Question

the cross sectional area of a horizontal tube increase along its length linearly show that:

Answer 1

$A=kl+c$

$p+fgh+\frac{1}{2}{fv}^2=constant$

$V_1{A}_1=V_2{A}_2$

$VAisconstant$

$V=\frac{C}{A}$

$△P+△fgh+\frac{1}{2}△{fv}^2=0$

$△P+\frac{1}{2}△f\frac{c^2}{\left\{{kx+c}^2\right\}}=0$

$△P+\frac{1}{2}f{c}^2△\frac{1}{kx+c^2}$

$p_2-p_1=-k△\frac{1}{kx+c^2}$

$p_2=p_1-k△\frac{1}{kx+c^2}$