Numerical Simulation of Compressor Stall Recovery Control

doi:10.16183/j.cnki.jsjtu.2020.261

Abstract

Abstract:

In order to explore the influencing factors of the effectiveness and stability for stall recovery and its flow mechanism when the antisurge valve is opened quickly, the dynamic stall recovery processes are simulated and the recovery processes at different discharging speeds are emphatically compared. Two numerical simulation methods, i.e., the distributed speed-changeable MG (Moore-Greitzer) model and the RANS (Reynolds Averaged Navier-Stokes) equation are used. The performance curves predicated by the two models agree well. The results of RANS show that the flow field changes are essentially the same when the valve is opened at different speeds. The disturbance, affected by the high-speed air flow generated at the inlet, moves downstream and finally reaches the leading edge of the rotor, whose scale will be further reduced with the impact of axial high-speed flow until completely dissipated. A comparison of different valve opening speeds indicates that the faster the valve is opened, the stronger the high-speed air flow generated at the inlet, shortening the stall recovery time. The greater the disturbance weakening degree, the faster the circumferential propagation speed of the disturbance, and the closer to the rotor speed. In the process of valve opening, the air flow fluctuation is more intense, and more energy is lost.

Key words: stall, antisurge valve control, distributed speed-changeable Moore-Greitzer model, numerical simulation, flow mechanism analysis

CLC Number:

V231.3

GAO Yuan, WU Yadong, OUYANG Hua. Numerical Simulation of Compressor Stall Recovery Control[J]. Journal of Shanghai Jiao Tong University, 2021, 55(11): 1343-1351.

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设计参数	数值
设计流量/(kg·s^-1)	6.8
额定转速/(r·min^-1)	3000
机匣半径/mm	300
轮毂半径/mm	225
转子叶片数	46
静子叶片数	44
动叶顶部间隙/mm	0.1

选取参数	数值
R/m	0.265
ρ/(kg·m^-3)	1.18
L_c/m	3
V_p/m³	0.424
A_c/m²	0.124
a_s/(m·s^-1)	340
I/(kg·m^-2)	0.211
U_d/(m·s^-1)	41.6

时刻	对应时间/转	时刻	对应时间/转
时刻	C=10	时刻	C=20
T₁	0.119	T'₁	0.119
T₂	0.494	T'₂	0.178
T₃	0.652	T'₃	0.405
T₄	0.711	T'₄	0.455
T₅	0.751	T'₅	0.484
T₆	0.968	T'₆	0.791