Differential U-tube manometer between two pipes

Setup

Let the mercury surface in the A limb be at depth $a$ below the centre of A. This is the datum plane. The mercury in the B limb stands $h = 0.3\ \text{m}$ higher, and B is $z = 0.5\ \text{m}$ above A.

                   o B
   o A             |
   |               | water
   | water      ===+=== Hg, B limb
   |               |   ^ h
 ==+== Hg, A limb  |   v
   |  datum -------|---
   +---- mercury --+

Pressure balance at the datum

Pressure at the datum, from the A side:

$$p_A + \rho_w g a$$

From the B side, the water column above the mercury is $(a + z - h)$ and the mercury column is $h$:

$$p_B + \rho_w g (a + z - h) + \rho_m g h$$

Equating and simplifying:

$$\begin{aligned} p_A - p_B &= \rho_w g (z - h) + \rho_m g h \\ &= \rho_w g z + (\rho_m - \rho_w) g h \end{aligned}$$

Substitution

$$\begin{aligned} p_A - p_B &= 1000 \times 9.81 \times 0.5 + (13600 - 1000) \times 9.81 \times 0.3 \\ &= 4905 + 37081.8 \\ &= 41\,986.8\ \text{Pa} \approx 41.99\ \text{kPa} \end{aligned}$$

Pressure in A

$$p_A = 50 + 41.99 = 91.99\ \text{kPa (gauge)}$$

Answer: $p_A - p_B = 41.99\ \text{kPa}$; $p_A = 91.99\ \text{kPa}$ (gauge)