title: MMME1048 // Thermodynamics

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What is Thermodynamics?

Thermodynamics deals with the transfer of heat energy and temperature.

Concepts and Definitions

System

A region of space, marked off by its boundary. It contains some matter and the matter inside is what we are investigating.

There are two types of systems:

Equilibrium

The system is in equilibrium if all parts of the system are at the same conditions, such as pressure and temperature.

The system is not in equilibrium if parts of the system are at different conditions.

Adiabatic

A process in which heat does not cross the system boundary

Perfect (Ideal) Gasses

A perfect gas is defined as one in which:

Perfect gases do not exist in the real world and they have two requirements in thermodynamics:

The Requirements of Perfect Gasses

Obey the Perfect Gas Equation

pV=nR̃TpV = n \tilde R T

where nn is the number of moles of a substance and R̃\tilde R is the universal gas constant

or

pV=mRTpV =mRT

where the gas constant R=R̃m̃R = \frac{\tilde R}{\tilde m}, m̃\tilde m is molecular mass

or

pv=RTpv = RT

(using the specific volume)

cpc_p and cvc_v are constant

This gives us the equations:

u2u1=cv(T2T1)u_2 - u_1 = c_v(T_2-T_1)

h2h1=cp(T2T1)h_2 - h_1 = c_p(T_2-T_1)

Relationship Between Specific Gas Constant and Specific Heats

cv=Rγ1c_v = \frac{R}{\gamma - 1}

cp=γγ1Rc_p = \frac{\gamma}{\gamma -1} R

Derivation

We know the following are true (for perfect gases):

cpcv=γ\frac{c_p}{c_v} = \gamma

u2u1=cv(T2T1)u_2 - u_1 = c_v(T_2-T_1)

h2h1=cp(T2T1)h_2 - h_1 = c_p(T_2-T_1)

So:

h2h1=u2u1+(p2v2p1v1)cp(T2T1)=cv(T2T1)+R(T2T1)cp=cv+Rcp=cvγcv+R=cvγcv=Rγ1cpγ=cvcp=cpγ+Rcp=γγ1R\begin{align*} h_2 - h_1 &= u_2 - u_1 + (p_2v_2 - p_1v_1) \\ c_p(T_2-T_1) &= c_v(T_2-T_1) + R(T_2-T_1) \\ c_p &= c_v + R \\ \\ c_p &= c_v \gamma \\ c_v + R &= c_v\gamma \\ c_v &= \frac{R}{\gamma - 1} \\ \\ \frac{c_p}{\gamma} &= c_v \\ c_p &= \frac{c_p}{\gamma} + R \\ c_p &= \frac{\gamma}{\gamma -1} R \end{align*}

The Specific and Molar Gas Constant

The molar gas constant is represented by R̃=8.31JK1mol1\tilde R = 8.31 \text{JK}^{-1}\text{mol}^{-1}.

The specific gas constant is R=R̃MR = \frac{\tilde{R}}{M}.
The SI unit for the specific gas constant is J kg1^{-1} mol1^{-1}.
The SI unit for molar mass is kg mol1^{-1}.

Thermodynamic Processes and Cycles

When a thermodynamic system changes from one state to another it is said to execute a process. An example of a process is expansion (volume increasing).

A cycle is a process or series of processes in which the end state is identical to the beginning. And example of this could be expansion followed by a compression.

Reversible and Irreversible Processes

During reversible processes, the system undergoes a continuous succession of equilibrium states. Changes in the system can be defined and reversed to restore the initial conditions

All real processes are irreversible but some can be assumed to be reversible, such as controlled expansion.

Constant _____ Processes

Isothermal

Constant temperature process

Isobaric

Constant pressure process

Isometric / Isochoric

Constant volume process

Heat and Work

Heat and Work are different forms of energy transfer.

They are both transient phenomena and systems never possess heat or work. Both represent energy crossing boundaries when a system undergoes a change of state.

