--- title: "Mercury manometer: absolute pressure of stream" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 1 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:mercury-manometer-97-5mm" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mercury-manometer-97-5mm site: BCT Engineering --- # Mercury manometer: absolute pressure of stream Fundamental of Thermodynamics & Heat Transfer · Chapter 1: Introduction - Asked in: 2072 Asoj, 6 marks - Video: [Thermodynamics - Pressure example 2 manometer](https://www.youtube.com/watch?v=NH73JxK_Zys) — Engineering Deciphered - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mercury-manometer-97-5mm - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-1 ## Question A mercury manometer is used to measure the pressure in steam pipe. The level of the mercury in the manometer is 97.5 mm. Find the absolute pressure of steam inside the pipe. If the reading of the manometer drops to 80 mm, what is the new pressure of steam? [Take sp.gr. of Hg = 13.6, Patm = 760 mm of Hg and g = 9.81 m/s²] ## Short answer Gauge pressure equals mercury column height; absolute pressure adds atmospheric pressure. At 97.5 mm, P_abs = 760 + 97.5 = 857.5 mm Hg (114.4 kPa); at 80 mm, it becomes 840 mm Hg (112.1 kPa). ## Answer **Given:** mercury column difference h₁ = 97.5 mm (steam pressure above atmosphere), later h₂ = 80 mm, SG of Hg = 13.6, P_atm = 760 mm Hg, g = 9.81 m/s². ρ_Hg = 13.6 × 1000 = 13 600 kg/m³. **Step 1: Atmospheric pressure** P_atm = ρ g h = 13600 × 9.81 × 0.760 = 101 396 Pa = 101.40 kPa **Step 2: Gauge pressure for 97.5 mm** P_gauge = 13600 × 9.81 × 0.0975 = 13 008 Pa = 13.01 kPa Absolute pressure P = P_atm + P_gauge = 101.40 + 13.01 = **114.40 kPa** (in mm Hg: 760 + 97.5 = 857.5 mm Hg) **Step 3: When the reading drops to 80 mm** P_gauge = 13600 × 9.81 × 0.080 = 10 673 Pa = 10.67 kPa P = 101.40 + 10.67 = **112.07 kPa** (840 mm Hg) **Answer: Absolute steam pressure = 114.4 kPa initially and 112.1 kPa after the reading falls to 80 mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The manometer reading of 97.5 mm Hg is a gauge pressure. How is the absolute steam pressure obtained? A. Subtract it from P_atm B. Use 97.5 mm Hg directly C. Add it to P_atm D. Multiply by P_atm Answer: C. Steam is above atmosphere, so P_abs = P_atm + P_gauge. (Key point: Absolute pressure) 2. With SG of Hg = 13.6, the density used is A. 13.6 kg/m^3 B. 1360 kg/m^3 C. 136 000 kg/m^3 D. 13 600 kg/m^3 Answer: D. rho = SG x 1000 kg/m^3 = 13 600 kg/m^3. (Key point: Density of mercury) 3. The initial absolute steam pressure is about A. 114.4 kPa B. 101.4 kPa C. 13.0 kPa D. 857.5 kPa Answer: A. 101.40 + 13.01 = 114.40 kPa (857.5 mm Hg). (Key point: Initial pressure) 4. When the reading drops to 80 mm, the new absolute pressure is A. 10.67 kPa B. 112.1 kPa C. 114.4 kPa D. 91.0 kPa Answer: B. Gauge part is 13600 x 9.81 x 0.08 = 10.67 kPa; adding 101.40 gives 112.07 kPa. 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