The First Law of Thermodynamics

By: Yidnekachew Messele

It is the law that relates the various forms of energiesfor system of different types.

It is simply the expression of the conservation ofenergy principle.

Based on experimental observations, the first law ofthermodynamics states that “Energy can be neither created nor destroyed during a

process; it can only change forms.”

For the system shown above, the conservation of energyprinciple or the first law of thermodynamics is expressed as

The total energy of the system, Esystem, is given as

Total Energy Total Energy The Change in Total=

Entering the System Leaving the System Energy of the System

−

in out systemE E E− = ∆

systemE = Internal Energy+ Kinetic Energy+ Potential Energy

systemE =U + KE + PE

If the system does not move with a velocity and has no changein elevation, it is called a stationary system, and theconservation of energy equation reduces to

in outE E U− = ∆

in outE E U + KE + PE− = ∆ ∆ ∆

Mechanisms of Energy Transfer, Ein and Eout Heat Transfer, (Q) Heat transfer to a system (heat gain)

increases the energy of the molecules and thus the internalenergy of the system and heat transfer from a system (heatloss) decreases.

Work Transfer, (W) Work transfer to a system (i.e., workdone on a system) increases the energy of the system, andwork transfer from a system (i.e., work done by the system)decreases .

Mass Flow, (m) When mass enters a system, the energy of thesystem increases because mass carries energy with it (in fact,mass is energy). Likewise, when some mass leaves the system,the energy contained within the system decreases .

The energy balance can be written more explicitly as

Or on a rate form, as

, ,( ) ( ) ( )in out in out in out mass in mass out SystemE E Q Q W W E E E− = − + − + − = ∆

{ } { } ( )Net energy transfer Change in internal, kinetic,in out Systemby heat, work and mass potential, etc..energies

E E E kJ− = ∆

{ } { } ( )Systemin outRate of net energy transfer Rate change in internal, kinetic, by heat, work and mass potential, etc..energies

E E E kW− = ∆

The first law and a closed system For the closed system where the mass never crosses the system

boundary, then the energy balance is

Closed system undergoing a cycle For a closed system undergoing a cycle, the initial and final states are

identical, and thus

( ) ( ) ∆in out in out systemQ -Q + W -W = E

2 1 0systemE E E∆ = − =

0in outE E− =

in outE E=

, ,net out net inW Q= , ,net out net inW Q=

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

2 112 12 2 1 2 1

( ) ( )2

m V VQ W U U mg Z Z−− = − + + −

( )12 2 1 12Q U U W= − +

Internal energy and Enthalpy Internal energy

The internal energy includes some complex forms of energy show updue 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 saturationpressure or temperature

Uum

=

f gU U U= + f f g gmu m u m u= +

f fgu u xu= +

Enthalpy It is another extensive property which has a unit of energy and

it is denoted by H. The enthalpy is a convenient grouping of the internal energy,

pressure, and volume and is given by

The enthalpy per unit mass is,H U PV= +

Hhm

=

h = u+ Pv( )12 2 1 12Q U U W= − + 12 2 1( )W Pdv P V V= = −∫ 12 2 1W PV PV= −

( ) ( )12 2 1 2 1Q U U PV PV= − + − ( ) ( )12 2 2 1 1Q U PV U PV= − + −

12 2 1Q H H= − f fgh h xh= −

Specific Heat It defined as; the energy required to raise the

temperature of a unit mass of a substance by onedegree.

It is an intensive property of a substance that willenable us to compare the energy storage capability ofvarious substances. The unit is .

In thermodynamics, we are interested in two kinds ofspecific heats: specific heat at constant volume andspecific heat at constant pressure.

KJ KJor KgKKg℃

The specific heat at constant volume can be viewedas the energy required to raise the temperature of theunit mass of a substance by one degree as the volumeis maintained constant.

Here the boundary work is zero because the volumeis constant

From first law Per unit mass

δQ dU=

q duδ = vq C dTδ =

vC dT du=

vv

duCdT

=

The specific heat at constant pressure Cp can beviewed as the energy required to raise thetemperature of the unit mass of a substance by onedegree as the pressure is maintained constant.

From first law Per unit mass

( )δQ dU PdV d U PV dH= + = + =

q dhδ = pq C dTδ =

pC dT dh=

pp

dhCdT

=

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 narrowtemperature difference

Pv RT=

vdu C dT=

2 1 vu u C dT− = ∫

2 1 , 2 1( )ave vu u C T T− = −

( )pdh C T dT= 2 1 ph h C dT− = ∫2 1 , 2 1( )ave ph h C T T− = −

Relation between CP and CV 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 pC dT du vC dT

p vC dT C dT RdT= +

p vC C R= +

p

v

CK

C=

p vC KC= v vKC C R= +

1vRC

K=

−p

p

CC R

K= +

1pKC R

K=

−

Internal Energy, Enthalpy, and Specific Heats of Solids and Liquids A substance whose specific volume (or density) is constant is

called an incompressible substance. The specific volumes of solids and liquids essentially remain

constant during a process. Therefore, liquids and solids can beapproximated as incompressible substances.

It can be mathematically shown that the constant-volume andconstant-pressure specific heats are identical forincompressible substances.

The specific heat can be expressed as

p vC C C= =

Example

1. The initial pressure and volume of a piston-cylinder arrangement is 200kPa and 1m3

respectively. 2000kJ of heat is transferred to thesystem and the final volume is 2m3. Determinethe change in the internal energy of the fluid.

