First Law of Thermodynamics: Principles, Applications, and Examples - Prof. Serdf, Schemes and Mind Maps of Mathematical Methods

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Chapter Three

The First Law of Thermodynamics

Engineering Thermodynamics I

The First Law of Thermodynamics

• (^) The first law is usually referred to as the law of

conservation of energy, i.e.

energy can neither be created nor destroyed, but rather

transformed from one state to another.

• (^) The energy balance is maintained within the system.
• (^) The various energies associated are then being

observed as they cross the boundaries of the system.

The First Law of Thermodynamics [Cont…]

• The^ total energy of the system,^ E system

, is given

as

system

E = Internal Energy + Kinetic Energy + Potential Energy

system E = U + KE + PE

• (^) Any change in the total energy of the system is

due to energy crossing the boundaries

The First Law of Thermodynamics [Cont…]

• (^) If the system does not move with a velocity and has

no change in elevation, it is called a stationary

system , and the conservation of energy equation

reduces to

in out E  E U

Energy Balance

• (^) The energy balance can be written more explicitly as , , ( ) ( ) ( ) in out in out in out mass in mass out System E  E  Q  Q  W  W  E  E E  (^) in out  (^) Net energy transfer  (^) System (^) Change in internal, kinetic,  by heat, work and mass (^) potential, etc..energies E  E  E kJ
• (^) Or on a rate form, as

The first law and a closed system

• (^) For the closed system where the mass never crosses the system boundary, then the energy balance is      in out in out system Q -Q + W -W = E

The first law and a closed system [Cont…]

• (^) If the total energy is a combination of internal energy, kinetic energy and potential energy
• (^) For negligible changes in kinetic and potential energy E U  KE  PE   2 2 2 1 12 12 2 1 2 1

m V V

Q W U U mg Z Z

  12 2 1 12 Q  U  U W

Internal energy and Enthalpy

• (^) Internal energy
  • (^) The internal energy includes some complex forms of energy show up due to translation, rotation and vibration of molecules.
  • (^) It is designated by U and it is extensive property.
  • (^) Or per unit mass as, specific internal energy,
• (^) If we take two phase as liquid and vapor at a given

saturation pressure or temperature

U u m  f g U U  U f f g g mu m u m u f fg u u xu

Specific Heat

• (^) It defined as; the energy required to raise the

temperature of a unit mass of a substance by one

degree.

• (^) It is an intensive property of a substance that will

enable us to compare the energy storage capability of

various substances. The unit is.

• (^) In thermodynamics, we are interested in two kinds of

specific heats: specific heat at constant volume and

specific heat at constant pressure.

KJ (^) or KJ Kg ℃ KgK

• (^) The specific heat at constant volume can be viewed

as the energy required to raise the temperature of the

unit mass of a substance by one degree as the volume

is maintained constant.

• (^) Here the boundary work is zero because the volume is

constant

• (^) From first law
• (^) Per unit mass δQ dU  q du v  q C dT v C dT du v v du C dT       

Internal Energy, Enthalpy, and Specific Heats of Ideal Gases

• (^) We defined an ideal gas as a gas whose temperature, pressure, and specific volume are related by
• (^) From the specific heat relation
• (^) Or taking average value of specific heat for narrow temperature difference Pv RT v du C dT

u 2  u 1 C dTv

u 2  u 1 C (^) ave v (^) , ( T 2  T 1 )

dh C p T dT

2 1 p h  h  C dT  h 2  h 1 C (^) ave p (^) , (T 2  T 1 )

Relation between C P and C V for Ideal Gases

• (^) Replacing by and by we have
• (^) At this point, we introduce another ideal-gas property called the specific heat ratio k, defined as h u  RT dh^ du^ RdT

dh

p

C dT du^ C dTv

p v C dT C dT RdT p v C C R p v C K C  p v C KC KCv^ C^ v^ R 1 v R C K   p p C C R K   1 p K C R K  

Examples

Examples

1. The initial pressure and volume of a piston-

cylinder arrangement is 200kPa and 1m

3

respectively. 2000kJ of heat is transferred to

the system and the final volume is 2m

3

Determine the change in the internal energy

of the fluid.