Saturday, May 16, 2009

PROPERTY DIAGRAMS AND STEAM TABLESThermodynamicsTemperature-Entropy (T-s) DiagramA T-s diagram is the type of diagram most frequently used to analyze energy transfer systemcycles. This is because the work done by or on the system and the heat added to or removedfrom the system can be visualized on the T-s diagram. By the definition of entropy, the heattransferred to or from a system equals the area under the T-s curve of the process. Figure 13 isthe T-s diagram for pure water. A T-s diagram can be constructed for any pure substance. Itexhibits the same features as P-u diagrams.Figure 13 T-s Diagram for WaterIn the liquid-vapor region in Figure 13, water and steam exist together. For example, at pointA, water with an entropy (s_f) given by point B, exists together with steam with an entropy (s_g)given by point C. The quality of the mixture at any point in the liquid-vapor region can be found by .s = xs_g + (1 - x)s

Amagat's law or the Law of Partial Volumes of 1880 describes the behaviour and properties of mixtures of ideal (as well as some cases of non-ideal) gases. Of use in chemistry and thermodynamics, Amagat's law states that the volume V_m of a gas mixture is equal to the sum of volumes V_i of the K component gases, if the temperature T and the pressure p remain the same: ^[1]

$V_m(T,p)=\sum_{i=1}^K V_i(T,p)$ .

In chemistry and physics, Dalton's law (also called Dalton's law of partial pressures) states that the total pressure exerted by a gaseousmixture is equal to the sum of the partial pressures of each individual component in a gas mixture. This empirical law was observed by John Dalton in 1801 and is related to the ideal gas laws.

Mathematically, the pressure of a mixture of gases can be defined as the summation

$P_{total} = \sum_{i=1} ^ n {p_i}$ or $P_{total} = p_1 +p_2 + \cdots + p_n$

where $p_{1},\ p_{2},\ p_{n}$ represent the partial pressure of each component.

It is assumed that the gases do not react with each other.

$\ P_{i} =P_{total}x_i$

where $x_i\ =$ the mole fraction of the i-th component in the total mixture of m components .

When a real gas, as differentiated from an ideal gas, expands at constant enthalpy (i.e., no heat is transfered to or from the gas, and no external work is extracted), the gas will be either cooled or heated by the expansion. That change in gas temperature with the change in pressure is called the Joule-Thomson coefficient and is denoted by µ, defined as:

µ = (dT/dP) at constant enthalpy

The value of u depends on the specific gas, as well as the temperature and pressure of the gas before expansion. For all real gases, µ will equal zero at some point called the "inversion point". If the gas temperature is below its inversion point temperature, µ is positive ... and if the gas temperature is above its inversion point temperature, µ is negative. Also, dP is always negative when a gas expands. Thus:

If the gas temperature is below its inversion temperature:

-- µ is positive and dP is always negative

-- hence, the gas cools since dT must be negative

If the gas temperature is above its inversion temperature:

-- µ is negative and dP is always negative

-- hence, the gas heats since dT must be positive

"Perry's Chemical Engineers' Handbook" provides tabulations of µ versus temperature and pressure for a number of gases, as do many other reference books. For most gases at atmospheric pressure, the inversion temperature is fairly high (above room temperature), and so most gases at those temperature and pressure conditions are cooled by isenthalpic expansion.

Helium and hydrogen are two gases whose Joule-Thomson inversion temperatures at atmospheric pressure are very low (e.g., about −222 °C for helium). Thus, helium and hydrogen will warm when expanded at constant enthalpy at atmospheric pressure and typical room temperatures.

It should be noted that µ is always equal to zero for ideal gases (i.e., they will neither heat nor cool upon being expanded at constant enthalpy).

the definition of entropy is given by

Rearranging the above equation gives

(1)

The entropy change during an internally reversible process (1-2) is

Only when the relation between δQ and T is known, the entropy change can be determined. The relations between δQ and T can be found by considering the energy balance of a closed system.

The differential form of the energy balance for a closed system, which contains a simple substance and undergoes an internally reversible process, is given by

dU = δQ_rev - δW_rev (2)

The boundary work of a closed system is

δW_rev = PdV (3)

Substituting equations (1) and (3) into equation (2) gives

dU = TdS- PdV
TdS = dU + PdV

Tds = du +Pdv (4)

where
s = entropy per unit mass

Equation (4) is known as the first relation of Tds, or Gibbs equation.

The definition of enthalpy gives

h = u + Pv

differential the above equation yields

dh = du +Pdv + vdP

Replacing du + Pdv with Tds yields

dh = Tds + vdP
Tds = dh -vdP (5)

Equation (5) is known as the second relation of Tds.