By convention, the transfer of energy into the system from the surroundings is positive (work is being done on the system by the surroundings).

Heat

Heat is defined as:

The form of energy that is transferred across the boundary of a system at a given temperature to another system at a lower temperature by virtue of the temperature difference between the two

Work

Work is defined as:

W=FdxW = \int\! F \mathrm{d}x

(the work, WW, done by a force, FF, when the point of application of the force undergoes a displacement, dx\mathrm{d}x)

Thermally Insulated and Isolated Systems

In thermally insulated systems and isolated systems, heat transfer cannot take place.

In thermally isolated systems, work transfer cannot take place.

1st Law of Thermodynamics

The 1st Law of Thermodynamics can be thought of as:

When a closed system is taken through a cycle, the sum of the net work transfer (WW) and net heat transfer (QQ) equals zero:

Wnet+Qnet=0W_{net} + Q_{net} = 0

1st Corollary

The change in internal energy of a closed system is equal to the sum of the heat transferred and the work done during any change of state

W12+Q12=U2U1W_{12} + Q_{12} = U_2 - U_1

2nd Corollary

The internal energy of a closed system remains unchanged if it thermally isolated from its surroundings

Properties of State

State is defined as the condition of a system as described by its properties. The state may be identified by certain observable macroscopic properties. These properties are the properties of state and they always have the same values for a given state.

A property can be defined as any quantity that depends on the state of the system and is independent of the path by which the system arrived at the given state. Properties determining the state of a thermodynamic system are referred to as thermodynamic properties of the state of the system.

Common properties of state are:

And these can be determined by simple measurements. Other properties can be calculated:

Intensive vs Extensive Properties

In thermodynamics we distinguish between intensive, extensive, and specific properties:

Units

Property Symbol Units Intensive Extensive
Pressure p Pa Yes
Temperature T K Yes
Volume V m3^3 Yes
Mass m kg Yes
Specific Volume v m3^3 kg1^{-1} Yes
Density ρ\rho kg m3^{-3} Yes
Internal Energy U J Yes
Entropy S J K1^{-1} Yes
Enthalpy H J Yes

Density

For an ideal gas:

ρ=pRT\rho = \frac{p}{RT}

Enthalpy and Specific Enthalpy

Enthalpy does not have a general physical interpretation. It is used because the combination u+pvu + pv appears naturally in the analysis of many thermodynamic problems.

The heat transferred to a closed system undergoing a reversible constant pressure process is equal to the change in enthalpy of the system.

Enthalpy is defined as:

H=U+pVH = U+pV

and Specific Enthalpy:

h=u+pvh = u + pv

Entropy and Specific Entropy

Entropy is defined as the following, given that the process s reversible:

S2S1=dQTS_2 - S_1 = \int\! \frac{\mathrm{d}Q}{T}

Change of Entropy of a Perfect Gas

Consider the 1st corollary of the 1st law:

dq+dw=du\mathrm dq + \mathrm dw = \mathrm du

and that the process is reversible:

ds=dqT|revdq=ds×Tdw=pdv\begin{align*} \mathrm ds &= \frac{\mathrm dq} T \bigg|_{rev} \\ \mathrm dq = \mathrm ds \times T \\ \mathrm dw &= -p\mathrm dv \\ \end{align*}

The application of the 1st corollary leads to:

Tdspdv=duT\mathrm ds - p\mathrm dv = \mathrm du

Derive the change of entropy

ds=duT+pdvTdu=cvdTpT=Rvds=cvdTTRdvvs2s1=cvln(T2T1)+Rln(v2v1)\begin{align*} \mathrm ds &= \frac{\mathrm du}{T} + \frac{p \mathrm dv}{T} \\ \\ \mathrm du &= c_v \mathrm{d}T \\ \frac p T &= \frac R v \\ \\ \mathrm ds &= \frac{c_v\mathrm{d}T}{T} \frac{R\mathrm dv}{v} \\ s_2 - s_1 &= c_v\ln\left(\frac{T_2}{T_1}\right) + R\ln\left(\frac{v_2}{v_1}\right) \end{align*}

