basic thermodynamic concepts system -...
TRANSCRIPT
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BASIC THERMODYNAMIC CONCEPTS SYSTEM Definition: Region of space which is under study Surrounding: the whole universe excluding the system Example: 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
BANK
Cash In Ci Cash Out Co
Cc
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The engineer using thermodynamics is an accountant. Instead of using
money we uses mass and energy terms.
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
mE-mB
mi mo
mc
(U +PE+KE)E - (U +PE+KE)B
(H+PE+KE)i (H+PE+KE)o
Q, W, Ec
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(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:
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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 otherin a systematic
manner.
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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.
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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) Sp = 0
adiabatic Q = 0
Therefore, for this case, entropy is constant: S = 0
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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
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Some Examples of Application:
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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).
Diketahui: h (23 psia, 270 oF) = 1175 Btu/lbm; v(23 psia, 270 oF) = 18.6 ft3/lbm h (620 psia, 700 oF) = 1350 Btu /lbm; s (620 psia, 700 oF) = 1.584 Btu /lbmoR h1(23 psia, s=1.584 Btu /lbmoR) = 1065 Btu /lbm Jawab:
System (1): Isi tangki
Constraints: Q = Pe = Ke =0
P= 23 psia, T =270 oF; frictionless piston & lines
1
2
3
Steam, 620 psia, 700 oF
Collector, 23 psia, 270 oF
Expander
Throttle valve
4
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Interactions: Aliran masuk, stream #3
Neraca massa: ( ) 312 mmm t =
Neraca Entrophy: ( ) pt SmsTQ
SS ++= 3312
atau ( ) 331122 msmsms t =
karena aliran #3 mempunyai variable yang sudah fix, P= 23 psia, T =270 oF, maka s1t = s2t, jadi:
( ) 3312 msmms tt = 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) = lbm..
).)((vV
0261818371
==
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: 312 mmm &&& =+
Neraca Energi: 331221 mhmhmh &&& =+
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Maka: )hh(
m)hh(m
21
3131
=
&&
System (3): Isi valve
Interaksi: aliran masuk #0
Aliran keluar #2
Constrains: Steady state; Pe = Ke = W = Q = 0
Neraca massa: 20 mm && =
Neraca energi: 2200 mhmh && =
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: ( ) pSmss += 1100 &
atau 10 ss =
s0 (620 psia, 700 oF) = 1.584 Btu /lbmoR, maka s1 = 1.584 Btu /lbm
oR
dan h1(23 psia, s=1.584 Btu /lbmoR) = 1065 Btu /lbm
Jadi: ftkenaikan
lbm.
)().)((
)hh(m)hh(
m1
24113501065
0213501175
21
3131 =
=
=&
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