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): CE – CB = ∑ Ci - ∑ Co ± Cc
The engineer using thermodynamics is an “accountant”. Instead of using money we uses mass and energy terms.
mi
mo
mE-mB
mc
mE – mB = ∑ mi - ∑ mo ± mc In term of Energy: Total Energy: U + PE + KE U
: Internal Energy
PE
: Potential Energy
KE
: Kinetic Energy
H (Enthalpy) = U + PV (H+PE+KE)i
(U +PE+KE)E
(U +PE+KE)B Q, W, Ec
(H+PE+KE)o
(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
: Heat
W
: Work
More complex system:
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.
A process may take place under conditions: Adiabatic
: no heat added to or removed from system
Isothermal : constant temperature Isobaric
: constant pressure
Isochoric
: constant volume
Isentropic
: 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 (1844-1906): S = k ln W - can be calculated from pressure and temperature - thermal process, term Q/Tb is used to calculate entropy crossing system boundary.
Actual Process: SE – SB > ∑ Si - ∑ So + ∑ Q/Tb ± Sc Perubahan entropy selalu positif untuk proses nyata.
Entropy balance: SE – SB = ∑ Si - ∑ So + ∑ Q/Tb + Sp ± Sc This is known as the Second Law of Thermodynamics.
Application: Compressor or Expander, reversible process (ideal) à Sp = 0 adiabatic
àQ=0
Therefore, for this case, entropy is constant: ∆S = 0
Unit of Properties Properties Length Volume Specific Volume Mass Mass density Temperature Pressure Work Heat Power Entropy Specific Heat Capacity Thermal conductivity viscosity
Unit (SI, British)
Intensive/Extensive
Some Examples of Application:
1.
Suatu aliran memasok steam pada 620 psia dan 700oF 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 oF, sebagaimana dapat dilihat pada gambar di bawah. Jika tangki collector mempunyai luas permukaan 37.18 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).
1
3 4
2
Collector, 23 psia, 270 oF
Expander Throttle valve
Steam, 620 psia, 700 oF
Diketahui: h (23 psia, 270 oF) = 1175 Btu/lbm; h (620 psia, 700 o F) = 1350 Btu /lbm;
v(23 psia, 270 o F) = 18.6 ft3/lbm
s (620 psia, 700 o F) = 1.584 Btu /lbmo R
h1(23 psia, s=1.584 Btu /lbmo R) = 1065 Btu /lbm
Jawab: System (1):
Isi tangki
Constraints:
Q = Pe = Ke =0 P= 23 psia, T =270 oF; frictionless piston & lines
Interactions:
Aliran masuk, stream #3
Neraca massa:
(m
2
− m 1 )t = m 3
(S
Neraca Entrophy:
atau
(s
2
2
− S1 )t = ∫
δQ + ∫ s 3 δ m3 + S p T
m2 − s 1 m1 ) t = s 3 m 3
karena aliran #3 mempunyai variable yang sudah fix, P= 23 psia, T =270 o
F, maka s1t = s2t, jadi:
s t (m 2 − m1 )t = s 3 m3
dan h (23 psia,270 oF) = 1175 Btu/lbm;
v(23 psia,270 oF) = 18.6 ft3/lbm
m4 (sehubungan dengan kenaikan piston 1 ft) =
V ( 1 )( 37.18 ) = = 2.0lbm v 18.6
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& 2 + m& 1 = m& 3 Neraca Energi: h1 m& 2 + h 2 m& 1 = h3 m& 3
m& 1 =
Maka:
( h3 − h1 )m& 3 ( h1 − h2 )
System (3): Isi valve Interaksi:
aliran masuk #0 Aliran keluar #2
Constrains:
Steady state; Pe = Ke = W = Q = 0
Neraca massa: m& 0 = m& 2 Neraca energi: h0 m& 0 = h2 m& 2 h0 (620 psia, 700 oF) = 1350 Btu /lbm, maka h2 = 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 0 − s 1 )m& 1 + S p atau
s0 = s1
s0 (620 psia, 700 oF) = 1.584 Btu /lbmoR, maka s1 = 1.584 Btu /lbmoR dan h1(23 psia, s=1.584 Btu /lbmoR) = 1065 Btu /lbm
Jadi: m& 1 =
( h3 − h1 )m& 3 ( 1175 − 1350 )( 2 .0 ) lbm = = 1.24 ( h1 − h 2 ) ( 1065 − 1350 ) kenaikan 1 ft
