Application of Thermodynamics

APPLICATION OF THERMODYNAMICS

THERMODYNAMIC OF GASES
(ADIABATIC CHANGE)

In thermodynamic of gases, we can study about the changes to the internal energy of gas and factors affecting the internal energy. Thermodynamic also involves work done and heat supplied or lost by a gas.

THERMODYNAMICS
The study of temperature, heat, and related macroscopic properties comprises the branch of physics.
Thermodynamics deals with processes which cause energy changes (internal energy) as a result of heat flow to or from a system and/or of work done on or by a system.
Internal energy, U is the sum of the kinetic energy and potential energies of the molecules of the system.
Internal energy, U = Kinetic Energy + Potential Energy of molecules

Work done by a gas when a gas expands. Conversely, work done is on a gas when a gas is compressed.

W = = p ∆V = Pressure x Increase in volume

Thermodynamic Equilibrium
We define this type of equilibrium by saying that when two system are placed together until no further change occurs in any macroscopic property, then two have reached thermodynamic equilibrium in each other.
TWO SYSTEM HAVE THE SAME TEMPERATURE IF THEY ARE IN THERMODYNAMIC EQUILIBRIUM.

First Law of Thermodynamics
The first law of thermodynamics is based on the principle of conservation of energy and state that the heat supplied to a system equals to the sum of the increase in internal energy of the system and the external work done by the system, ∆Q= ∆U + W
Heat supplied = Increase in internal energy + Work done by gas

+ (positive)
- (negative)
∆Q
Heat supplied to system
Heat loss by system
∆U
Increase in internal energy
Decrease in internal energy
W
Work done by system
Work done on system

Heat Capacities
The specific heat capacity, c of a material is the heat required to raise the temperature of a unit mass of the material by one kelvin (1 K).
Example of specific heat capacity, c of some materials:
Water = 4180 J kg-1K-1, Copper = 400 J kg-1 K-1
The heat ∆Q absorbed by a material of mass m when its temperature increases by ∆T is ∆Q = mc ∆T .

The heat capacity C of the body is the heat required to raise the temperature of the body by 1 K. Heat capacity, C = mc where m = mass of body, c = specific heat capacity.
The molar heat capacity of a material, Cm is the heat required to raise the temperature of one mole of the material by 1 K. (Unit: J mol-1 K-1). The heat gained by n moles of a material when its temperature increases by ∆T is ∆Q = n Cm ∆T
The heat required to raise the temperature of a gas depends on whether the gas is allowed to expand when it is heated.
A gas has two heat capacities, one at constant volume, and another at constant pressure.
The molar heat capacity of a gas at constant volume CV, m is the heat required to raise the temperature of one mole of the gas by 1 K at constant volume.
The molar heat capacity of a gas at constant pressure Cp, m is the heat required to raise the temperature of one mole of the gas by 1 K at constant pressure.
The heat supplied to increase the temperature of n moles of gas by ∆T
a) at constant volume is ∆Q = n CV, m ∆T
b) at constant pressure is ∆Q = n Cp, m ∆T

Relation between CV, m and Cp, m

Figure shows one mole of an ideal gas at a pressure p, temperature T and volume V m. This state of the gas is presented by the point A on the p-V graph.
When the temperature of the gas is increased to T + ∆T at constant volume, the pressure increases to p +∆p. This change of state is represented by AB on the p-V graph.
Heat supplied, ∆Q = n CV, m ∆T
Work done by gas, W = p ∆T = 0 (∆V=0)
Using the first law of thermodynamics. ∆Q = ∆U + W ∆Q = ∆U + 0 Hence ∆U = ∆Q = CV, m ∆T
When the temperature of gas is increased from T to T + ∆T at constant pressure p, the volume of the gas increases from V m to V m + ∆V. This represented by the change AC on the p-V graph.
Since the gas is heated at constant p,
The heat supplied, ∆Q = n Cp, m ∆T
Work done by gas, W = p ∆V pV m= RT = R ∆T pV m = R ∆T
Using the first law of thermodynamics ∆Q = ∆U + W Cp, m ∆T = CV, m ∆T + R ∆T Cp, m= CV, m + R Cp, m - CV, m = R The Ratio of C p, m C v, m

