05first_law - Sept 2005: changes reflect text: Woud Sept...

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First Law Sept 2005: changes reflect text: Woud Sept 2006: added examples first law: during any cycle a system undergoes, the cyclic integral of the heat is proportional to the cyclic integral of the work pg 83 van Wylen & Sonntag Fundamentals of Classical Thermodynamics 3rd Edition SI Version first law for cycle 1 dQ = 1 dW (5.2) The net energy interaction between a system and its environment is zero for a cycle executed by the system. pg 2 Cravalho and Smith 1 dQ 1 dW = 0 where integral are cyclic and dQ = δ Q dW = δ W plot data Pressure Volume plot for Processes 0.5 1 1.5 2 2.5 p 1 2 1 0.5 1 1.5 2 2.5 v A process B process 1 2 2 2 2 C process 2 1 2 1 2 2 1 dQ = 1 dW apply first law to cycle A B 1 dQ A 1 dQ B 1 dW A 1 dW B + + = 2 1 2 1 1 1 2 1 apply first law to cycle A C 1 dQ A 1 dQ C 1 dW A 1 dW C + + = 1 1 subtract A C from A B 1 dQ B 1 dQ C 1 dW B 1 dW C = rearrange . .. 1 i.e. δ Q - δ W is a point function . .. only dependent 2 9/21/2006 1 9/21/2006 dE δ Q 1 dQ_W B 1 dQ_W C = upon the end points => define as . .. energy, point function − δ W = 2 1
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first law for system (Woud: Closed system) - change of state N.B. Woud starts with rate equation and obtains this assuming steady state rearrange and integrate . .. Q 1_2 is the heat transferred TO system = + δ Qd E + δ W Q 1_2 = E 2 E 1 W 1_2 E 1 E 2 are intial and final values of energy of system and . .. (5.5) W 1_2 is work done BY the system energy E consists of internal energy + kinetic energy + potential energy E = U + K + PE dE = dU + dKE + dPE and first law can be restated . .. δ + dU + dPE + (5.4) = δ W = + dKE δ W Closed System d U = Q_dot W_dot dU = δ Q − δ W m_dot e = m_dot i = 0 (W 2.3) dt d and δ : VW&S: page 62 Woud page 11 d = differential of point functions state variables difference between d and δ δ = differential of path functions - amount depends on path/process: diminutive see discussion of cyclic process below cycle may be considered a closed system; initial state and final state are identical, For example (detailed discussion later) set up limits and calculations p-v plot of Brayton cycle 0 1 2 3 4 5 0 2 4 6 pressure volume adiabatic compression heat addition adiabatic expansion in turbine heat rejection closed (cycle) t d U d = 0 Q_dot W_dot = Q_dot = W_dot Q cycle W cycle = (W 2.6) this is where we started above using different approach to first law 9/21/2006 2 9/21/2006
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example 5.3 Sonntag example 5.3: vessel with volume 5 m 3 contains 0.05 m 3 of saturated liquid water and 4.95 m 3 of saturated water vapor at 0.01 MPa. Heat is added until the vessel is filled with saturated vapor. Determine Q. State 1: V := 5m 3 V := 4.95m 3 MPa := 10 6 Pa 3 vap kJ := 10 J 3 V liq := 0.05m p := 0.1MPa 3 m kJ constant volume and := 0.001043 := 417.36 page 616 1 Q 2 mass => constant v v f kg u f kg Sonntag steam tables at p = m 3 kJ kJ 0.1 MPa v g := 1.694 u g := 2506.1 u fg := 2088.7 kg kg kg State 2: V 2 = V v = v g u = u g Liq H 2 O Vap H 2 O first law: Q 1_2 = E 2 E 1 + W 1_2 W 1_2 = 0 Q 1_2 = U 2 U
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05first_law - Sept 2005: changes reflect text: Woud Sept...

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