Construction method of aero-engine thermodynamic model based on QAR data
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摘要:
为了能够建立更加接近真实运行环境的航空发动机热力学模型,提出了一种依据航空发动机运行QAR(quick access recorder)数据的热力学模型构建方法。依据发动机传统设计点热力学方程,利用最小二乘辨识原理,修正25站位压强求解方程,获得干空气热力学模型;根据混合气体熵值的可加性,构建了降雨工况下的湿空气热力学模型;最后结合遗传算法优化的粒子群算法,依据QAR数据进行干空气和湿空气热力学模型参数计算与验证。结果表明该热力学模型计算得到的干空气和湿空气热力学参数与QAR数据间最大误差小于13%,较为接近实际绝热参数。证实了基于QAR数据构建发动机热力学模型的可行性和该热力学模型构建方法求解热力参数的有效性。
Abstract:In order to establish a thermodynamic model of aero-engine in an approximate real operating environment, a thermodynamic model construction method based on QAR (quick access recorder) data of aero-engine operation was proposed. According to the thermodynamic equation of the traditional design point of the engine, and using the least square identification principle, the equation for solving the pressure at the 25 station was corrected, and the dry air thermodynamic model was obtained; according to the additivity of the entropy value of the mixed gas, a thermodynamic model of wet air under rainfall conditions was constructed; finally, combined with particle swarm optimization optimized by genetic algorithm, the parameters of dry air and wet air thermodynamic model were calculated and verified according to QAR data. The results showed that the maximum error between the dry air and wet air thermodynamic parameters calculated by the thermodynamic model and the QAR data was less than 13%, which was closer to the actual adiabatic parameters. The feasibility of constructing a thermodynamic model of an engine based on QAR data and the effectiveness of the thermodynamic model construction method for solving thermodynamic parameters were proved.
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表 1 CFM56-7B发动机站位信息
Table 1. CFM56-7B engine station information
站位序号 站位名 站位序号 站位名 0 发动机远前方 12 风扇进口 25′ 低压压气机出口 25″ 高压压气机进口 30 高压压气机出口 40 燃烧室出口 40′ 高压涡轮进口 45 高压涡轮出口 45′ 低压涡轮进口 50 低压涡轮出口 表 2 QAR记录热力参数
Table 2. QAR records thermal parameters
参数 变量符号 风扇进口温度/K $ {T_{12}} $ 风扇进口压强/Pa ${p_{12} }$ 高压压气机进口温度/K $ {T_{25}} $ 高压压气机出口温度/K $ {T_{30}} $ 高压压气机出口压强/Pa ${p_{30} }$ 燃油流量/($ {\text{kg}}/{\text{s}} $) $q_{m,{\rm{f}}}$ 尾喷口温度/K $ {T_{50}} $ 表 3 辨识方程系数
Table 3. Identification equation coefficients
模型阶数 ${p_{ {{{\text{25} } } }^{\prime \prime }}}$系数 ${p_{ {{{\text{25} } } }^\prime}}$系数 $ k - 1 $ 2.8628 0.4244 $ k - 2 $ −4.5039 −0.5258 $ k - 3 $ 5.1570 0.1169 $ k - 4 $ −4.4539 0 $ k - 5 $ 2.9498 0 $ k - 6 $ −1.3913 0 $ k - 7 $ 0.3557 0 表 4 数据截取条件
Table 4. Data interception conditions