Although the Tds equations are obtained through an internally reversible process, the results can be used for both reversible or irreversible processes since entropy is a property.

Rewriting equations (4) and (5) in the following form

ds = du/T + Pdv/T
ds = dh/T + vdP/T

The entropy change during a process can be determined by integrating the above equations between the initial and the final states.

The vaporization curves of most liquids have similar shape. The vapour pressure steadily increase as the temperature increases. A good approach is to find a mathematical model for the pressure increase as a function of temperature. Experiments showed that the pressure P, enthalpy of vaporization, DH_vap, and temperature T are related,P = A exp (- DH_vap / R T)where R (= 8.3145 J mol^-1 K^-1) and A are the gas constant and unknown constant. This is known as the Clausius- Clapeyron equation. If P₁ and P₂ are the pressures at two temperatures T₁ and T₂, the equation has the form:

     P₁    DH_vap    1     1 ln (---) = ----  (--- - ---)      P₂     R      T₂    T₁

The Clausius-Clapeyron equation allows us to estimate the vapor pressure at another temperature, if the vapor pressure is known at some temperature, and if the enthalpy of vaporization is known.

As we have seen, the fundamental thermodynamic relation ${\rm d}E=T{\rm d}S-P{\rm d}V$ implies that the natural variable in which to express are and : .

That means that on purely mathematical grounds, we can write

$\begin{displaymath} {\rm d}E=\left({\partial E\over\partial S}\right)_V{\rm d}S+\left({\partial E\over\partial V}\right)_S{\rm d}V \end{displaymath}$

But comparison with the fundamental thermodynamic relation, which contains the physics, we can make the following identifications:

$\begin{displaymath} T=\left({\partial E\over\partial S}\right)_V\qquad\hbox{and}\qquad P=-\left({\partial E\over\partial V}\right)_S \end{displaymath}$

These (especially the second) are interesting in their own right. But we can go further, by differentiating both sides of the first equation by and of the second by :

$\begin{displaymath} \left({\partial T \over\partial V}\right)_S =\left({\partia... ...\partial S}\left({\partial E\over\partial V}\right)_S\right)_V \end{displaymath}$

Using the fact that the order of differentiation in the second derivation doesn't matter, we see that the right hand sides are equal, and thus so are the left hand sides, giving

$\begin{displaymath} \left({\partial T \over\partial V}\right)_S=-\left({\partial P\over\partial S}\right)_V \end{displaymath}$

By starting with , and , we can get three more relations.

$\begin{figure}\begin{center}\mbox{\epsfig{file=maxwell.eps,width=14truecm,angle=0}} \end{center}\end{figure}$

The two equations involving derivatives of are particularly useful, as they provide a handle on which isn't easily experimentally accessible.

For non-hydrodynamic systems, we can obtain analogous relations involving, say, and instead of and ; for instance by starting with ${\rm d}E=T{\rm d}S+ m{\rm d}B$ we get $(\partial T /\partial B)_S=(\partial m/\partial S)_B$ .

The third law of thermodynamics is a statistical law of nature regarding entropy and the impossibility of reaching absolute zero of temperature. The most common enunciation of third law of thermodynamics is:

“	As a system approaches absolute zero, all processes cease and the entropy of the system approaches a minimum value.

In thermodynamics, the Helmholtz free energy is a thermodynamic potential which measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature and volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant. Under these conditions, it is minimized at equilibrium.

The Helmholtz energy is defined as:

$A \equiv U-TS\,$ ^[2]

where

A is the Helmholtz free energy (SI: joules, CGS: ergs),
U is the internal energy of the system (SI: joules, CGS: ergs),
T is the absolute temperature (kelvins),
S is the entropy (SI: joules per kelvin, CGS: ergs per kelvin).

Gibbs Function

The Gibbs function is also known as the Gibbs free energy and is defined in terms of temperature, T, the enthalpy, H, and entropy, S, such that :
G = H - TS.

In a thermodynamic process with no composition change, the change in Gibbs function is given by:
dG = dH - TdS - SdT.

Using the combined first and second laws, dU = TdS - PdV and dH = TdS + VdP gives:

dG = VdP -SdT

For a constant pressure process: (dG/dT)_p = - S, and for a constant temperature process:(dG/dP)_T = V, where V is the volume of the system.

A law of physics, the second law of thermodynamics, states that the total entropy of any system cannot decrease except insofar as it flows outward across the boundary of the system. As a corollary, in an isolated system, the entropy cannot decrease (the second law places no restrictions on the increase of entropy).

