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Chapter 5 – The First Law of Thermodynamics

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Energy conservation law Need to consider conservation of matter in

advance5.1 The first law of thermodynamics for

a control mass undergoing a cycle

(system) Back to original state

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The first law of thermodynamics states that during any cycle a system (control mass) undergoes, the cyclic integral of the heat is proportional to the cyclic integral of the work. 保持相同物質

Cyclic integral

J the net work during the cycle

The net heat transfer during the cycle.Proportionality factor (depends on unit)

See Fig. 5.1 (P.97) Ex: Q: CalorieW: Joule

通常令Q和W都使用相同單位，則J = 1 3機械工程系 陳俊勳老師

5.2 The First Law of Thermodynamics for a Change in State of a Control MassFig. 5.2

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In thermodynamicsE = internal energy (U) + kinetic energy (KE)

+ potential energy (PE)E = U + KE + PE

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dE = dU + d(KE) + d(PE)δQ = dU + d(KE) + d(PE) + δ w

The first law of thermodynamics for a change of state (control mass)

The form of K.E. for a control massNo heat transfer δQ = 0No change in internal energy dU = 0No change in potential energy dPE = 0

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δW = - d(KE) = - F dxWork done on the system

F = ma = m = m = mv

d KE = F dx = mv dv = md = d(m )Control mass (量不變)

KE =m

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δQ = 0 d(KE) = 0 (no kinetic energy)dU = 0δW = -d(PE) = -F dzIf F = mg, then dz is the variation in elevationd(PE) = mg dz

= m Constant;independent of z

(PE)2 – (PE)1 = mg(z2 – z1)

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• dE = dU + mv dv + mg dzstate 1 state 2E2 – E1 = U2 – U1 +mv22/2 – mv12/2 + mgz2 –

mgz1 = (U + mv2/ 2 + mgz)2- (U + mv2/ 2 + mgz)1

• δQ = dU + + d(mgz) +δW

1Q2 = U2 – U1 + m(v22-v12)/2 + mg(z2 – z1) + 1W2

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整理：

(1) 1Q2 - 1W2 = (U + mv2/ 2 + mgz)2 - (U + mv2/ 2 + mgz)1 = E2 – E1

For control mass(2) Conservation of energy

Read the book (joint checking account shared by husband and wife)由上式觀察; E為property☆Q and W must cross the boundary

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(3) The changes in internal energy, kinetic energy and potential energy.Not absolute values in U, KE and PE.If want to refer the absolute value in U, KE and PE reference state

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5.3 Internal Energy (U)– A Thermodynamic PropertyExtensive property (so does the KE and PE)u = U / m (specific internal energy)U: total internal energyu: internal energy per unit mass像其他性質一樣，U (或u) 亦可描述狀態Ex: Table B.1.1 (P.664) 及 B.1.2 (P.668)

uf (internal energy of saturated liquid)ug (internal energy of saturated vapor)

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U = Uliq + Uvapm u = mliq uf + mvap ugu = uf + x ufg where ufg = ug – uf

Ex: 5.1 (P.102) 注意正負號 (練習了解)Ex: 5.2 (P.102)☆In steam table, u = 0 for saturated liquid at

0.01℃. (designated)

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5.4 Problem Analysis and Solution Technique

1. Control mass? Control volume?(fixed mass) (mass can cross the boundary)

2. Initial state properties3. Final state properties4. Process5. Draw a diagram in steps 2 to 46. What kind model can we use? Ideal gas equ. / Steam table

7. Analysis of the problem8. Solution techniqueHw 5.1 5.10 5.11Ex.5.3 (P.103) 第一次小考

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5.5 Enthalpy (H) ------A Thermodynamic Property

For a control mass, KE = 0 and PE = 0 1Q2 = U2 – U1+ 1W2

1W2 = If P = const. (P1 = P2 = P) (Fig.5.5;

P.106)

1W2 = P = P (V2 – V1) = P2V2 – P1V11Q2 = U2 – U1+ P2V2 – P1V1= (U2 + P2V2) - (U1 + P1V1)

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1Q2 = H2 – H1 (const. pressure process)

For saturated stateh = (1 – x) hf + x hg

= hf + x hfg (hfg = hg - hf)

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作業： 5.20 5.28 5.29

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5.6 The Constant-Volume and Constant Pressure Specific Heats

☆no phase change (may be solid or liquid or gas) homogeneous phaseSpecific heat(C): the amount of heat required

per unit mass to raise the temperature by one degree.

d(KE) = 0d(PE) = 0δQ = dU + δW = dU + P dV

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δQ = dU + P dV= dU + d(PV) – d(U + PV) = dH

Cv and Cp are thermodynamic properties(independent of processes) Fig. 5.7 (P.111)

Ex: 5.5 (P.111)For solid and liquid, v =const.(incompressible)

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5.7 The Internal Energy, Enthalpy, and Specific Heat of Ideal Gases

Equation of state of ideal gasP v = R T (R = / M)

☆for an ideal gasu = f(T) internal energy is a function of

temperature only和壓力 (pressure) 無關

1843, Joule experiment (P.114)

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Questions: (1) Is air an ideal gas?(2) Can it be really sure that no

change in temperature was found?

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5.8 The First Law as A Rate Equation

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5.9 Conservation of MassControl mass: fixed quantity of mass

Q: energy change (for a control mass)Whether the mass change?

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