Lecture 04: STKM3212

LECTURE NOTES 04/07 STKM3212: FOOD PROCESSING

TECHNOLOGY THERMODYNAMICS

(Termodinamik)

SAIFUL IRWAN ZUBAIRI PMIFT, Grad B.E.M. B. Eng. (Chemical-Bioprocess) (Hons.), UTM

M. Eng. (Bioprocess), UTM

ROOM NO.: 2166, CHEMISTRY BUILDING,TEL. (OFF.): 03-89215828,

FOOD SCIENCE PROGRAMME,CENTRE OF CHEMICAL SCIENCES AND FOOD TECHNOLOGY,

UKM BANGI, SELANGOR

1.1 OUTLINES

1.2 DEFINITION OF THERMODYNAMICS.

1.3 INTERNAL ENERGY.1.4 1st LAW OF THERMODYNAMICS.1.5 IMPORTANT TERMS IN

THERMODYNAMICS. 1.6 2nd LAW OF THERMODYNAMICS.

1.2 DEFINITION OF THERMODYNAMICSTHERMODYNAMICS “Science regarding on the relationship

between all forms of energy (e.g.: kinetic energy, internal energy, electrical energy, heat energy etc.)”

HISTORY

(a) 1798 : Benjamin Thompson suggested that cannon (meriam) can be hot due to the WORK that have been done during the construction of the cannon channel (saluran meriam).

(b) 1840 : James Prescott Joule had successfully made calculation on the HEAT & WORK - simple experimental work.

Example of thermodynamics in Food Sciences are follows:

(a) Heating of the food in the microwave oven.

(b) Freezing of the food.

(c) Drying of the fruits.

(d) Pasteurization of milk.

1.3 INTERNAL ENERGY

Experiment done by Joules = “He placed known amounts of H2O (kg) in an insulated (ditebat) container and agitated (dikacau) the H2O with a rotating stirrer (Pengaduk)”

SO ENERGY is added to the H2O as WORK, BUT it is extracted (diambil) from the H2O as HEAT.

Q arises ----- ???? = What happen to this energy between the time it is added to the H2O as WORK and the time it is extracted as HEAT?

ANS: The energy is restored by the transfer of heat in the H2O. The energy restored in the H2O = “INTERNAL ENERGY”.

The addition of heat to a substance increases this molecular activity and thus causes an increase in its internal energy.

CONTINUE:

Internal energy is defined as the energy associated with the random, disordered motion of molecules. It is separated in scale from the macroscopic ordered energy associated with moving objects; it refers to the invisible microscopic energy on the atomic and molecular scale. For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething (FURIOUS) mass of high speed molecules traveling at hundreds of meters per second. If the water were tossed (dilempar) across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole.

CONTINUE:

Potential energy (TENAGA KEUPAYAAN) exists whenever an object which has mass (kg) has a position within a force field (tarikan graviti). The most everyday example of this is the position of objects in the earth's gravitational field. The potential energy of an object in this case is given by the relation:

PE = mgh Where:

PE = Energy (in Joules)

m = mass (in kg)

g = gravitational acceleration of the earth (9.8 m/s2)

h = height above earth's surface (in meters)

CONTINUE:

Kinetic Energy (TENAGA KINETIK) exists whenever an object which has mass (kg) is in motion with some velocity (bergerak dengan sesuatu halaju [m/s]). Everything you see moving about has kinetic energy. The kinetic energy of an object in this case is given by the relation:

KE = (1/2)mv2 Where:

KE = Energy (in Joules)

m = mass (in kg)

v = velocity (in m/s)

CONTINUE:

WHY specific heat (Cp) of H2O is much larger that copper? When the

sample of water and copper are both heated by 1°C, the addition to the kinetic energy (KE) is the same (because of the same temperature given). But to achieve this increase of 1 0C for water, a much larger proportional energy must be added to the potential energy (PE) portion of the internal energy (U) in water (molekul air adalah berselerak, jadi wujud tenaga yg berkaitan dgn tarikan graviti iaitu [PE]). So the total energy required to increase the temperature of the water is much larger.

