Question
Water flows through a horizontal pipe of which the cross-section is not constant. The pressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point where the velocity is 0.65m/s.
✓
Answer
For streamlined flow, the sum of the pressure head, velocity head and gravitational head is a constant, i.e.,$\frac{\text{P}}{\rho\text{g}}+\frac{\text{v}^2}{2\text{g}}+\text{h}=\text{Constant}$
Taking h the same, we have,$\frac{\text{P}_1}{\rho\text{g}}+\frac{\text{v}^2_1}{2\text{g}}=\frac{\text{P}_2}{\rho\text{g}}+\frac{\text{v}^2_2}{2\text{g}}$
$1+\frac{(0.35)^2}{2\text{g}}=\frac{\text{P}_2}{\rho\text{g}}+\frac{(0.65)^2}{2\text{g}}$
$\text{P}_2=\frac{(0.35)^2-(0.65)^2}{2\text{g}}+1$
$=1-\frac{0.3}{2\text{g}}$
$=1-0.015=0.985$
$\therefore$ Pressure = 0.985cm
Taking h the same, we have,$\frac{\text{P}_1}{\rho\text{g}}+\frac{\text{v}^2_1}{2\text{g}}=\frac{\text{P}_2}{\rho\text{g}}+\frac{\text{v}^2_2}{2\text{g}}$
$1+\frac{(0.35)^2}{2\text{g}}=\frac{\text{P}_2}{\rho\text{g}}+\frac{(0.65)^2}{2\text{g}}$
$\text{P}_2=\frac{(0.35)^2-(0.65)^2}{2\text{g}}+1$
$=1-\frac{0.3}{2\text{g}}$
$=1-0.015=0.985$
$\therefore$ Pressure = 0.985cm
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