theta;}={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_H={\left(\frac{T_g-T_w}{T_w-T_c}\right)}_L\&CenterDot;\&CenterDot;\&CenterDot;$ (32)

Obtain local cooling effect A by low-temperature condition _nWith average cooling effect

Can obtain condition of high temperature parameter substitution (31) the corresponding topical wall temperature and the mean wall temperature of the condition of high temperature; Promptly

T _w，H，n＝T _g，H-(T _g-T _c) _H·A _n………………………………………………(32)

$\stackrel{\&OverBar;}{T_{w,H}}=T_{g,H}-{(T_g-T_c)}_H\&CenterDot;\stackrel{\&OverBar;}{A}\&CenterDot;\&CenterDot;\&CenterDot;$ …(33)

Utilize formula (32) (33) to be converted into the wall temperature value under the condition of high temperature wall surface temperature that records under the low-temperature condition (comprising the gentle mean wall temperature of partial wall); Relatively under the identical test state; Not with thermal barrier coating with the band thermal barrier coating the wall temperature value, its difference is heat insulation temperature value.

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