By implication, the entropy of the whole universe, assumed to be an isolated system, cannot decrease; it is always increasing. This is because there are processes that produce entropy from scratch, and the second law defines that these increases cannot be undone elsewhere.

Two important consequences are that heat cannot of itself pass from a colder to a hotter body: i.e., it is impossible to transfer heat from a cold to a hot reservoir without at the same time converting a certain amount of work to heat. It is also impossible for any device that can operate on a cycle to receive heat from a single reservoir and produce a net amount of work; it can only get useful work out of the heat if heat is at the same time transferred from a hot to a cold reservoir. This means that there is no possibility of a "perpetual motion" system. Also, from this it follows that a reduction in the increase of entropy in a specified process, such as a chemical reaction, means that it is energetically more efficient.

In general, according to the second law, the entropy of a system that is not isolated may decrease. An air conditioner, for example, cools the air in a room, thus reducing the entropy of the air. The heat, however, involved in operating the air conditioner always makes a bigger contribution to the entropy of the environment than the decrease of the entropy of the air. Thus, the total entropy of the room and the environment increases, in agreement with the second law.

Entropy is the measure of disorder of a system. In thermodynamics, entropy is a measure of how close a thermodynamic system is to equilibrium. A thermodynamic system is any physical object or region of space that can be described by its thermodynamic quantities such as temperature, pressure, volume and density. In simple terms, the second law of thermodynamics states that for a system, the differences in intensive thermodynamic quantities such astemperature, pressure, and chemical potential tend to even out as time goes by, unless there is an outside influence which works to maintain the differences. The end point of this evening-out process is called equilibrium, and as the system moves toward equilibrium, the entropy of the system increases, becoming a maximum at equilibrium.

A process that is not reversible is termed irreversible. In an irreversible process, finite changes are made; therefore the system is not at equilibrium throughout the process. At the same point in an irreversible cycle, the system will be in the same state, but the surroundings are permanently changed after each cycle

In thermodynamics, a reversible process, or reversible cycle if the process is cyclic, is a process that can be "reversed" by means ofinfinitesimal changes in some property of the system without loss or dissipation of energy.^[1] Due to these infinitesimal changes, the system is at rest throughout the entire process. Since it would take an infinite amount of time for the process to finish, perfectly reversible processes are impossible. However, if the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle.^[2]

Clausius inequality

Let system I be an arbitrary thermodynamic system, which is able to exchange heat with thermal reservoirs R1 and R2, having the temperatures T1 and T2 respectively. In this derivation we will not specify which of these two reservoirs is a heater and which is a cooler. We will consider the heat lost by a thermal reservoir to be positive and the heat gained by a reservoir to be negative. This will allow formulating the final result symmetrically with respect to both reservoirs.

Let the system I perform a cyclic process – reversible or irreversible – during which it received heat Q1 from reservoir R1 and Q2 from reservoir R2. Since the system has returned to its initial state the total amount of heat gained by the system Q1+Q2 is equal to the work it has performed.

Lets now take a Carnot engine and make it work using the same heat reservoirs R1 and R2. In order to the presence of the Carnot engine not to influence the amounts of heat Q1 and Q2 the system I has received from the heat reservoirs during its cycle, we can run the Carnot engine after the system I has completed its cycle. If at this point we thermally isolate the system I, the reservoirs R1 and R2 will exchange heat with the Carnot engine only. The presence of the Carnot engine does not affect the cyclic process in the system I since it has already finished by the time the Carnot engine was introduced.

Let the Carnot engine perform a cyclic process, during which it receives heat Q1’ from reservoir R1 and Q2’ from R2 respectively. It is essential for the further derivation to note that the Carnot engine is reversible. It can work either as an engine or a refrigerator. Apart from that the isotherm 12 in Carnot cycle can be taken as short as desired (see Fig. 1). Therefore, the work done by the Carnot engine can be as small as desired. One can also make the Carnot engine perform as much work as desired via performing many cycles. Thus the Carnot engine can perform any specified amount of positive or negative work. This allows choosing the value of either Q1’ or Q2’. It is always possible to make one of these heats have a chosen positive or negative value.

Figure 1. Carnot cycle

According to Carnot theorem

(Q1’/T1) + (Q2’/T2) = 0

Let us combine the system I and the Carnot engine into one complex system. The cycles performed consecutively by the system I and Carnot engine can be combined into one cyclic process. During this process the combined system

has received the heat Q1 + Q1’ from reservoir R1,

has received the heat Q2 + Q2’ from reservoir R2,

has performed the work W = Q1 + Q1’ + Q2 + Q2’.