Much heat (Q) must be supplied to the H2O molecules than the copper to increase to 10C - REASON: the H2O has a lot of scattered molecules as compared to copper. Scattered molecules take time to transfer the energy to the other molecules. Copper has an uniform, tight & static molecules. When the heat is supplied, the molecules besides will vibrate/transfer the energy to the other molecules at instantly.

1.4 1st LAW OF THERMODYNAMICS 1st LAW OF THERMODYNAMICS “Energy cannot be MADE &

DESTROYED but it can change its form to the others”. e.g.: Ball rolls down the hill - Changes of KE PE e.g.: Concentration process using HEAT:

Heat supplied = Heat used for H2O molecules to start evaporated + Heat used for increasing the temperature

Can be interpreted also as (SPECIFIC DEFINITION): “Changes of INTERNAL ENERGY (U) for a system is equal to the HEAT added/supplied (q) and WORK (W) done by the system.

U (Joules) = Q (Joules) + W (Joules) -------- (a)

OR

(Energy of the system) + (Energy of surrounding) = 0

CONTINUE:

HOW EQUATION (a) APPEARS? (Energy of the system) = Ut + EK + EP ------------- (1)

(Energy of surrounding) = Q W -------------------- (2) In the thermodynamics logic = HEAT & WORK refer to energy IN

TRANSIT (in OR out) ACROSS THE BOUNDARY which divides the system from its surrounding.

These forms of energy are not STORED & never contained in a body or system.

Energy restored with the material objects is KE & PE because of the position/configuration/motion of substance.

IF in CLOSED SYSTEM = (1) mass is constant, (2) heat and work exchange occurs in/out the system & (3) no transport of internal energy cross the boundary of the system.

CONTINUE:

SO: (1) = (2) Ut + EK + EP = Q W ------------- (3)

But, closed system often undergo processes that cause NO CHANGE in KE and PE ------- [EK + EP = 0].

Only change in INTERNAL ENERGY. SO; Final equation will be:

U (Joules) = Q (Joules) + W (Joules)

CONTINUE:

HEAT (Q) “Energy that can be transferred between SYSTEM & SURROUNDING because of the temperature differentiation”.

sign convention:

Qsystem < 0 Heat is transferred from the system to the surroundings.

Qsystem > 0 Heat is transferred to the system from the surroundings WORK (W) “Energy between the system and surrounding and

independent with the temperature differentiation. When a force acts to move an object, we say that Work was done on the object by the Force”.

Types of WORK are:

(a) ELECTRIC: e.g.: current through a battery solution.

(b) CHEMICAL: e.g.: molecules synthesis, formation of

bonding (Van der Waal etc.).

(c) MECHANICAL: e.g.: Volume change, mass movement .

CONTINUE:

RECALL:FORCE (F) = m.a = mass × acceleration = kg.m/s2 = Newton (N)

WORK (W) = FORCE (N) × DISTANCES (m) = N.m (Joules)

CONTINUE:

sign convention:

Wsystem < 0 Work is done by the system on the surroundings.

Wsystem > 0 Work is done on the system by the surroundings. INTERNAL ENERGY (U) “Total HEAT & WORK that have been done to

the system” e.g.: CONCENTRATION PROCESS:

Heat is given to the process.

No work done.

Incremental of Internal Energy (e.g.:H2O molecules start to change to other phase).

e.g.: REFRIGERATION PROCESS:

Work is done = pump, coolant medium is moving

Heat is removed.

Changing in Internal Energy.

1.5 IMPORTANT TERMS IN THERMODYNAMICS(1) The Concept of a “SYSTEM” A thermodynamic system is a quantity of matter of fixed identity,

around which we can draw a boundary. The boundaries may be fixed or moveable. Work or heat can be

transferred across the system boundary. Everything outside the boundary is the surroundings.

CONTINUE:

When working with devices such as engines it is often useful to define the system to be an individual volume with flow in and out. This is termed a control volume.

A closed system is a special class of system with boundaries that matter cannot cross. Hence the principle of the conservation of mass is automatically satisfied whenever we employ a closed system analysis. This type of system is sometimes termed a control mass.

