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

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

(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 à Q = 0

Therefore, for this case, entropy is constant: ∆S = 0

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

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).

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

Interactions: Aliran masuk, stream #3

Neraca massa: ( ) 312 mmm t =−

Neraca Entrophy: ( ) pt SmsTQ

SS ++=− ∫∫ 3312 δδ

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 &&& =+

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 /lbmoR

dan h1(23 psia, s=1.584 Btu /lbmoR) = 1065 Btu /lbm

Jadi: ftkenaikan

lbm.

)().)((

)hh(m)hh(

m1

24113501065

0213501175

21

3131 =

−−

=−

−=

&&