Let us choose Q1’ so that Q1’ = -Q1 then

Q2’ = T2 (Q1/T1).

As a result of the cycle the state of the reservoir R1 remains unchanged. The reservoir R2 looses the amount of heat

Q2 + Q2’ = Q2 + t2 (Q1/T1) = T2 [(Q1/T1) + (Q2/T2)]

An equivalent work W = Q2 + Q2’ has been performed using this amount of heat. If this work were positive, we would have the Thomson-Plank process, which is forbidden by the II law of thermodynamics. Therefore, the work should be negative or equal to zero (W ≤ 0). Since the absolute temperature T2 is always positive this leads to

(Q1/T1) + (Q2/T2) ≤ 0.

This inequality is called Clausius inequality.

On the basis of the non-occurrence of perpetual motion, Carnot was able to demonstrate that: (1) no heat engine operating between two reservoirs can be more efficient than a Carnot engine operating between the same reservoirs (Carnot’s theorem); (2) the efficiency of a Carnot engine depends only on the temperatures of the reservoirs and not on the nature of the working substance (the corollary to Carnot’s theorem).

A heat pump is a machine or device that moves heat from one location (the 'source') to another location (the 'sink' or 'heat sink') usingmechanical work. Most heat pump technology moves heat from a low temperature heat source to a higher temperature heat sink.^[1] Common examples are food refrigerators and freezers, air conditioners, and reversible-cycle heat pumps for providing thermal comfort.

Heat pumps can be thought of as a heat engine which is operating in reverse. One common type of heat pump works by exploiting the physical properties of an evaporating and condensing fluid known as a refrigerant. In heating, ventilation, and air conditioning (HVAC) applications, a heat pump normally refers to a vapor-compression refrigeration device that includes a reversing valve and optimized heat exchangers so that the direction of heat flow may be reversed. Most commonly, heat pumps draw heat from the air or from the ground. Some air-source heat pumps do not work as well when temperatures fall below around −5°C(23°F).

A heat engine is a physical or theoretical device that converts thermal energy to mechanical output. The mechanical output is called work, and the thermal energy input is called heat. Heat engines typically run on a specific thermodynamic cycle. Heat engines can be open to the atmospheric air or sealed and closed off to the outside (Open or closed cycle).

Equivalence of Two Statement

The Clausius statement and the Kelvin-Planck statement of the second law of thermodynamics are equivalent in their consequences. Any device that violates the Kelvin-Planck statement also violates the Clausius statement.

Consider two devices A and B working between a high-temperature reservoir and a low-temperature reservoir. Device A is assumed to transfer Heat (QA,H) to work ( WA, net) and have 100 percent efficiency. That is,

WA, net = QA,H

Device A violates the Kelvin-Planck statement.

Device B is a heat pump which receives heat (QB,L) from the low-temperature reservoir, receives work (WA, net) from device A and rejects the total energy of WA, net and QB,L to the high-temperature reservoir. Heat rejected by device C, which is the combination of devices A and B, to the high-temperature reservoir equals

(QB,L + WA,net) - QA,H

= (QB,L + WA, net) - WA, net

= QB,L

Device C receives heat (QB,L) from the low-temperature reservoir and rejects the same amount of heat (QB,L) to the high-temperature reservoir. It violates the Clausius statement.

It can also be shown in a similar way that if a device violates the Clausius statement, it also violates the Kelvin-Planck statement. Hence, the Clausius statement and the Kelvin-Planck statement are two equivalent expressions of the second law of thermodynamics.

Kelvin-Planck statement

No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work.

It is impossible to convert heat completely into work in a cyclic process.

That is, it is impossible to extract energy by heat from a high-temperature energy source and then convert all of the energy into work. At least some of the energy must be passed on to heat a low-temperature energy sink. Thus, a heat engine with 100% efficiency is thermodynamically impossible.

Also, due to Rudolf Clausius, is the simplest formulation of the second law, the heat formulation or Clausius statement:

Heat generally cannot spontaneously flow from a material at lower temperature to a material at higher temperature.

Informally, "Heat doesn't flow from cold to hot (without work input)", which is obviously true from everyday experience. For example in a refrigerator, heat flows from cold to hot, but only when aided by an external agent (i.e. the compressor). Note that from the mathematical definition of entropy, a process in which heat flows from cold to hot has decreasing entropy. This can happen in a non-isolated system if entropy is created elsewhere, such that the total entropy is constant or increasing, as required by the second law. For example, the electrical energy going into a refrigerator is converted to heat and goes out the back, representing a net increase in entropy.

