BASIC THERMODYNAMIC CONCEPTS

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1 BASIC THERMODYNAMIC CONCEPTS SYSTEM Definition: Region of space which is under study Surrounding: the whole universe excluding the system Example: Cash In Ci Cash Out Co BANK Cc Ci : all deposits Co : all withdrawals Cc : others (service charge, interest, etc.) The bank is the system chosen for study. The money flows to, and from the surroundings. Equations (for a given period of time E and B): C E C B = Ci - Co ± Cc

2 The engineer using thermodynamics is an accountant. Instead of using money we uses mass and energy terms. mi mo m E -m B mc m E m B = mi - mo ± mc In term of Energy: Total Energy: U + PE + KE U PE KE : Internal Energy : Potential Energy : Kinetic Energy H (Enthalpy) = U + PV (H+PE+KE)i (U +PE+KE) E - (U +PE+KE) B (H+PE+KE)o Q, W, Ec

3 (U+PE+KE) E (U+PE+KE) B = (H+PE+KE)i - (H+PE+KE)o +Q W ± Ec This is the First Law of Thermodynamics Q W : Heat : Work More complex system:

6 Thermodynamic properties are variables depending only on the state of substance. Properties are functions of the state and in no way dependent on its history. It follows that a change in a property is dependent only on the initial and final states and in no way dependent upon the method or path followed in going from one state to another. Heat and Work are not functions of the state and, therefore are not properties. Intensive properties : not dependent upon the mass of substance Examples : temperature, pressure, fugacity, etc. Extensive properties : dependent upon the mass of substance Examples : volume, enthalpy, entropy, etc. Thermodynamics utilizes concepts that may be related to pressure, volume, and temperature (measurable variables) and to each other in a systematic manner.

7 A process may take place under conditions: Adiabatic Isothermal Isobaric Isochoric Isentropic : no heat added to or removed from system : constant temperature : constant pressure : constant volume : constant entropy Isenthalpic : constant enthalpy Entropy & the Second Law of Thedrmodynamics First Law: - merely keeps track of the energy and mass quantities if the process does proceed. - does not able to describe how a process must proceed. Entropy (S): - measure of randomness, Boltzmann ( ): S = k ln W - can be calculated from pressure and temperature - thermal process, term Q/Tb is used to calculate entropy crossing system boundary.

8 Actual Process: S E S B > Si - So + Q/Tb ± Sc Perubahan entropy selalu positif untuk proses nyata. Entropy balance: S E S B = Si - So + Q/Tb + Sp ± Sc This is known as the Second Law of Thermodynamics. Application: Compressor or Expander, reversible process (ideal) Sp = 0 adiabatic Q = 0 Therefore, for this case, entropy is constant: S = 0

9 Unit of Properties Properties Unit (SI, British) Intensive/Extensive Length Volume Specific Volume Mass Mass density Temperature Pressure Work Heat Power Entropy Specific Heat Capacity Thermal conductivity viscosity

10 Some Examples of Application:

22 1. Suatu aliran memasok steam pada 620 psia dan 700 o F untuk turbin (expander) adiabatic reversible yang mengeluarkannya ke suatu collector terinsulasi yang dipasang suatu piston tanpa friksi dan tekanannya dijaga konstan 23 psia. Tambahan steam dimasukkan ke kolektor melalui throttling valve sehingga temperature di dalam collector tetap 270 o F, sebagaimana dapat dilihat pada gambar di bawah. Jika tangki collector mempunyai luas permukaan ft2, berapa lb (pounds) steam yang melalui turbin, diperlukan untuk mengangkat piston setinggi 1 ft. (Abaikan perubahan energi potensial dan kinetic pada steam serta panas dan friksi sepanjang jalur perpipaan). Expander Throttle valve 4 Collector, 23 psia, 270 o F Steam, 620 psia, 700 o F Diketahui: h (23 psia, 270 o F) = 1175 Btu/lbm; v(23 psia, 270 o F) = 18.6 ft 3 /lbm h (620 psia, 700 o F) = 1350 Btu /lbm; s (620 psia, 700 o F) = Btu /lbm o R h1(23 psia, s=1.584 Btu /lbm o R) = 1065 Btu /lbm Jawab: System (1): Isi tangki Constraints: Q = Pe = Ke =0 P= 23 psia, T =270 o F; frictionless piston & lines

23 Interactions: Aliran masuk, stream #3 Neraca massa: ( m m ) = m 2 1 t 3 δq T Neraca Entrophy: ( S S ) = + s m + S t δ p atau ( s m s m ) = s m t 3 3 karena aliran #3 mempunyai variable yang sudah fix, P= 23 psia, T =270 o F, maka s 1t = s 2t, jadi: s ( m m ) s m 2 1 t 3 3 t = dan h (23 psia,270 o F) = 1175 Btu/lbm; v(23 psia,270 o F) = 18.6 ft 3 /lbm m 4 (sehubungan dengan kenaikan piston 1 ft) = V v ( 1)( ) = = 2. 0lbm Jumlah massa di atas berasal dari dua aliran: dari turbin dan dari valve. System (2): Isi dari mixing T Interaksi: Aliran masuk #1 dan #2 Aliran keluar #3 Constrains: Steady state; Pe = Ke = W = Q = 0 Neraca massa: m & + m& = m& Neraca Energi: h m& + h m& = h m&

24 Maka: m& 1 ( h h )m& = ( h h ) System (3): Isi valve Interaksi: aliran masuk #0 Aliran keluar #2 Constrains: Steady state; Pe = Ke = W = Q = Neraca massa: m & = m& 0 2 Neraca energi: h m& = h m& h 0 (620 psia, 700 o F) = 1350 Btu /lbm, maka h 2 = 1350 Btu /lbm System (3): Isi turbin Interaksi: aliran masuk #0 Aliran keluar #1 Constrains: Steady state, reversible; Pe = Ke = Q = 0 Neraca entrophy: 0 = ( s s ) m& + S p atau s = s 0 1 s 0 (620 psia, 700 o F) = Btu /lbm o R, maka s 1 = Btu /lbm o R dan h 1 (23 psia, s=1.584 Btu /lbm o R) = 1065 Btu /lbm Jadi: m& ( h h )m& = ( h h ) 1 2 ( )( 2. 0 ) = ( ) 1 = lbm kenaikan 1 ft

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