CONTINUE:

(2) The Concept of a “State” (Keadaan) THERMODYNAMIC PROPERTIES: [Pressure (atm), density (g/ml) and

temperature (oC)] reflects the THERMODYNAMIC STATE OF THE SYSTEM reflects changes at the molecular /microscopic level (INTERNAL ENERGY).

e.g.: N2 at T=300K, P=1 bar, has a fixed specific gas density (g/ml) and internal energy (J). If there are changes among these 2 properties, it will give set of thermodynamics properties. If the gas heated/cooled/compress/expanded & return to its initial properties it is found to have exactly the same set of properties as before.

Properties may be extensive (general) or intensive (specific). EXTENSIVE PROPERTIES: depend on the QUANTITY of matter present

e.g.: Density (g/ml). INTENSIVE PROPERTIES: do not depend on the QUANTITY of matter

present Temperature (oC) and pressure (atm).

CONTINUE:

(3) The Concept of “Equilibrium” Is word denoting a “STATIC CONDITION” The absence of change

(tiada perubahan). A system at equilibrium may be described as one in which all

FORCES are in exact balance. e.g.: Hot food is exposed in the ambient temperature room

eventually, the temperature of the hot food will be in equilibrium with the ambient room temperature (surrounding).

System is in equilibrium when its state function [P (atm), T (oC), V (ml)] is CONSTANT at any locations in the system.

A system in thermodynamic equilibrium satisfies:

(a) Mechanical equilibrium (no unbalanced forces)

(b) Thermal equilibrium (no temperature differences)

(c) Chemical equilibrium (no amount differences).

CONTINUE:

(4) Heat capacity (C) (of a system) the heat required to raise the temperature of the system by one Kelvin (or one °C)

The numerical value of the heat capacity depends on the manner in which the heat transfer is made. The principal heat capacities are:

CV = heat capacity at constant volume

CP = heat capacity at constant pressure

e.g.: For water, Cp = 4.184 kJ/kg.0C

(5) Specific heat (c) (of a substance) the heat required to raise the temperature of one gram of the substance by one Kelvin (or one °C)

(6) Latent heat [LH] (Haba pendam) Amount of heat that required for the phase changing at constant temperature.

Ice cube Liq H2O Vapor LH, T=ambient LH, T=1000C

EXAMPLE 1:

Water is flows over a waterfall 100 m in height. Consider 1 kg of the water and assume that no energy is exchanged between 1 kg and its surroundings.

(a) What is the potential energy (PE) of the water at the top of the falls with respect to the base of the falls?

(b) What is the kinetic energy of the water just before it strikes bottom?

(c) After 1 kg of water enters the river below the falls, what change has occurred in its state (keadaan)?

ANS:

Taking the 1 kg water as the system and noting that it exchanges NO ENERGY WITH ITS SURROUNDING ------ Q + W = 0

(Energy of the system) = Ut + EK + EP

(Energy of surrounding) = Q W 0

CONTINUE:

If; (Energy of the system) + (Energy of surrounding) = 0

[Ut + EK + EP] + [Q + W] = 0

So; Ut + EK + EP = 0 ----- This equation applies to each part of the process.

(a)PE = mzg = 1 kg × 100 m × 9.8066 m.s-2 = 980. 66 kg.m2.s-2

(b)During the free fall of the water, NO MECHANISM exists for the conversion of potential/kinetic into internal energy. So, Ut = 0 and

Ut + EK + EP = 0 ----------- EK + EP = EK2 - EK1 + EP2 - EP1 = 0

EK2 - EP1 = 0 ----- EK2 = EP1 = 980.66 J

0

Conversion factor: 1 J = 1 kg.m2.s-2 = 1 N.m

0 0

CONTINUE:

(c) As the 1 kg of water strikes bottom and mixes with other falling water to form a river, the resulting turbulence (GELORA) has the effect of converting (KE) Internal Energy. During this process,

EP = 0 and becomes:

Ut + EK + EP = 0 ----------- Ut + EK = 0 OR

Ut + (EK3 - EK2) = 0 ----------- Ut = - EK3 + EK2

However, the river velocity is assumed small, and therefore EK3 is negligible. SO;

Ut = EK2 = 980.66 J

0

CONTINUE:

The overall result of the process is the CONVERSION OF:

{PE of the water INTERNAL ENERGY of the water}

This change in internal energy is manifested by a temperature rise of the water.