When an ideal gas is heated at constant volume, the heat supplied is used to increase the internal energy of the gas. No external work is done by the gas because there is no change in volume.
For one mole of an ideal gas at a temperature T, internal energy, U = RT where f= degree of freedom.
The molar heat capacity at constant volume CV, m = CV, m = R
Hence, the molar heat capacity at constant pressure Cp, m = CV, m + R = R + R Cp, m = R

The ratio of Cp, m to CV, m is denoted by the symbol ƴ. ƴ = C p, m C v, m = ƴ =

It can be summarized the above discussion by using this table.

Monatomic
Diatomic
Polyatomic
Degrees of freedom f = 3
3 translational f = 5
3 translational
2 rotational f = 6
3 translational
3 rotational
Mean kinetic energy of one molecule

3KT
Internal energy of 1 mole, U

3RT
CV, m =

3R
Cp, m = CV, m + R

4R ƴ = C p, m C v, m

Reversible Change

The state of a gas is determined by the three variables, pressure p, volume V and temperature T. A gas is said to be in equilibrium if its state does not change with time and satisfies the equation of state of a real gas, = constant.

A reversible process or reversible change in the state of a gas is a process where the gas is taken from an equilibrium state to another equilibrium state through a number of small steps, where the gas is always equilibrium at every intermediate step (Figure a). The process can be reversed to its original state via the same small steps, but in the opposite direction (Figure b).

(a) Change from original state to final state via small steps 1, 2, 3….

(b) Process is reversed via the same steps but in the opposite direction
A reversible change can be represented by a smooth continuous curve as shown in figure.
A change in the state of a gas becomes irreversible if the gas is not in equilibrium at any of the intermediate steps. An irreversible change cannot be represented by a smooth curve.

ADIABATIC CHANGE

IT IS THE CHANGE OF STATE OF A FIXED MASS OF GAS (THERMODYNAMIC SYSTEM) WHEREBY NO HEAT ENTERS OR LEAVES THE SYSTEM.

Suppose the gas expands, and pushes the piston upward thereby doing external work. If no energy is allowed to enter the system, then it is forced to use its own internal energy to do the external work. As a result its temperature will fall. On the other hand, during an adiabatic compression, the temperature of the gas/ system will rise. dQ = du + dW (first law of thermodynamics)
But dQ = 0,
Hence dU = -dw

Relationship between P, V and T for an Adiabatic Change
a) PV= RT for 1 mole of gases
b) When the volume of the gas changes by ΔV, its temperature change by ΔT. Change in internal energy, ΔU= CV, m ΔT Work done W = p ΔV = () ΔV ΔQ =ΔU + W ΔQ = 0 (ADIABATIC) 0 = Cv, m ΔT + () ΔV R= Cp, m- CV, m γ=

PVγ = constant P1V1γ= P2V2γ TV γ-1 = constant T1V1γ-1= T2V2γ-1 TγP1-γ = constant T1γP11-γ= T2γP21-γ

Graphical Method
The graph of against V for adiabatic change is called an adiabatic curve. Every point on an adiabatic curve represents a given of the thermodynamic system under different pressure (P), different volume (V) and different temperature (T).

The graph shows the variation of pressure p with volume for an adiabatic compression (AB), and adiabatic expansion (AD).

Adiabatic Work Done

a) From First Law of Thermodynamics dQ = dU + dW
But dQ = 0 and dU = CV, m dT dW=-dU=-CV,mdT

W=∫ DW = -CV, m W = CV, m (T1-T2)

b) From W = ∫ P dV P Vγ = constant P =

W = Constant dV = Constant x (V1γ-1-V2γ-1)

But: P1V1γ= P2V2γ =Constant W = (P2V2 - P1V1)
Application of Thermodynamics
(Adiabatic Transformations in Diesel Engines)

