2300_les04

2300_les04 - II. Energy and the First Law of Thermodynamics...

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Unformatted text preview: II. Energy and the First Law of Thermodynamics A. Energy Transfer by Work. Energy is Conserved. 1. Energy can cross the boundary of a closed system by only two mechanisms: heat transfer and work transfer. 2. The change in energy of a closed system is equal to the net heat transferred to the system minus the net work performed by the system. Net work out, Wout,net = Wout - Win ∆Etotal = ∆Esystem + ∆Esurroundings = 0 ∆Esystem = −∆Esurroundings Closed system ∆Esys = Qin,net − Wout ,net (2-34); (4-17) Net heat in, Qin,net = Qin - Qout Lesson 4, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics B. Mechanical work and power 1. Introduction a. Work done by force F on a body displaced distance s is Constant force F W = Fs ( kJ ) 1 W = ∫ F ds ( kJ ) 1 2 Variable force (2-22) W = area under curve 2 s b. Power supplied to body moving with velocity V: s1 s2 W = FV ( kW ) Lesson 4, Geof Silcox, Chemical Engineering, University of Utah 1 II. Energy and the First Law of Thermodynamics c. Work and heat are processes, not properties. Work and heat are represented as areas on a graph. E, e, V, T, P, m are properties and are represented by points on a graph. d. Work and heat are also known as path functions. ∫ dE = E − E = ∆E ∫ dE = 0 or ∫ dV = 0 ∫ δW = W 1 2 1 2 1 12 2 (any path between states 1, 2) (any cycle) (not ∆W) P 2 Shaded area = ∫ δW out = Wout ,net 1 V Lesson 4, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 2. Expansion and compression work a. Transfer of energy to a system by boundary work requires that a force act on the boundary and that the boundary move P b. Differential work done by the system is, for a piston of area A, 1 δ Wb,out = F ds = PA ds = P dV 2 c. Replacing F with pA is strictly correct only for a quasi-equilibrium (reversible) process d. The total boundary work done by system is area under curve P = f(V) V1 V2 V Wb,out = ∫ P dV V1 V2 (4-2) Lesson 4, Geof Silcox, Chemical Engineering, University of Utah 2 II. Energy and the First Law of Thermodynamics 3. Evaluation of integral (4-2) Wb,out = ∫ P dV V1 V2 Remember that this equation only applies to a quasi-equilibrium (reversible) process. How can we evaluate this integral? We look at three examples: - P may be constant (see Example 4-2). - P may be defined by ideal gas law for an isothermal process (see Example 4-3). - Process may be polytropic (Section 4-1). Lesson 4, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 3. Evaluation of integral. Example - Air is contained in a pistoncylinder device at 500 kPa at an initial volume of 0.040 m3. The air expands to a final volume of 0.075 m3. Calculate the work output under conditions of (a) constant pressure, (b) constant temperature. Data V1 = 0.040 m3 V2 = 0.075 m3 P1 = 500 kPa Model 1) Closed system. 2) Quasi-equilibrium process. 3) Ideal gas: PV = mRT Lesson 4, Geof Silcox, Chemical Engineering, University of Utah 3 II. Energy and the First Law of Thermodynamics 3. Example (cont.) (a) Analysis (constant pressure) Wout = ∫ PdV = P (V2 − V1 ) V2 V1 Wout = 500 kPa ( 0.075 − 0.040 ) m 3 = 18 kJ (b) Analysis (constant temperature) V2 dV mRT dV = mRT ∫ V1 V1 V1 V V ⎛V ⎞ ⎛V ⎞ = mRT ln ⎜ 2 ⎟ = PV1 ln ⎜ 2 ⎟ 1 ⎝ V1 ⎠ ⎝ V1 ⎠ V2 V2 Wout = ∫ PdV = ∫ Wout ⎛ 75 ⎞ Wout = 500 kPa 0.04 m 3 ln ⎜ ⎟ = 13 kJ ⎝ 40 ⎠ ( ) Lesson 4, Geof Silcox, Chemical Engineering, University of Utah II. Energy and the First Law of Thermodynamics 4. Polytropic process (Section 4-1 of text) Many compression and expansion processes can be modeled as polytropic processes. The boundary work for such a process can be calculated as follows, where n is typically 1.2 or 1.3. PV n = C or P = CV − n Wout = ∫ PdV V1 V2 n = 0 and n = 1 correspond to isothermal and constant pressure cases considered above. V2 Wout P V − PV1 V 1−n 1 , n ≠1 = C ∫ V dV = C = 22 V1 1− n V 1− n V2 −n 1 (4-9) For an ideal gas, PV = mRT and Wb,out = P2V2 − PV1 mR(T2 − T1 ) 1 = , n ≠1 1− n 1− n (4-10) Lesson 4, Geof Silcox, Chemical Engineering, University of Utah 4 ...
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