Since energy in the amount of 4,184 J.kg-1 {For water, Cp = 4.184 kJ/kg.oC} is required for a temperature to rise 1 oC in water, the temperature increase is:

4.184 J/kg --------- 1 0C

If; 980.66 J/kg -------- ??? 1/4.184 × 980.66 = 0.234 oC (if there is no heat transfer with the surroundings)

CONTINUE:

(7) Enthalpy is used to solved a problems that involved the FLUID/GAS FLOW PROCESSES (steady state - no accumulation of material/energy at any point) & NON CONSTANT MASS.

When the WORK (w) is done to the system with P = constant, the equation of ENTHALPY must be used instead of 1st law equation. The HEAT (q) is replaced by Enthalpy (H).

U + PV = H = enthalpy ------- (1)Where: (SI unit) U = internal energy (Joule)

H = enthalpy (Joule)

P = absolute pressure (N.m2)

V = m3

If there is a differential change occurs in the system, THIS FINAL EQUATION is used. After the integration:

H = U + P.V --------- (2) (flowable process & non-constant mass)

CONTINUE:

RECALL:FORCE (F) = m.a = mass × acceleration = kg.m/s2 = Newton (N)

WORK (W) = FORCE (N) × DISTANCES (m) = N.m (Joules)

For specific WORK (W) of THERMODYNAMICS:

WORK = W = - P.dV = Joules (J)

If, P = constant:

W = - P.dV = - P.(V2-V1) = - P.V --------- (1)

If, P and V = constant: Generally ---- dU = dQ + dW = dQ - PdV

dU = dQ - P(0) = dQ ---- after integration

U = Q ------------ (2)

Conversion factor: 1 J = 1 Pa.m3 = 10-5 Bar.m3

1 bar = 105 Pa

U (Joules) = Q (Joules) + W (Joules) --- (non-flow process & constant mass)

EXAMPLE 2:

Calculate H and U for 1 kg of water when it is vaporized at the constant temperature of 100 OC and constant pressure of 101.33 kPa. The specific volumes of liquid and vapor water at these conditions are 0.00104 and 1.673 m3 kg-1. For this change, heat in the amount of 2,256.9 kJ is added to the water.

ANS:

GIVEN BASIS: 1 kg of water is taken as the system.

IMAGINE: The fluid contained in a cylinder by a frictionless piston

exerts/uses a constant P = 101.33kPa.

As heat is added (q = +), the water is expands from its initial to its final volume, doing WORK on the piston:

W = - P.V = - 101.33kPa × (1.673 - 0.001) m3 =

W = -169.4 kPa.m3 = -169.4 kNm-2.m3 = - 169.4 kJ

0.001 m3 kg-1 --- given basis 1 kg, so to get the volume only --- 0.001 m3 kg-1 × 1 kg = 0.001 m3

CONTINUE:

TAKE NOTE: The specific volumes of liquid (saturated liquid) and vapor water (saturated vapor) at these conditions are 0.00104 and 1.673 m3 kg-1 can be obtained using the “STEAM TABLES -PROPERTIES OF SATURATED STEAM ” (the values have been given).

Since Q = 2,256.9 kJ so,

U = Q + W ----- (1st law) = 2,256.9 - 169.4 = 2,087.5 kJ

with P = constant ------ H = U + P.V

But P.V = - W, therefore:

H = U + P.V = U - W = Q = 2,256.9 kJ

CONTINUE:

1.6 2nd LAW OF THERMODYNAMICS

RECALL: 1st LAW

CONTINUE:

(1) “Heat flows spontaneously from a hot to a cold body” tells us that an ice cube must melt on a hot day, rather than becoming colder.

Example of VIOLATION of the 2nd law ------ “REFRIGERATOR” To make a cool object in a warm place cooler this is what a

refrigerator does but this involves the input of some external energy. As such, the flow of heat is not spontaneous in this case.

