chapter4 - Chapter 4 The First Law of Thermodynamics for a...

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Chapter 4 The First Law of Thermodynamics for a Closed System 1. Define boundary work 2. Apply the first law to a variety of different closed systems 3. Introduce the concept of specific heat
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The First Law of Thermodynamics Energy can not be created or destroyed. A system will gain or lose energy through heat transfer, work, or mass transfer. Energy of the system = E = U +KE + PE Heat transfer = Q Work = Wb + Wsh + We Mass transfer brings in mass that has energy E
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Closed Systems Consider a system that has undergone a process from state 1 to state 2: Change in energy of the system must be equal to the heat and work transfer during the process. in net in net W Q E E , , 1 2 + = - system on net work system heat to net , , 1 2 = - = = - = + + = = - in out out in in net in out out in in net W W W Q Q Q PE KE U E E E ) ( 2 1 2 1 2 2 V V m - ) ( 1 2 z z mg -
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Forms of 1st Law for Closed Systems in net in net W Q E , , + = During a very small time dt, this becomes in net in net W Q dE , , δ + = Which can be converted to a rate form: in net in net W Q dt dE , , + = Can normalize the 1st law by the system mass to get in net in net w q e , , + =
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Forms of 1st Law for Closed Systems Often the changes in KE and PE are negligible (but not always) in net in net W Q U , , + = in net in net w q u , , + = Now that we know how to find internal energy for a substance, we can apply the first law to more complicated systems to obtain useful quantitative information. in net in net w q dt du , , + =
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Common Modes of Closed System Work Shaft work (power) rotatio n Torque on system due to motor (environment) t T W T W T W in sh in sh in sh = = = θ , , , Electrical work (power) t VI W VI W VI W in e in e in e = = = , , , “paddle wheel” motor + - Remember, electrical work can be out of the system!
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Boundary Work P Expansion and compression of gases: Force of piston on gas Force of gas on piston These forces are equal and opposite Gas expansion F d s 0 , = Fds W out b δ Work is done by gas on the surroundings (Wout is positive) 0 , < = Fds W out b F d s Gas compression Work done on gas by the surroundings (Wout is negative) Lets use this force Lets calculate the boundary work done by the gas (i.e. Wb,out)
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Boundary Work P d s = = = = 2 1 , , V V out b out b PdV W PdV PAds Fds W δ P is always positive (absolute pressure) dV > 0 & expansion & Wb,out > 0 dV < 0 & compression & Wb,out < 0 If Wb,out < 0, means positive work was done on the system Force on the piston related to the pressure of the system is only valid for quasi- equilibrium processes! One of the most important equations for MAE 204!
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Consider a Piston-Cylinder System Neglect PE, KE Constant pressure process Assume quasi-equilibrium process (i.e. work is PdV) Process starts at state 1, ends up at state 2 in Q out b W , - = = 2 1 ) ( 1 2 , V V out b V V P PdV W We know how to find the work: First law of thermodynamics: out b in W Q E E , 1 2 - = -
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Consider a Piston-Cylinder System out b in W Q E E , 1 2 - = - System energy E only contains internal energy ) ( 1 2 1 2 V V P Q U U in - - = - in Q PV U PV U = - - + 1 1 2 2 P P P = = 2 1 Constant pressure process: ( 29 ( 29 in Q V P U V P
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chapter4 - Chapter 4 The First Law of Thermodynamics for a...

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