Download Thermodynamics Assignment Solutions: Pressure Concepts and more Lecture notes Thermodynamics in PDF only on Docsity! ANSWER KEY IN THERMODYNAMICS 1 (ASSIGNMENT 2) Prepared by: Adam C. Macapili Problem 1-34C. What is the difference between gage pressure and absolute pressure? Answer: The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure. Problem 1-40. A vacuum gage connected to a chamber reads 35 kPa at a location where the atmospheric pressure iS 92 kPa. Determine the absolute pressure in the chamber. Analysis The absolute pressure in the chamber is determined from iis) muy Ppc = Pam —Pige = 92-35 =57 kPa P.. 35 kPa Paim = 92 kPa Problem 1-41E. A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the manometer arm with the (a) higher and (b) lower fluid level being attached to the tank. Assumptions The fluid in the manometer is incompressible. Properties The specific gravity of the fluid is given to be SG = 1.25. The density of water at 32°F is (Table A-3E) Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, p=SGX py, = (1.25)(62.4lbmvft’) = 78.0lbm/ft” Aur | , 28) in The pressure difference | oes corresponding to a differential height of 28 in. between the two arms of the manometer is Pitm = 12.7 psia 2 AP = och = thay s2.740/s* 08120] 1 Ibf | lft = 1.26psia 32.174 Ibm. fils” || 144in? | Pt Prater SI + Po Sly a Pmercury 23 = Pot Air Solving for P; Py = Patn ~ Prater $1 ~ Poi Sz + Pmercury $3 hl or, hs P, — Patra = &(Pmercury!3 — Pwater! — Poi!) Water h, Noting that Pj sase = P1 - Pam and substituting. P, cage = (9.81 m/s” )[(13.600 kg/m’ )(0.46 m) — (1000 kg/m? )(0.2 m) —(850kg/m?)(0.3 »| | are | lkg-m/s~ |. 1000 N/m~ =56.9kPa Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid |_simplifies the analysis greatly. Problem 1-43. Determine the atmospheric pressure at a location where the barometric reading is 750 mm Hg. Take the density of mercury to be . Properties The density of mercury is given to be . Analysis The atmospheric pressure is determined directly from Prt = Pgh = (13.600 kg/m? )(9.81 mv/s?)(0.750 m) ——~— | —**Pe =| lkg-m/s~ | 1000 N/m~ =100.1 kPa Problem 1-44. The gage pressure in a liquid at a depth of 3 m is read to be 28 kPa. Determine the gage pressure in the same liquid at a depth of 9 m. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as Pt = P- pen : Patm “5 | kPa | = (145 kPa) —(1000 kg/m? \(9.81 m/s*)(5 m) | | 1000 N/m? | = 96.0 kPa : hi (ob) The absolute pressure at a depth of 5m in the other liquid is P= Pim + Pgh i ee = Tl 1 kP% = (96.0 kPa) + (850 kg/m 3\(9.81 m/s” 5 m ) —_ {1000 N/m* | =137.7 kPa Note that at a given depth, the pressure in the lighter fluid is lower, as expected. Problem 1-46E. Show that . Analysis Noting that, , and , we have 1 kgf =9.80665 N = (9.80665 N ) ae) = 2.20463 lbf and ? ; ) | 254m ) ; Lkgfiem* = 2.20463 Ibfem~ = (2.20463 Ibf/em* ) 7 = 14.223 Ibt/in = 14.223 psi in THANK YOU