The exception to this is in statistically unlikely events where hot particles will "steal" the energy of cold particles enough that the cold side gets colder and the hot side gets hotter, for an instant. Such events have been observed at a small enough scale where the likelihood of such a thing happening is large enough.^[2] The mathematics involved in such an event are described by fluctuation theorem.

There are many ways of stating the second law of thermodynamics, but all are equivalent in the sense that each form of the second law logically implies every other form.[1] Thus, the theorems of thermodynamics can be proved using any form of the second law and third law.

The formulation of the second law that refers to entropy directly is as follows:

In a system, a process that occurs will tend to increase the total entropy of the universe.

Thus, while a system can undergo some physical process that decreases its own entropy, the entropy of the universe (which includes the system and its surroundings) must increase overall. (An exception to this rule is a reversible or "isentropic" process, such as frictionless adiabatic compression.) Processes that decrease total entropy of the universe are impossible. If a system is at equilibrium, by definition no spontaneous processes occur, and therefore the system is at maximum entropy.

Carnot Cycle

Nicholas Sadi Carnot devised a reversible cycle in 1824 called the Carnot cycle for an engine working between two reservoirs at different temperatures. It consists of two reversible isothermal and two reversible adiabatic processes. For a cycle 1-2-3-4, the working material

Undergoes isothermal expansion in 1-2 while absorbing heat from high temperature reservoir
Undergoes adiabatic expansion in 2-3
Undergoes isothermal compression in 3-4, and
Undergoes adiabatic compression in 4-1.

Heat is transferred to the working material during 1-2 (Q₁) and heat is rejected during 3-4 (Q₂). The thermal efficiency is thus η_th = W/Q₁. Applying first law, we have, W = Q₁ − Q₂, so that η_th = 1 − Q₂/Q₁.

Carnot's principle states that

No heat engine working between two thermal reservoirs is more efficient than the Carnot engine, and
All Carnot engines working between reservoirs of the same temperature have the same efficiency.

The proof by contradiction of the above statements come from the second law, by considering cases where they are violated. For instance, if you had a Carnot engine which was more efficient than another one, we could use that as a heat pump (since processes in a Carnot cycle are reversible) and combine with the other engine to produce work without heat rejection, to violate the second law. A corollary of the Carnot principle is that Q₂/Q₁ is purely a function of t₂ and t₁, the reservoir temperatures. Or,

$\frac{Q_1}{Q_2} = \phi \left( t_1, t_2 \right)$

Friday, May 15, 2009

Heater.. The steady flow equation as applied to a fluid heater..

Potential energy (z) assumed to be constant..
Kinetic energy changes (1 to 2) assumed to be very small

Heater

w = 0 therefore
q = h₂-h₁

Turbine ..(Assumed Adiabetic Expansion)..The steady flow equation as applied to a turbine..

Potential energy (z) assumed to be constant..
Kinetic energy changes (1 to 2) assumed to be very small

Turbine

q = 0 therefore
w = h₂-h₁

Throttling ..(Assumed Adiabetic )...The steady flow equation as applied to a orifice..

Potential energy (z) assumed to be constant..
The higher velocity at orifice section is dissipated in tube downstream of the orifice and therefore the kinetic energies at 1 and 2 are similar

Orifice

q = w = 0 therefore
therefore h₁ = h₂

Nozzle..(Assumed Adiabetic )...The steady flow equation as applied to smooth nozzle..

Potential energy (z) assumed to be constant..
Kinetic energy changes are assumed to be significant

Nozzle

q = w = 0 therefore
(v₂² - v₁² ) /2 = (h₁ - h₂ )

The steady flow energy equation relates to open systems working under steady conditions i.e in which conditions do not change with time.

The boundary encloses a system through which fluid flows at a constant rate, whilst heat transfer occurs and external work is done all under steady conditions ,that is , the rates of mass flow and energy flow are constant with respect to time.

The equation for steady flow ( the steady flow energy equation ) is generally written per unit mass as

q = heat transfer across boundary per unit mass
w = external work done by system per unit mass
z = fluid height
v = fluid velocity
h = fluid enthalpy ( u (internal energy + pv (pressure.specific volume)

Enthalpy (H) - The sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume:

After a series of rearrangements, and if pressure if kept constant, we can arrive at the following equation:

where H is the H_final minus H_initial and q is heat

Thermodynamics

Saturday, May 16, 2009

Clausius inequality

Carnot Cycle

Friday, May 15, 2009

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