Thermodynamics

Chapter 1 | 31EXAMPLE 1–8Measuring Atmospheric Pressurewith a BarometerEngineLungsDetermine the atmospheric pressure at a location where the barometric readingis 740 mm Hg and the gravitational acceleration is g 9.81 m/s 2 .Assume the temperature of mercury to be 10C, at which its density is13,570 kg/m 3 .Solution The barometric reading at a location in height of mercury columnis given. The atmospheric pressure is to be determined.Assumptions The temperature of mercury is 10C.Properties The density of mercury is given to be 13,570 kg/m 3 .Analysis From Eq. 1–26, the atmospheric pressure is determined to beP atm rgh1 N 113,570 kg>m 3 219.81 m>s 2 210.74 m2 a1 kg # ba 1 kPam>s21000 N>m b 2 98.5 kPaDiscussion Note that density changes with temperature, and thus this effectshould be considered in calculations.FIGURE 1–53At high altitudes, a car enginegenerates less power and a person getsless oxygen because of the lowerdensity of air.EXAMPLE 1–9Effect of Piston Weight on Pressure in a CylinderThe piston of a vertical piston–cylinder device containing a gas has a massof 60 kg and a cross-sectional area of 0.04 m 2 , as shown in Fig. 1–54. Thelocal atmospheric pressure is 0.97 bar, and the gravitational acceleration is9.81 m/s 2 . (a) Determine the pressure inside the cylinder. (b) If some heat istransferred to the gas and its volume is doubled, do you expect the pressureinside the cylinder to change?Solution A gas is contained in a vertical cylinder with a heavy piston. Thepressure inside the cylinder and the effect of volume change on pressure areto be determined.Assumptions Friction between the piston and the cylinder is negligible.Analysis (a) The gas pressure in the piston–cylinder device depends on theatmospheric pressure and the weight of the piston. Drawing the free-bodydiagram of the piston as shown in Fig. 1–54 and balancing the verticalforces yieldSolving for P and substituting,P P atm mgA 0.97 bar 160 kg2 19.81 m>s2 210.04 m 2 2 1.12 barPA P atm A W1 Na1 kg # ba 1 barm>s210 5 N>m b 2(b) The volume change will have no effect on the free-body diagram drawn inpart (a), and therefore the pressure inside the cylinder will remain the same.Discussion If the gas behaves as an ideal gas, the absolute temperaturedoubles when the volume is doubled at constant pressure.P atm = 0.97 barm = 60 kgA = 0.04 m 2P = ?P atmPW = mgFIGURE 1–54Schematic for Example 1–9, and thefree-body diagram of the piston.