Manometer reading in pipe with minor losses

Reading of the figure. Water flows along the horizontal 75 mm pipe from A to B (3 m, $f = 0.02$) and discharges at B into a large tank whose water surface is 2.4 m above the pipe axis. The mercury manometer joins the pipe at A to the tank wall at C, which is 0.9 m below B. Velocity in the pipe $V = 4.5$ m/s. Minor loss: exit (sudden enlargement into the tank), $K = 1.0$.

Pressure at A

$$\frac{V^2}{2g} = \frac{4.5^2}{2(9.81)} = 1.0321\ \text{m},\qquad h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02\left(\frac{3}{0.075}\right)(1.0321) = 0.8257\ \text{m},\qquad h_{exit} = \frac{V^2}{2g} = 1.0321\ \text{m}$$

Energy equation from A (pipe axis, datum) to the tank water surface (2.4 m above the axis, velocity zero, atmospheric pressure):

$$\frac{p_A}{\gamma} + \frac{V^2}{2g} = 2.4 + h_f + h_{exit} \;\Rightarrow\; \frac{p_A}{\gamma} = 2.4 + 0.8257 + 1.0321 - 1.0321 = 3.2257\ \text{m}$$

The exit loss cancels the velocity head, so only the friction loss changes the pressure.

Pressure at C

C is in the quiet water of the tank, so the pressure is hydrostatic: $\dfrac{p_C}{\gamma} = 2.4 + 0.9 = 3.3$ m.

Manometer reading

Piezometric heads (datum at the pipe axis):

$$\left(\frac{p_A}{\gamma} + z_A\right) - \left(\frac{p_C}{\gamma} + z_C\right) = 3.2257 - (3.3 - 0.9) = 0.8257\ \text{m of water}$$

For a mercury-water differential manometer: $\left(S_{Hg} - 1\right)h = 12.6\,h = 0.8257$

$$h = \frac{0.8257}{12.6} = 0.0655\ \text{m} = 65.5\ \text{mm of mercury}$$

Direction. The piezometric pressure at A is greater than at C, so mercury is pushed down in the limb connected to A and rises in the limb connected to C. The mercury level in the C limb is higher than in the A limb by about 65.5 mm (the reading equals the friction head $h_f$, converted to mercury).

Answer: Manometer reading $\approx 66$ mm of Hg, with mercury higher on the C side (the A side is at the higher pressure).