There are two other forms of the equation that can be derived:

s2s1=cvln(p2p1)+cpln(v2v1)s_2 - s_1 = c_v\ln\left(\frac{p_2}{p_1}\right) + c_p\ln\left(\frac{v_2}{v_1}\right) s2s1=cpln(T2T1)Rln(p2p1)s_2 - s_1 = c_p\ln\left(\frac{T_2}{T_1}\right) - R\ln\left(\frac{p_2}{p_1}\right)

Heat Capacity and Specific Heat Capacity

Heat capacity is quantity of heat required to raise the temperature of a system by a unit temperature:

C=dQdTC = \frac{\mathrm{d}Q}{\mathrm{d}T}

Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass substance by a unit temperature:

c=dqdTc = \frac{\mathrm{d}q}{\mathrm{d}T}

Heat Capacity in Closed Systems and Internal Energy

The specific heat transfer to a closed system during a reversible constant volume process is equal to the change in specific internal energy of the system:

cv=dqdT=dudTc_v = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}u}{\mathrm{d}T}

This is because if the change in volume, dv=0\mathrm{d}v = 0, then the work done, dw=0\mathrm{d}w = 0 also. So applying the (1st Corollary of the) 1st Law to an isochoric process:

dq+dw=dudq=du\mathrm{d}q + \mathrm{d}w = \mathrm{d}u \rightarrow \mathrm{d}q = \mathrm{d}u

since dw=0\mathrm{d}w = 0.

Heat Capacity in Closed Systems and Enthalpy

The specific heat transfer to a closed system during a reversible constant pressure process is equal to the change in specific enthalpy of the system:

cp=dqdT=dhdTc_p = \frac{\mathrm{d}q}{\mathrm{d}T} = \frac{\mathrm{d}h}{\mathrm{d}T}

This is because given that pressure, pp, is constant, work, ww, can be expressed as:

w=12pdv=p(v2v1)w = -\int^2_1\! p \,\mathrm{d}v = -p(v_2 - v_1)

Applying the (1st corollary of the) 1st law to the closed system:

q+w=u2u1q=u2u1+p(v2v1)q=u2+pv2(u1+pv1)=h2h1=dhdq=dh\begin{align*} q + w &= u_2 - u_1 \rightarrow q = u_2 - u_1 + p(v_2 - v_1) \\ q &= u_2 + pv_2 - (u_1 + pv_1) \\ &= h_2 - h_1 = \mathrm{d}h \\ \therefore \mathrm{d}q &= \mathrm{d}h \end{align*}

Ratio of Specific Heats

cp>cvc_p > c_v is always true.

Heating a volume of fluid, VV, at a constant volume requires specific heat qvq_v where

qv=u2u1cv=qvΔTq_v = u_2 - u_1 \therefore c_v = \frac{q_v}{\Delta T}

Heating the same volume of fluid but under constant pressure requires a specific heat qpq_p where

qp=u2u1+p(v2v1)cp=qpΔTq_p =u_2 - u_1 + p(v_2-v_1) \therefore c_p = \frac{q_p}{\Delta T}

Since p(v2v1)>0p(v_2-v_1) > 0, qpqv>1qp>qvcp>cv\frac{q_p}{q_v} > 1 \therefore q_p > q_v \therefore c_p > c_v.

The ratio cpcv=γ\frac{c_p}{c_v} = \gamma

Process and State Diagrams

Reversible processes are represented by solid lines, and irreversible processes by dashed lines.

Isentropic and Polytropic Processes

Polytropic Processes

A polytropic process is one which obeys the polytropic law:

pvn=k or p1v1n=p2v2npv^n = k \text{ or } p_1v_1^n = p_2v_2^n

where nn is a constant called the polytropic index, and kk is a constant too.

A typical polytropic index is between 1 and 1.7.

Example 1

Derive

p2p1=(T2T1)nn1\frac{p_2}{p_1} = \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}}

p1v1n=p2v2npv=RTv=RTpp2p1=(v1v2)n=(p2T1T2p1)n=(p2p1)n(T1T2)n(p2p1)1n=(T1T2)np2p1=(T1T2)n1n=(T2T1)nn1\begin{align*} p_1v_1^n &= p_2v_2^n \\ pv &= RT \rightarrow v = R \frac{T}{p} \\ \frac{p_2}{p_1} &= \left( \frac{v_1}{v_2} \right)^n \\ &= \left(\frac{p_2T_1}{T_2p_1}\right)^n \\ &= \left(\frac{p_2}{p_1}\right)^n \left(\frac{T_1}{T_2}\right)^n \\ \left(\frac{p_2}{p_1}\right)^{1-n} &= \left(\frac{T_1}{T_2}\right)^n \\ \frac{p_2}{p_1} &= \left(\frac{T_1}{T_2}\right)^{\frac{n}{1-n}} \\ &= \left(\frac{T_2}{T_1}\right)^{\frac{n}{n-1}} \\ \end{align*}

How did you do that last step?

For any values of xx and yy

xy=(yx)1(xy)n=(yx)n(xy)n1n=(yx)n1n=(yx)nn1\begin{align*} \frac x y &= \left(\frac y x \right) ^{-1} \\ \left(\frac x y \right)^n &= \left(\frac y x \right)^{-n} \\ \left(\frac x y \right)^{\frac{n}{1-n}} &= \left(\frac y x \right)^{\frac{-n}{1-n}} \\ &= \left(\frac y x \right)^{\frac{n}{n-1}} \\ \end{align*}

Isentropic Processes

Isentropic means constant entropy:

ΔS=0 or s1=s2 for a precess 1-2\Delta S = 0 \text{ or } s_1 = s_2 \text{ for a precess 1-2}

A process will be isentropic when:

pvγ=constantpv^\gamma = \text{constant}

This is basically the polytropic law where the polytropic index, nn, is always equal to γ\gamma.

Derivation

0=s2s1=cvln(p2p1)+cpln(v2v1)0=ln(p2p1)+cpcvln(v2v1)=ln(p2p1)+γln(v2v1)=ln(p2p1)+ln(v2v1)γ=ln[(p2p1)(v2v1)γ]e0=1=(p2p1)(v2v1)γp2v2γ=p1v1γpvγ=constant\begin{align*} 0 &= s_2 - s_1 = c_v \ln{\left(\frac{p_2}{p_1}\right)} + c_p \ln{\left( \frac{v_2}{v_1} \right)} \\ 0 &= \ln{\left(\frac{p_2}{p_1}\right)} + \frac{c_p}{c_v} \ln{\left( \frac{v_2}{v_1} \right)} \\ &= \ln{\left(\frac{p_2}{p_1}\right)} + \gamma \ln{\left( \frac{v_2}{v_1} \right)} \\ &= \ln{\left(\frac{p_2}{p_1}\right)} + \ln{\left( \frac{v_2}{v_1} \right)^\gamma} \\ &= \ln{\left[\left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma\right]} \\ e^0 = 1 &= \left(\frac{p_2}{p_1}\right)\left( \frac{v_2}{v_1} \right)^\gamma \\ &\therefore p_2v_2^\gamma = p_1v_1^\gamma \\ \\ pv^\gamma = \text{constant} \end{align*}

During isentropic processes, it is assumed that no heat is transferred into or out of the cylinder. It is also assumed that friction is 0 between the piston and cylinder and that there are no energy losses of any kind.

This results in a reversible process in which the entropy of the system remains constant.

An isentropic process is an idealization of an actual process, and serves as the limiting case for real life processes. They are often desired and often the processes on which device efficiencies are calculated.

Heat Transfer During Isentropic Processes

Assume that the compression 1-2 follows a polytropic process with a polytropic index nn. The work transfer is:

W=11n(p2V2p1V1)=mRn1(T2T1)W = - \frac{1}{1-n} (p_2V_2 - p_1V_1) = \frac{mR}{n-1} (T_2-T_1)

Considering the 1st corollary of the 1st law, Q+W=ΔUQ + W = \Delta U, and assuming the gas is an ideal gas we know that ΔU=mcv(T2T1)\Delta U = mc_v(T_2-T_1) we can deduce:

Q=ΔUW=mcv(T2T1)mRn1(T2T1)=m(cvRn1)(T2T1)\begin{align*} Q &= \Delta U - W = mc_v(T_2-T_1) - \frac{mR}{n-1} (T_2-T_1) \\ &= m \left(c_v - \frac R {n-1}\right)(T_2-T-1) \end{align*}

Now, if the process was isentropic and not polytropic, we can simply substitute nn for γ\gamma so now:

Q=m(cvRγ1)(T2T1)Q = m \left(c_v - \frac R {\gamma-1}\right)(T_2-T-1)

But since we know cv=Rγ1c_v = \frac R {\gamma - 1}:

Q=m(cvcv)(T2T1)=0Q = m (c_v-c_v)(T_2-T_1) = 0

This proves that the isentropic version of the process adiabatic (no heat is transferred across the boundary).

2nd Law of Thermodynamics

The 2nd Law recognises that processes happen in a certain direction. It was discovered through the study of heat engines (ones that produce mechanical work from heat).

Heat does not spontaneously flow from a cooler to a hotter body.

~ Clausius’ Statement on the 2nd Law of Thermodynamics

It is impossible to construct a heat engine that will operate in a cycle and take heat from a reservoir and produce an equivalent amount of work.

~ Kelvin-Planck Statement of 2nd Law of Thermodynamics

Heat Engines

A heat engine must have:

Steam Power Plant

Thermal Efficiency

For heat engines, Qout>0Q_{out} > 0 so Wout<QinW_{out} < Q_{in} as Wout=QinQoutW_{out} = Q_{in} - Q_{out}

η=WoutQin=1QoutQin\eta = \frac{W_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}

Early steam engines had efficiency around 10% but large diesel engines nowadays have efficiencies up to around 50%, with petrol engines around 30%. The most efficient heat engines we have are large gas-steam power plants, at around 60%.

Carnot Efficiency

The maximum efficiency for a heat engine that operates reversibly between the heat source and heat sink is known as the Carnot Efficiency:

ηcarnot=1T2T1\eta_{carnot} = 1 - \frac{T_2}{T_1}

where TT is in Kelvin (or any unit of absolute temperature, I suppose)

Therefore to maximise potential efficiency, you want to maximise input heat temperature, and minimise output heat temperature.

The efficiency of any heat engine will be less than ηcarnot\eta_{carnot} if it operates between more than two reservoirs.

Reversible and Irreversible Processes

Reversible Processes

A reversible process operate at thermal and physical equilibrium. There is no degradation in the quality of energy.

There must be no mechanical friction, fluid friction, or electrical resistance.

Heat transfers must be across a very small temperature difference.

All expansions must be controlled.

Irreversible Processes

In irreversible processes, the quality of the energy degrades. For example, mechanical energy degrades into heat by friction and heat energy degrades into lower quality heat (a lower temperature), including by mixing of fluids.

Thermal resistance at both hot sources and cold sinks are an irreversibility and reduce efficiency.

There may also be uncontrolled expansions or sudden changes in pressure.

Energy Quality

Quantifying Disorder (Entropy)

S=klogeWS = k\log_eW

where SS is entropy, k=1.38×1023k = 1.38\times10^{-23} J/K is Boltzmann’s constant, and WW is the number of ways of reorganising energy.s

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