A useful analogy in this regard is to think of heat flowing from hot to cold objects as running “down hill”, which is what objects naturally do in Newtonian mechanics.

It is possible to make objects go up hill, but only by doing external work on them.

This movement of heat from a cool to a warm reservoir through some external work is the basis of the following three devices.

CONTINUE:

In a (A) refrigerator, the cool reservoir is the inside of the refrigerator, and the warm reservoir is the room itself. From this, one can see that leaving a refrigerator door open will not cool off the room that it is in.

In an (B) air conditioner, the cool reservoir is the inside of a house, and the warm reservoir is the outside. This is used to cool a house in the summer.

In a (C) heat pump, the cool reservoir is the outside of a house, and the warm reservoir is the inside. This can be used to warm a house in the winter. The heat pump is thus just the reverse of an air conditioner, and indeed some heat pumps have a switch which allows them to function as an air conditioner in the summer.

CONTINUE:

A generic heat pump

CONTINUE:

(2)“Heat cannot be completely converted into other forms of energy” (A) steam powered engines (B) internal combustion.

we can understand that in principle, instead of going into kinetic energy to raise the temperature of the cooler substance, heat can be harnessed to do useful work.

In this context, the 2nd law makes the very surprising statement some of the heat energy must always be lost, so that the conversion from heat to work is never 100% efficient.

However, this form of the 2nd law places no restriction on converting other forms of energy into heat it is the conversion of heat into other forms of energy that turns out never to be 100% efficient, even in principle.

CONTINUE:

An machine which converts heat into other forms of energy is called (A) “heat engine/steam powered engines”

The important aspect here is that some “waste heat” is always expelled into the cooler reservoir; no heat engine could operate without such expulsion/discharged (pembuangan).

This is why, for example, the winter near a steam powered electrical generating plant that nearby ice on a river is melted this comes from the waste heat of the plant being expelled into the river.

CONTINUE:

A generic heat engine

CONTINUE:

For an (B) “internal combustion engine” the fuel mixture explodes, either from a spark plug for a gas engine or from the high pressure for a diesel engine.

This drives the piston downwards, which subsequently turns the crankshaft and eventually the ROTATION of wheels this is the part which converts the energy of heat into useful work.

The piston then rises, expelling the exhaust gases which carry away the waste heat.

The cycle then goes on to draw in more fuel mixture to repeat the cycle.

The major point here is that the exhaust gases carry with them excess heat which could not be converted into useful work.

CONTINUE:

A simplified internal combustion engine

CONTINUE:

(3)“Systems become more disorganized over time” EXPERIMENT: Take a jar full of pennies that have carefully been arranged so that all

the heads are facing up. Then tip the jar so that all the pennies fall to the ground theoretically,

it is possible that all the pennies will land with heads facing up. Our experience though is that most of the time, some of the pennies will

have heads up and some will have tails up. This example illustrates clearly, and quite accurately what is meant by

‘ORDER’, and why nature prefers ‘DISORDER’. Having 50 pennies in a jar all with heads up is clearly an ordered state. Having 50 pennies on the ground with some, unspecified number

heads, and the rest tails is clearly disordered state.

CONTINUE: This example also shows that ordered states are by no means

impossible to achieve in Nature, they just require WORK: someone must put a lot of effort into arranging the pennies so that they are all facing the same way.

“Heat transfer by conduction” works is also an illustration of this statement of the 2nd Law.

When a hot and a cold body are initially in contact, the system is somewhat ordered, in that we know most of the molecules in the hot side are moving faster than those in the cool side.

However, this degree of ORDER is “LOST” after the system has attained a uniform temperature (SUHU SEIMBANG) eventually it became DISORDER .

CONTINUE:

The concept of entropy is introduced to characterize the order of a system.

Here, state is a very general term, and could include position, speed, etc.

This is equivalent to saying that the number of states available to a system increases in general, by which the system thus becomes more disordered. Therefore:

[ STATE in the system; DISORDERED the system] In the coin example, the initial state (all heads up) was unique, whereas

the likely finally state (roughly half heads, half tails) could be achieved in a very large number of different ways The entropy increased.