# A manometer containing mercury is connected to two points A and B 15 m apart, on an inclined pipeline conveying water as shown.?

## The pipeline is straight and slopes at an angle of 15 o with the horizontal. The manometer gives a reading of 150 mm. Calculate the pressure difference between the two points A and B. (Specific weight of water = 9.81 kN and S.G of Hg= 13.6) Mar 20, 2018 If pressure at point $A$ is $P$ and at $B$ is $P '$,then considering pressure at a horizontal level passing through $s$ we can write,

$P + \left(p q + q r + r s\right) \rho \cdot g = P ' + q r \cdot \rho \cdot g + r s \cdot 13.6 \cdot 1000 \cdot g$ ($1 \frac{g}{c {m}^{3}} = 1000 \frac{K g}{m} ^ 3$)

where, $\rho \cdot g = 9.8 k N$

Now,from the diagram $p q = 15 \sin 15 = 3.87 m$

given, $r s = \frac{150}{1000} m$

So,putting the values we get,

$P + 3.87 \cdot \rho \cdot g + q r \cdot \rho \cdot g + r s \cdot \rho \cdot g = P ' + q r \cdot \rho \cdot g + 19992$

or,$P + 3.87 \cdot 9.8 \cdot {10}^{3} + \left(\frac{150}{1000}\right) \cdot 9.8 \cdot {10}^{3} = P ' + 19992$

or, $P - P ' = 19992 - 39396 = - 19404 P a$