The cross sectional of a horizontal tube increase along its length linearly,

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Solution

$A = k l + c$
$p + f g h + \frac{1}{2} {f v}^{2} = c o n s t a n t$
$V_{1} A_{1} = V_{2} A_{2}$
$V A i s c o n s t a n t$
$V = \frac{C}{A}$
$△ P + △ f g h + \frac{1}{2} △ {f v}^{2} = 0$
$△ P + \frac{1}{2} △ f \frac{c^{2}}{{{k x + c}^{2}}} = 0$
$△ P + \frac{1}{2} f c^{2} △ \frac{1}{k x + c^{2}}$
$p_{2} - p_{1} = - k △ \frac{1}{k x + c^{2}}$
$p_{2} = p_{1} - k △ \frac{1}{k x + c^{2}}$

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The cross sectional of a horizontal tube increase along its length linearly,

A=kl+cp+fgh+12fv2=constantV1A1=V2A2VAisconstantV=CA△P+△fgh+12△fv2=0△P+12△fc2{kx+c2}=0△P+12fc2△1kx+c2p2−p1=−k△1kx+c2p2=p1−k△1kx+c2