参数名称 截取条件 相对转速/% >90 燃油流量/$ ({\text{kg} }/{\text{s} } ) $ >0.252 空地状态 AIR 油门杆角度/(°) >70 油门杆角度稳定时间/s >10 高度/m <1500 表 5 热力参数计算结果与QAR数据对比
Table 5. Comparison of thermal parameter calculation results with QAR data
时刻 热力
参数PSO-GA
计算数据QAR记录
原始数据相对
误差/%1 T25/K 387.58 398.768 2.81 T30/K 767.90 811.22 5.34 p30/Pa 1964952 2247770 12.58 2 T25/K 384.47 397.90 3.38 T30/K 758.58 813.04 6.70 p30/Pa 1985821.06 2233980 11.11 3 T25/K 404.98 392.78 3.01 T30/K 741.20 809.05 8.39 p30/Pa 1862370 2096080 11.15 表 6 燃烧室及涡轮部件计算参数
Table 6. Calculation parameters of combustor and turbine components
时刻 热力参数 PSO-GA计算数据 1 T40/K 1709.21 ${T_{{{40} }^\prime}}$/K 1661.44 p50/Pa 73648.43 ${q_{m,\text{a} } }$/$ (\text{kg}/{\text{s} }) $ 28.18 f 0.0330 2 T40/K 1686.67 ${T_{{{40} }^\prime }}$/K 1639.69 p50/Pa 65817.25 ${q_{m,\text{a} } }$/$ (\text{kg}/{\text{s} }) $ 58.73 f 0.0247 3 T40/K 1752.18 ${T_{{{40} }^\prime}}$/K 1701.83 p50/Pa 60030.30 ${q_{m,\text{a} } }$/$ (\text{kg}/{\text{s} }) $ 50.82 f 0.0581 表 7 湿空气热力学模型热力参数计算结果与QAR数据对比
Table 7. Comparing the calculation results of thermodynamic parameters of wet air thermodynamic model with QAR data
时刻 热力
参数/KPSO-GA
计算数据QAR记录
原始数据相对
误差/%1 T25 387 379.36 2.01 T30 757 773.51 2.13 2 T25 374 374.14 0.03 T30 774 767.15 0.89 3 T25 366 387.20 5.47 T30 749 803.14 6.74 表 8 湿空气热力学模型燃烧室及涡轮部件计算参数
Table 8. Calculation parameters of combustor and turbine components of wet air thermodynamic model
时刻 热力参数 PSO-GA计算数据 1 ${D_{ { {25} }^{\prime \prime } } }/ ({\text{kg} }/{\text{kg} } ) $ 0.0262 ${\varphi _{{{25} }^{\prime \prime }}}$/% 0.3263 D30/$ (\text{kg}/{\text{kg} }) $ 0.3762 $ {\varphi _{30}} $/% 0.0011 ${p_{ {{{\text{25} } } }^{\prime \prime }}}$/Pa 146895 T40/K 1667 ${T_{{{40} }^\prime }}$/K 1616 P50/Pa 68116 ${q_{ {m,\text{wet} } } }/ (\text{kg}/{\text{s} }) $ 36 f 0.0206 2 ${D_{{{25} }^{\prime \prime }}}$/$ (\text{kg}/{\text{kg} }) $ 0.0472 ${\varphi _{{{25} }^{\prime \prime }}}$/% 0.5363 D30/$ (\text{kg}/{\text{kg} }) $ 0.1044 $ {\varphi _{30}} $/% 0.1086 ${p_{ {{{\text{25} } } }^{\prime \prime } }}$/Pa 164219 T40/K 1548 ${T_{{{40} }^\prime }}$/K 1502 p50/Pa 74369 ${q_{ {m,\text{wet} } } }/ (\text{kg}/{\text{s} }) $ 39 f 0.0321 3 ${D_{{{25} }^{\prime \prime } }}$/$ (\text{kg}/{\text{kg} }) $ 0.0028 ${\varphi _{{{25} }^{\prime \prime } }}$/% 0.5460 D30/$ (\text{kg}/{\text{kg} }) $ 0.0287 $ {\varphi _{30}} $/% 0.1590 ${p_{ { { {\text{25} } } }^{\prime \prime } } }$/Pa 144000 T40/K 1601 ${T_{{{40} }^\prime}}$/K 1557 p50/Pa 64222 ${q_{ {m,\text{wet} } } }$/$ (\text{kg}/{\text{s} }) $ 34 f 0.0315 -
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