Differential manometer between CCl4 and oil pipes

Use the manometer principle: pressures at the same level in the same continuous liquid are equal. Take the datum at the lower mercury level, which lies $x$ below Z–Z.

Given: $p_M = 105$ kN/m², $S_1 = 1.594$ (CCl₄); $p_N = 175$ kN/m², $S_2 = 0.8$ (oil); $S_{Hg} = 13.6$; $g = 9.81$ m/s².

Since $p_N > p_M$, mercury is pushed down in the limb under N and up in the limb under M. So Z–Z (higher mercury level) is in M's limb and the lower level is in N's limb.

• Height of M above Z–Z $= 2.5 + 1.5 = 4.0$ m.
• Oil column in N's limb $= 1.5 + x$.

Pressure balance at the datum (lower mercury level)

Left limb (M side):

$$p_M + \rho_1 g (4.0) + \rho_{Hg} g\, x$$

Right limb (N side):

$$p_N + \rho_2 g (1.5 + x)$$

Equating (in kN/m², $\rho g$ for water $= 9.81$ kN/m³):

$$\begin{aligned} 105 + 1.594(9.81)(4.0) + 13.6(9.81)x &= 175 + 0.8(9.81)(1.5 + x) \\ 105 + 62.549 + 133.416x &= 175 + 11.772 + 7.848x \\ 125.568\,x &= 186.772 - 167.549 = 19.223 \\ x &= 0.153\ \text{m} \end{aligned}$$

Answer: the difference between the mercury levels is $x \approx 0.153$ m $= 153$ mm (mercury higher in the limb below pipe M).