The diesel engine is a well-known application of an adiabatic transformation. In its operation, a cylinder is filled with air at atmospheric pressure and temperature, and then compressed to less than 1/20 of its original volume. This compression is nearly adiabatic, if it were truly adiabatic and if the air were an ideal gas, then the ratio of the final to initial temperature would be Tf/Ti = (Vi/Vf) (γ-1).Although some heat is lost to the cylinder wall, piston, and cylinder head, and the ideal gas equation of state is not exact for air, the final temperature of most hydrocarbons. When, at this point, a combustible hydrocarbon is introduced into the pressurized air, it will burn, transferring its chemical energy into heat energy and raising the pressure of the gases in the cylinder .The hot gas is allowed to expand, driving down the piston and doing work. In practice, diesel engines are very tolerant to the type of fuel burned, everything from flour to powdered coal has been used. Note that because the end temperature of the compressed gas is proportional to the incoming temperature diesel engines do not work as well in cold climates.

DIESEL ENGINE
Example
In a diesel engine is injected into a cylinder where the temperature of the air has been increased by adiabatic compression to exceed the ignition temperature of diesel, 630. Air enters the cylinder at a pressure of 1.0x 105 Pa and temperature of 28 .The initial volume of the air is 5.0x10-4m3.

A) Calculate the ratio for the air cylinder so that its final temperature is the ignition temperature of diesel.
B) What is the work done to compress the air?( For air ,γ= 1.40) Solution

A. T1=28=301K V1 = 5.0X10-4m3
T2=630=903K V2 = final volume Using T1 V1γ- = T2V2γ-1 (301) V11.40-1 = (930) V21.40-1 ( ) 0.40 == 3 =15.6

B. Work done = (T1-T2) n = = x x (301-903) = -250J

Conclusion

To conclude, thermodynamics is vital and common to apply in engineering in our normal life. It is the study of energy and its various inter conversions from one form to another. Thermodynamics has various types of applications in our daily life. For instance: Fossil-fueled steam power plants, Spark-ignition engines and Jet engines. All types of vehicles that we use, for example motorcycles, cars, trucks, ships, airplanes, and many other types work on the thermodynamic concept.

They may be using petrol engine or diesel engine, but the law remains the same. Even cooling machines, such as refrigerators and air conditioners, actually use heat, simply reversing the usual process by which particles are heated. The refrigerator pulls heat from its inner compartment-the area where food and other perishables are stored-and transfers it to the region outside. This is why the back of a refrigerator is warm.
One of the important fields of thermodynamics is heat transfer, which relates to transfer of heat between two media. There are three modes of heat transfer: conduction, convection and radiation. The concept of heat transfer is used in wide range of devices like heat exchangers, evaporators, condensers, radiators, coolers, heaters, etc.
Thermodynamics also involves study of various types of power plants like thermal power plants, nuclear power plants, hydro-electric power plants, power plants based on renewable energy sources like solar, wind, geothermal, tides, water waves and other else.
Based on what we stated, we can see how important does the thermodynamics work in our real life.

References

Richard Wolfson & Jay M. Pasachoff. (1999). Physics for Scientists and Engineers Third Edition. ADDISON-WESLEY.
Giancoli. (1998). Physic Volume 1 Fifth Edition. PRENTICE HALL.
Ohanian,Hans C.(2007). Physics for Engineers and Scientists. New York : W.W.Norton
Poh Liong Yong. (2010). Physics Volume 1. Pelangi. https://www.khanacademy.org/science/physics/thermodynamics/v/thermodynamics--part-4 http://www.scienceclarified.com/everyday/Real-Life-Physics-Vol-2/Thermodynamics-Real-life-applications.html

References: Richard Wolfson & Jay M. Pasachoff. (1999). Physics for Scientists and Engineers Third Edition. ADDISON-WESLEY. Giancoli. (1998). Physic Volume 1 Fifth Edition. PRENTICE HALL. Ohanian,Hans C.(2007). Physics for Engineers and Scientists. New York : W.W.Norton Poh Liong Yong. (2010). Physics Volume 1. Pelangi. https://www.khanacademy.org/science/physics/thermodynamics/v/thermodynamics--part-4 http://www.scienceclarified.com/everyday/Real-Life-Physics-Vol-2/Thermodynamics-Real-life-applications.html