--- title: "Differential manometer between CCl4 and oil pipes" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:manometer-ccl4-oil-pipes" source: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-ccl4-oil-pipes site: BCT Engineering --- # Differential manometer between CCl4 and oil pipes Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2082 Chaitra, 6 marks - Video: [Pressure Measurement Manometers](https://www.youtube.com/watch?v=oXXGIFSmZRE) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/manometer-ccl4-oil-pipes - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question As shown in the figure below, pipe M contains carbon tetrachloride of specific gravity 1.594 under a pressure of 105 kN/m² and pipe N contains oil of specific gravity 0.8. If the pressure in the pipe N is 175 kN/m² and the manometric fluid is mercury, find the difference x between the levels of mercury. [Figure: pipe M (carbon tetrachloride, sp. gr. 1.594) on the left and pipe N (oil, sp. gr. 0.8) on the right are joined by a U-tube containing mercury; the centre of M is 2.5 m above the centre of N, the centre of N is 1.5 m above the higher mercury level (Z–Z), and x is the difference between the two mercury levels.] ## Short answer Equate pressures at lower mercury level: 105 + 1.594×9.81×4 + 13.6×9.81x = 175 + 0.8×9.81(1.5+x). Solving gives x = 0.153 m (153 mm), with mercury higher in the limb under pipe M. ## Answer 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The method equates pressures: A. At any two points in the U-tube B. At the same level in the same continuous liquid C. At the pipe centres only D. At the free surface Answer: B. Pressures at the same level within a continuous static fluid are equal, so the datum is the lower mercury level. (Key point: Manometer principle) 2. Since p_N (175) > p_M (105), the mercury: A. Rises in N's limb and falls in M's limb B. Is pushed down in N's limb and up in M's limb C. Stays level D. Moves equally in both limbs Answer: B. The higher pressure at N pushes mercury down in its limb, so the higher level Z-Z is in M's limb. (Key point: Direction of mercury movement) 3. On the M side the CCl4 column above the datum is 4.0 m, because it is: A. 2.5 m only B. 2.5 + 1.5 m above Z-Z (the datum is x lower, mercury adds rho_Hg g x) C. 1.5 + x D. 4.0 + x Answer: B. M is 2.5 + 1.5 = 4.0 m above Z-Z; the mercury column x is added separately. (Key point: Column heights) 4. The balance 105 + 62.549 + 133.416x = 175 + 11.772 + 7.848x gives x of about: A. 0.153 m B. 0.0153 m C. 1.53 m D. 0.35 m Answer: A. 125.568x = 19.223, so x = 0.153 m, i.e. 153 mm. 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