Answer:
[tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
Explanation:
first write the newtons second law:
F[tex]_{s}[/tex]=δma[tex]_{s}[/tex]
Applying bernoulli,s equation as follows:
∑[tex]δp+\frac{1}{2} ρδV^{2} +δγz=0\\[/tex]
Where, [tex]δp[/tex] is the pressure change across the streamline and [tex]V[/tex] is the fluid particle velocity
substitute [tex]ρg[/tex] for {tex]γ[/tex] and [tex]g_{0}-cz[/tex] for [tex]g[/tex]
[tex]dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0[/tex]
integrating the above equation using limits 1 and 2.
[tex]\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
there the bernoulli equation for this flow is [tex]p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant[/tex]
note: [tex]ρ[/tex]=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular