Thermodynamics and Types

One or more forms of works are involved at the same time in a control system. The control volume boundary must be stationary and the moving boundary work must be zero. The work involved in the form of electrical work or shaft work. And also the other form of work must be from fluid, which is flow work.

Flow Energy:

When a mass flow is involved, the boundaries of the control volume will push the fluid in and out of the boundary. The work result is known as flow energy or flow work. In a control volume work flow is necessary for marinating a continuous flow.

Consider a system having a fluid elements of pressure “P”, volume “V”, and a cross sectional area “A”.

In the system, there is a particular layer fluid where the element is flows from one section to the other section through a duct and enters into the control volume. It must be viewed as an imaginary piston and the force applied by with the help of imaginary piston will be

F = PA

So the work done at the time of fluid element enters through the boundary and to the control volume must be

$W_{flow}=FL=PAL=PV$

where,

F= force

L = Displacement

P= pressure (Constant)

V= volume

During the work done, the fluid element that enters into the control volume is equal to the fluid element leaves the control volume.

Total energy of the flowing fluid:

The total energy of the compressible system consists of three quantities. They are Kinetic energy, potential energy and internal energy

E = KE+PE+U

E = energy

U = internal energy

KE= kinetic energy

PE = potential energy

By substituting the values in the equation it turns into the

$\theta=Pv+\frac{v^2}{2}+gz+u$

We know that h= Pv+u

By replacing the values the obtained equation must be

$\theta=h+\frac{v^2}{2}+gz$

Pressure Volume Diagram

In the pressure volume diagram the pressure is given by the height of the diagram and the volume is given as length of the diagram. In this diagram any point will represent the state of the gas and this is known as the state point. The p-v diagram is also known as the indicator diagram.

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Process:

When a system under goes change from one condition to other condition or from one state point to another state point is known as process.

Cycle:

When a system under goes number of process and it is able to attain its original condition, then it is said to be complete a cycle. If the cycle is not completed then the system will not have continuous work output. The requirements for a complete cycle include the heat added, heat rejected and the work done. The order in which the different operation should take place during a cycle is the heat added.

Thermal equilibrium

When there is no heat transfer that takes place between the two bodies then they are said to be in thermal equilibrium with one another. When the two bodies are in thermal equilibrium then the temperature of the bodies is in the same state.

Zeroth law of thermo dynamics:

When the two bodies are separately in thermal equilibrium with the third body, then the first bodies is also in the thermal equilibrium with one another.

Quasistatic Process:

When a system undergoes a change in such a manner that the final condition is nearest to the original or equilibrium conditions, then the change is known as the Quasistatic process.

dw = Pdv

$\int dw=\int P dv$

$W=\int_1^2 P dv$

Reversible and Irreversible process

Reversible process:

When a system undergoes changes in such a manner it is able to retain its original condition by following the same Thermodynamic path in the reverse direction, it is then said to have undergone a reversible process

Irreversible process:

When the system is unable to reach the original condition by retracing its path or attain the original conditions along other thermo dynamic paths, then the process is said to be an irreversible process

A process becomes irreversible due to the friction.
A Quasistatic process can be applied to a reversible process only.

Work done during a Quasistatic process:

When a closed system undergoes a process, no energy is available for the change in volume due to the mass flow externally. Hence there is a drop in pressure with the increase in the volume of the system. Hence we consider a Quasistatic process for which the change in pressure is negligible and change is differently small.

If the assumed constant pressure during the quasistatic process is P, then the work done for the quasistatic process is given by

dw = p dv

The work done for the entire process is formed by the summation of all the differentially small work done.

$\int dw=\int P dv$

Total work done = $W=\int_1^2 P dv$

W = work done

P = constant for closed system

Types of process:

Constant volume process

The constant volume process is also known as isochoric process or isometric process or isovolumetric process. This process is a thermodynamic process in which the volume of the closed system goes through such a method remains constant. The isochoric process is represented by the cooling or the heating of the sealed contents in the inelastic container. Mainly the thermodynamic process is removal or addition of the heat. In the closed system the container establishes the isolation of the contents; the incompetence of the container is bent to force the constant-volume condition. At this point of time the isochoric process one must be a quasistatic process.

V= constant

Ddv = 0

$W=\int_1^2 P dv$

$=\int_1^2 P \times 0$

$=\int_1^2\:\:\:\: 0$

Constant pressure process:

The constant pressure process is also known as isobaric process. When the pressure of ideal gas is kept constant then the temperature must be increased and the gas expands. During the expansion process heat must be added.

$W=\int_1^2 P dv$

$W=P\int_1^2 \:\: dv$

Constant temperature process:

Constant temperature process is also known as isothermal process. In this process the temperature must remain constant. This process occurs when the system is in contact with the outside reservoirs and the change take place to allow the system work continuously by adjusting the temperatures of the reservoir through heat exchange.

In this process temperature must be constant (T=Constant)

Then pV = c

$P=\frac{C}{V}$

$W=\int\:_1^2 \:\:pdv$

$W=\int\:_1^2 \:\:\frac{C }{V}dv$

$W=\int\:_1^2 \:\:\frac{dv }{v}dv$

$W=pV(\log_e \:\:V)_1^2$

$W=pV \:\log_e (V_2-V_1)$

$W=pV \:\log_e\:\:\: \frac{V_2}{V_1}$

The adiabatic process is happens without the transfer of matter or heat between a system and surroundings. A rapid contraction or expansion of a gas is almost adiabatic. The process which occurs within the container is a good thermal insulator is also known as adiabatic. By increasing the entropy adiabatic process are characterized as, when there is a change in entropy then the process is irreversible and if there is no change in entropy then the process is reversible. Entropy does not decrease with the adiabatic process.

$P=\frac{c}{v^{\gamma}}=c\times V^{\gamma}$

$W=\int \:\:_1^2 p\:\:dv$

$W=\int \:\:_1^2c v^{-\gamma}dv$

$W=c\int \:\:_1^2 v^{-\gamma}dv$

$W=\Bigg[pv^{\gamma}[\frac{v^{-\gamma+1}}{-\gamma+1}]\Bigg]_1^2$

$W=\frac{p_1v_1-p_2v_2}{\gamma-1}$

In this process pressure ‘p’, specific volume ‘v’, Polytropic index ‘n’ and ‘c’ is constant. Mainly the Polytropic process is suitable for characterizing compression and expansion process which includes the heat transfer. The equation can exactly describe a very wide range of thermodynamics processes. They ranges from n= 0 to ∞ which covers, n=0 for isobaric process, n = 1 for isothermal process, n = ∞ for isochoric process.

$W=\frac{p_1v_1-p_2v_2}{n-1}$

Properties of gases:

Gas:

The process of complete evaporation of the liquid, to form gas is known as gaseous state.

Vapour:

The state existing due to incomplete evaporation of liquid is defined as vapor

There are different types of vapours namely:

Wet vapour
Dry and saturated vapour
Super-heated vapour
Super saturated temperature

Ideal gas:

Ideal gas which obeys the all the gas laws

PV= nRT

P = pressure

V= volume

n= moles of gas

R= ideal gas law constant

T= temperature

Real gas:

A real gas does not obey any gas laws. Under certain conditions temperature and pressure of some real gases must behave like ideal gases

Boyles law:

Boyle’s law states that, the point at which the temperature remain same and the volume of the given mass of the gas is inversely proportional to its absolute pressure.

$V \: \alpha \:\:\: \frac 1 p$

$V =C\times \frac 1 p$

Charles law

According to the Charles law the pressure and volume of the gas remains constant.

The volume of the given mass of the gas is directly proportional to absolute gas temperature. In this process the pressure must be constant and the temperature and the volume change.

V α T

V = CT

$\frac V T=C$

$\frac{V_1}{T_1}=\frac{V_2}{T_2}=\frac{nR}{p}$

Where,

V= volume

T = temperature

P = pressure

n= moles of gas

R= ideal gas law constant

Now volume remains constant then the pressure of the given mass of the gas is directly proportional to the absolute temperature of the gas

P α T

P = CT

$\frac P T=C$

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_2}{T_1}=\frac{nr}{V}$

Charles law II statement:

According to the Charles law the pressure remain constant and the volume of the given mass of the gas increases or decreases by $\frac1 {273}$ of its original volume, this is observed for every degree increases or decreases in the

scale of temperature.

Where p = constant

$V=v_0+\Bigg(\frac{t^oC}{273}\:\:\:\times v_0\Bigg)$

$V=v_0+\Bigg[1+\frac{t^oC}{273}\Bigg]$

$V=v_0+\Bigg[\frac{273+t^oC}{273}\Bigg]$

When the temperature of the gas is -273

, from the above equation the volume of the gas must be zero. So -273

is the lowest possible temperature of the gas. To observe zero volume, the temperature of the gas is also zero. A new scale temperature is introduced such that zero in this scale that corresponds to zero volume. This scale is known as the absolute or Kelvin scale of temperature. The change in temperature is the same in both kelvin and centigrade scale of time is as:

From the above equations we can get

$V = V_0 \Big(\frac{T^k}{273}\Big)$

$V = V_0 \Big(\frac{T_1^k}{273}\Big)$

$V = V_0 \Big(\frac{T_2^k}{273}\Big)$

From the above equations we obtain an equation

$\frac{V_2}{V_1}=\frac{T_2^K}{T_1^K}$

Universal gas law:

Let us consider, a system under going change from state 1 to state 2 in such a manner that the pressure temperature and volume change from

. On considering the final condition, it obtains a constant volume path (1-a) and then a constant pressure path (a-2) we have for the constant volume path.

$\frac {T_a}{T_1}=\frac {P_a}{P_1}$

$T_a=\frac {P_a}{P_1} \times T_1$

$\frac {T_a}{T_1}=\frac {V_a}{V_2}$

$T_a=\frac {V_a}{V_2} \times T_2$

From equating the above equations

$\frac {V_a}{V_2} \times T_2=\frac {P_a}{P_1} \times T_1$

$\frac {P_1V_1}{T_1}=\frac {P_2 V_2}{T_2}$ or $\frac {PV}{T}=R$

The above equation must be known as the gas equation and ‘R’ is known as the gas constant.

In the equation if “m” mass of the gas is introduced then the equation must be

In the above equation, the

is known as the characteristic gas constant

If the mass of the gas is taken equal to its molecular weight, then we have

The above equation is known as universal gas equation, and

is known as the universal gas constant. This is same for all the gases.

Heat:

One unit of heat is the amount of heat required to rise the temperature of 1 gram of air through 1

, this is by keeping the pressure constant through 1

by keeping the pressure constant throughout the process. In SI unit’s one heat is defined as one joule.

Specific heat:

Specific heat of a substance is the amount for heat required to raise the temperature of 1gm of the substance through 1

Let us consider a substance of mass m kg and ‘q’ amount of heat is supplied to it. If the temperature increases by ‘t’, then

$C=\frac{Q}{mt}$

$Heat \:\:\:transfer = [mass\:\times specific \:\:\: heat\:\times change \:\:\:in \:\:\: temperature]$

Specific heat of gases:

The specific heats are expressed as molar specific heats. For monoatomic ideals gases the internal energy must be in the form of the kinetic energy. The equation for the internal energy must be

$U=\frac 3 2 nRT$