Using sympy to simplify following equation with regards to γ:
RdL∗[g(1+LqcvRdT)+qcvLεγδ+γ]=(RvRdT2γ+gRvT)(Cp+qcvL2RvT2)
RdL∗[g(RdT+Lqcv)(δ+γ)+qcvLεRdTγ]∗RvT2=(RvRdT2γ+gRvT)(CpRvT2+qcvL2)(δ+γ)RdT
RdL∗[g(RdT+Lqcv)(δ+γ)+qcvLεRdTγ]∗RvT2−(RvRdT2γ+gRvT)(CpRvT2+qcvL2)(δ+γ)RdT=0
python code:
from sympy import *
x, Rd, Rv, L, g, qvc, Cp, epsilon, sigma, T = symbols('x Rd Rv L g qvc Cp epsilon sigma T')
expr = Rd*L*((g*Rd*T + g*L*qvc)*(sigma + x) + qvc*L*epsilon*Rd*T*x)*Rv*T**2 - (Rv*Rd*T**2*x + g*Rv*T)*(Cp*Rv*T**2 + qvc*L**2)*(sigma + x)*Rd*T
expr2 = expr.expand() #expand expression
expr3 = collect(expr2, x) # collect expression with respect to x
simplify(expr3) # simplify expression with respect to x
result:
Rd∗Rv∗T3∗(−Cp∗Rv∗T∗g∗sigma+L∗Rd∗g∗sigma−Rd∗x2∗(Cp∗Rv∗T2+L2∗qvc)−x∗(Cp∗Rd∗Rv∗T2∗sigma+Cp∗Rv∗T∗g−L∗∗2∗Rd∗epsilon∗qvc+L∗∗2∗Rd∗qvc∗sigma−L∗Rd∗g))Note, in the code, γ is set to x!
last update: 2022/06/30