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:

RdRvT3(CpRvTgsigma+LRdgsigmaRdx2(CpRvT2+L2qvc)x(CpRdRvT2sigma+CpRvTgL2Rdepsilonqvc+L2RdqvcsigmaLRdg))

Note, in the code, γ is set to x!

last update: 2022/06/30