I. Concepts and Definitions. I. Concepts and Definitions

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Meryl Andrews
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1 F. Properties of a syste (we use the to calculate changes in energy) 1. A property is a characteristic of a syste that can be given a nuerical value without considering the history of the syste. Exaples include T, P, ρ, velocity, E, U, volue (we can only easure T, P, velocity, ass, volue). Exaples and definitions a. pressure, P force exerted by fluid P = unit area N P ( = ) = 1pascal = 1Pa 1 at = 1015 Pa = bar = psia 5 1bar = 10 Pa b. density, ρ ass kg ρ = =, volue V (1-4) c. atospheric pressure (absolute) = P at d. gage pressure = P gage = P abs -P at (1-15) e. absolute pressure = P abs = P at + P gage f. vacuu pressure = P vac = P at -P abs (1-16) g. specific volue = v = 1/ρ = Vol/ass, /kg h. intensive properties are independent of the size of the syste: T, e, P, ρ, specific volue (v), u. The units on e are kj/kg. i. extensive properties depend on the size of the syste: actual volue (V), E,, U. The units on E are kj. Exaple. The pressure gage attached to the air storage tank of a copressor reads 100 psi. If the atospheric pressure of the surroundings is 14.7 psi, what is the absolute pressure in the tank? (Answer: psia) Stiff etal tube p Bourdon tube gage gives P gage. 1

2 Exaple. You are designing a snorkel that will allow you to reain suberged in water at a depth of 1. Discuss any probles you ight experience in using this device. At depth h = 1, the net pressure resisting the expansion of your lungs will be (see Eq. 1-19, p. 4 of text) kg N P = ρgh = psi = = s If we approxiate that part of the trunk that expands during breathing by a cylinder with diaeter d = 0 c and height L = 0 c, the curved surface will have area A = πdl = The force acting on your poor trunk, as you try to breath, will be P*A = 774 N. This corresponds to the force exerted by a ass of 8 kg or 64 lb. Your snorkel will not work at this depth because you will not be able to breath against the force exerted by the water surrounding your body. j. teperature T(K) = T(ºC) (1-9) T(ºR) = T(ºF) = 1.8 T(K) (1-10, 11) 1K = 1.8ºR T(ºF) = 1.8 T(ºC) + (1-1) T(K) = T(ºC) (1-1) T(ºR) = T(ºF) (1-14) (1-1) eans that a teperature change of 10 degrees K = teperature change of 10 degrees C.

3 . The properties of a syste at equilibriu do not change with tie when the syste is isolated fro its surroundings. Properties are only defined in equilibriu states. A glass of pure water at roo teperature. Therodynaics can copletely describe the state of the water at equilibriu. Add an ice cube to the water. Therodynaics cannot describe how long it will take for the ice to elt nor the phenoena that occur during elting. If we wait long enough the ice will elt and the syste will return to a condition of equilibriu. 4. A set of properties that copletely describe a syste specify the state of the syste. 5. The State Postulate (an experiental observation, see p. 14) The intensive state of a pure substance is copletely specified by two independent, intensive properties. a. for exaple, u = f(t,v). T and v (teperature and specific volue) are always independent, but T and p are not necessarily so. T and p are both intensive properties but in a two-phase region they cannot be varied independently. We will coe back to this. b. independent eans that one property can be varied while the other is held constant. c. pure eans that we have a single pure cheical species (dry air is considered pure).

4 6. Properties of pure substances a. Solids and liquids pressure, kpa Solids and liquids are approxiately incopressible and that is how we odel the. For exaple, the densities of water at 0 C, and 1 and 00 bar, are 998 and 1011 kg/. Table A- gives the properties of solids and liquids. b. Ideal gases (low pressure, see p. 17 of text) olar basis PV = NR T u ass basis PV = RT volue, nuber of koles teperature, K universal gas constant, 8.14 kj/(kol K) nuber of kg olecular weight specific gas constant, R u R = MW There are no tables in the book that give the density of ideal gases. That s because they are so easy to calculate fro the ideal gas law. c. Density of an ideal gas ass P P( MW ) ρ = = = = volue V RT R T Exaple: the density of air at 5 C and 1 at is 101. kpa kg 1at 9 P(MW ) 1at kol kg ρ = = = 1.19 RT u kpa 8.14 ( 5 + 7) K kol K Exaple: the density of air at 0 C and 0.85 at is kg 0.85 at 98 K kg ρ = = 1at 7 K u 4

5 I. Concepts and definitions d. Exaple one ore application of the ideal gas law Find the ass (kg) and weight (N) of the air contained in a by by roo if the teperature and pressure of the air are.0 C and 0.85 at. Assuption: the ideal-gas law applies Data: R = (kpa )/(kg K) (Table A-1) g = 9.81 /s (inside front pages of book) T = = K Calculations: PV = RT kpa 0.85 at (4 )7(8 ) PV 1at kg = = ρv = = = 1 kg RT kpa (95.15 K ) kg K weight = g = 1( kg ) N s = G. Processes and Cycles 1. A process changes a syste fro one equilibriu state to another. Reoval of energy Liquid water Ice 5

6 G. Processes and Cycles. A cycle is a process that returns a syste to its original state. A stea power cycle is an exaple of a process in which a fluid is alternatively vaporized and condensed. Q in Boiler Vapor Turbine W T,out W P,in Liquid Pup Liquid Condenser Vapor-liquid ixture Q out G. Processes and Cycles. A quasi-equilibriu or reversible process is one in which the syste reains infinitesially close to equilibriu at all ties. This is an idealization that is approxiated by any real processes. 4. The intensive properties of a single-phase syste undergoing a quasiequilibriu process are spatially unifor. Suppose we have a frictionless, piston-cylinder device that is filled with a copressed gas. We flick the particles of sand off the piston one at a tie. The gas in the cylinder undergoes a reversible expansion. sand 6

Answers to assigned problems from Chapter 1

Answers to assigned problems from Chapter 1 Answers to assigned probles fro Chapter 1 1.7. a. A colun of ercury 1 in cross-sectional area and 0.001 in height has a volue of 0.001 and a ass of 0.001 1 595.1 kg. Then 1 Hg 0.001 1 595.1 kg 9.806 65