2. Describe the second law in terms of adiabatic and reversible processes.

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1 Lecture #3 1 Lecture 3 Objectives: Students will be able to: 1. Describe the first law in terms of heat and work interactions.. Describe the second law in terms of adiabatic and reversible processes. 3. Identify the difference between internal and total energy. 4. Identify specialized thermodynamic equations and tell when they can be applied. 5. Describe ideal gas thermodynamic properties. 1. Heat, work and the first law. We observe that for a closed system the only way to change the energy of a system is by heat and work interactions with the system. We write the differential first law as de = δw +δq where E is the total energy of the system. Note that de is an exact differential, whereas δw and δq are inexact. The integrated form is E = w+q For an adiabatic process E = w ad and Q = 0, so w ad is path independent, but you could also take a different path to get the same E, so w ad = w +Q. Likewise, if the work interactions are zero then there must be some path independent heat interaction, de = δq rev where Q rev is the reversible heat. Now suppose that the whole system is translating and accelerating: So, w = w stationary +mg y+ m }{{} (v v1 }{{} PE KE E PE KE = w }{{} stationary +Q state function Define the internal energy U = E PE KE so that U is also a state function, then, U = w stationary +Q

2 Lecture #3. The second law. Why does a basket ball bounce? If I drop a basket ball from a height of 1 m, will it return to the same height when it bounces? Why not? Analyze the basket ball bounce in terms of the first law: U + PE + KE = w +Q We are converting potential and kinetic energy into internal energy of the gas molecules in the ball, and the material of the ball, along with heat and work. The first law does not preclude the ball bouncing to the same height, and bouncing forever. However, we observe that this never happens. Likewise, you can drop a weight into water and convert the potential energy of the weight into internal energy of the water, however, we never observe the converse, although the first law says that it is possible. Hence the need for the second law of thermodynamics. There is a directionality to spontaneous processes, this directionality gives rise to another state function, which we call entropy. We define entropy as ds = δq rev T the differential entropy is the reversible heat divided by the absolute temperature. Note that reversible heat requires that there be no temperature gradients, just as reversible work required no force imbalances. The second law can be written as ds = ds sys +ds surr 0 The Clausius statement of the second law: Heat cannot be converted into an equal amount of work through a cyclical process. Proof: For a cyclical process U = 0 and S = 0. Add heat to the system reversibly, then extract work from the system reversibly and adiabatically. U = 0 = Q rev +w ad w ad = Q rev OK, by 1st law. Now check second law S = S rev + S ad = 0 but S ad = 0 and S rev = Q rev /T 0, so the second law is violated. 3. Generalized equations: (a Closed system, no translations du = δq+δw

3 Lecture #3 3 (b Open systems where de = δq+δw + in [( de = d [ out U +mgz + mv [( Ĥ out +gz out + v out (c Steady state, d[ ] cv = 0 [ ] δm i Ĥ i + v i +gy i = δq+δw i ] Ĥ in +gz in + v in ] δm out (d Compressors & turbines (pressure change work H = Q+w s where w s is shaft work. (e Throttling devices (pressure drop H = Q (f Nozzles ( KE P Ĥ = 1 (v v 1 (g Heat exchangers H = Q 4. Ideal Gas State Properties Ũ = Ũ(T ( Ũ T V H = H(T ( H T P = C V = C P C P = C V +R δm in ] κ = C P C V Both C P and C V are only functions of temperature.

4 Lecture # Open system example: Filling a storage tank. A storage tank is pressurized by rapid filling from a much larger storage reservoir. The tank is well-insulated. The tank volume is m 3 and the gas in the tank is initially at 1 bar and 80 K. The air in the reservoir is at 100 bar and 300 K. What is the temperature of the gas in the tank when the pressure reaches 5 bar? Approach: This is an open system with negligible change in kinetic and potential energy. There is one inlet stream and no outlet streams. Heat and work interactions are zero. de = du = H in δn in, with H in a constant and δn in = dn. du = ndũ +Ũdn = H in δn. We can rearrange this to dũ H in Ũ = dn n. We can integrate this between states 1 and : ( ( Ũ ln( Hin Ũ = lnn n Hin Ũ1 n Ũ n 1 ln = ln. 1 H in Ũ n 1 Rearranging and assuming ideal gas behavior of the air, H in Ũ1 H in Ũ = n n 1 = P V RT RT 1 P 1 V 1 = P T 1 P 1 T. Let us assume that the heat capacities are constant. The enthalpy and internal energy can then be written as referenced to some temperature T 0. H = H0 + C P (T T 0 and Ũ = Ũ0 + C V (T T 0. We can now write H in Ũ1 = H 0 + C P (T in T 0 [Ũ0 + C V (T 1 T 0 ] = H 0 Ũ0 + C P T in C V T 1 ( CP C V T 0 But, H 0 Ũ0 = PṼ = RT 0 and C P C V = R, for an ideal gas. Therefore, H 0 Ũ0 RT 0 = 0, and Therefore, H in Ũ1 = C P T in C V T 1 C P T in C V T 1 C P T in C V T = P T 1 P 1 T.

5 Lecture #3 5 Solving for T, ( CP T in C V T 1 P 1 T = [ T P 1 ( CP T in C ] V T 1 +P T 1 CV T = ( CP T in C V T P T 1 = P T 1 CP T in (P /P 1 T 1 T in κ κt in T 1 +(P /P 1 T 1 where κ = C P / C V. Using C V = 5/R and κ = 1.4 we get T = K. The number of moles at this point is n = P V RT = 315 moles.

Thermodynamics ENGR360-MEP112 LECTURE 7

Thermodynamics ENGR360-MEP112 LECTURE 7 Thermodynamics ENGR360-MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).