CHAPTER 1
Basic Principles
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on elevation; in Denver, Colorado, it is about 84 kPa; in a mountain city with elevation 3000 m, it is only 70 kPa. In Table B.1, the variation of atmospheric pressure with elevation is listed. EXAMPLE 1.4 Express a pressure gage reading of 20 mm Hg in absolute pascals at an elevation of 2000 m. Use gHg = 13.6gwater. Solution First we convert the pressure reading into pascals. We have P = γ Hg h = (13.6 × 9810)
N N × 0.020 m = 2668 2 3 m m
or 2668 Pa
To find the absolute pressure we simply add the atmospheric pressure to the above value. Referring to Table B.1, Patm = 0.7846 × 101.3 = 79.48 kPa. The absolute pressure is then P = Pgage + Patm = 2.668 + 79.48 = 82.15 kPa Note: We express an answer to either 3 or 4 significant digits, seldom, if ever, more than 4. Information in a problem is assumed known to at most 4 significant digits. For example, if the diameter of a pipe is stated as 2 cm, it is assumed that it is 2.000 cm. Material properties, such as density or a gas constant, are seldom known to even 4 significant digits. So, it is not appropriate to state an answer to more than 4 significant digits. EXAMPLE 1.5 A 10-cm-diameter cylinder contains a gas pressurized to 600 kPa. A frictionless piston is held in position by a stationary spring with a spring constant of 4.8 kN/m. How much is the spring compressed? Solution The pressure given is assumed to be absolute (this is the case unless otherwise stated). A force balance on the piston provides PA − Patm A = K spring Δ x (600 000 − 100 000) π × 0.052 = 4800 Δ x
∴ Δ x = 0.818 m
Make sure you check the units. Pressure typically has units of pascals when input into equations and area has units of square meters.