2. A piston–cylinder device initially contains 0.8 m3 ofsaturated water vapor at 250 kPa. At this state, thepiston is resting on a set of stops, and the mass of thepiston is such that a pressure of 300 kPa is required tomove it. Heat is now slowly transferred to the steamuntil the volume doubles. Show the process on a P-vdiagram with respect to saturation lines anddetermine (a) the final temperature, (b) the workdone during this process, and (c) the total heattransfer.

3.A piston-cylinder contains steam initially at 1Mpa, 450 oC and 2.5m3. Steam ia allowed to coolat constant pressure until it first start condensing.Show the process on a T-v diagram with respectto saturatin lines and detrmine.

a) The mass of the steamb) The final temperaturec) The amount of heat tarnsfer.

4. A piston–cylinder device initially contains steam at 200 kPa,200°C, and 0.5 m3. At this state, a linear spring is touching thepiston but exerts no force on it. Heat is now slowlytransferred to the steam, causing the pressure and the volumeto rise to 500 kPa and 0.6 m3, respectively. Show the processon a P-v diagram with respect to saturation lines and determine(a) the final temperature, (b) the work done by the steam, and(c) the total heat transferred.

The First Law and the Control Volume

The conservation of mass and the conservation of energyprinciples for open systems or control volumes apply tosystems having mass crossing the system boundary or controlsurface.

In addition to the heat transfer and work crossing the systemboundaries, mass carries energy with it as it crosses the systemboundaries.

Hence the conservation of mass principle can be used to relate mass which entering and leaving a system. It can be expressed as

The net mass transfer to or from a control volume during aprocess (a time interval t) is equal to the net change (increaseor decrease) in the total mass within the control volume duringthat process (t). That is,

Thermodynamic processes involving control volumes can beconsidered in two groups: steady-flow processes and unsteady-flow processes.

Total mass entering Total mass leaving Net change in mass- =

the CV during Δt the CV during Δt within the CV during Δt

in out CVm m m− = ∆ ( )kg

/in out CVm m dm dt− =

( / )kg s m Vρ=

Steady state process The flow through a control volume is at steady state if, “the

property of the substance at a given position within or at theboundaries of the control volume do not change with time”.

During a steady-flow process, the total amount of masscontained within a control volume does not change with time(mCV= constant).

0CVCVdm m

dt= ∆ =

/in out CVm m dm dt− =

in outm m=

in outm m=∑ ∑

in outin outV Vρ ρ=

outin incompressible assumptionρ ρ=

in outV V=

in in out outV A V A=

Unsteady state process The properties within the control volume change with time but

remain uniform at any instant of time. Typical example:- filling and empting processes where most of

the cases average value of properties must be used.

0cvdmdt

≠

i ecv mdm

dm

t= −∑ ∑

cvi e

dm m mdt

= −

Flow Work and The Energy of a Flowing Fluid

Unlike closed systems, control volumes involve mass flowacross their boundaries, and some work is required to push themass into or out of the control volume.

This work is known as the flow work, or flow energy.

F PA=

flowW FL= PAL= PV= ( )kJ

floww Pv=

, , flow in i i flow exit e ew Pv and w P v= =

, ,( ) ( ) flow in i flow exit ei i e eW Pv m and mP vW= =

( ) ( )flow e ie e i iW mP v Pv m= −

flow cvW W W= −

Development of energy balance The general representation of the first law of thermodynamics

The first law for open system will also have the same form, but

The fluid entering or leaving a control volume possesses anadditional form of energy, the flow energy Pv

Then the total energy of a flowing fluid on a unit-mass basis(denoted by) becomes

12 12 2 1Q W E E= + −

12 flow cvW W W= +

E = Internal Energy+ Kinetic Energy+ Potential EnergyE =U + KE + PE

2

2Ve u ke pe u gz= + + = + +

Pv eθ = + ( )Pv u ke pe= + + +

But the combination Pv + u has been previously defined as theenthalpy h. So the relation in the above equation reduces to

For inlet

For outlet

General equation

2

2Vh ke pe h gzθ = + + = + + ( / )kJ kg

2

2i

i i i i i i iVe Pv u gz Pv+ = + + +

2

2i

i i i i iVe Pv h gz+ = + +

2

2e

e e e e eVe P v h gz+ = + +

/in out systemE E dE dt− =0( )

0steady

=

in in out outin out

Q W m Q W mθ θ+ + = + +∑ ∑

2 2

( ) ( )2 2in in out out

in out

V VQ W m h gz Q W m h gz+ + + + = + + + +∑ ∑

In such cases, it is common practice to assume heat to betransferred into the system (heat input) at a rate of , and workproduced by the system (work output) at a rate of , and thensolve the problem. The first-law or energy balance relation inthat case for a general steady-flow system becomes

When the fluid experiences negligible changes in its kineticand potential energies (that is,ke = 0, pe = 0), the energybalance equation is reduced further to

Q

W

2 22 1

2 1 2 1( )2

V VQ W m h h g z z −

− = − + + −

2 22 1

2 1 2 1( )2

V Vq w h h g z z−− = − + + −

2 1q w h h− = −

Some Steady-Flow Engineering Devices

Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines,

rockets, spacecraft, and even garden hoses. A nozzle is a device that increases the velocity of a fluid at the

expense of pressure. A diffuser is a device that increases the pressure of a fluid by

slowing it down.

in outm m=

1 2m m m= =

in outE E=

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

2 2

2 2i e

i i e eV Vm h m h

+ = +

22( )e i e iV h h V= − +

Turbines In steam, gas, or hydroelectric power plants, the device that

drives the electric generator is the turbine. As the fluid passes through the turbine, work is done against

the blades, which are attached to the shaft. As a result, theshaft rotates, and the turbine produces work.

in outm m=

1 2m m m= =

in outE E=

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

i i e e outm h m h W= +

( )out i eW m h h= −

Compressors Compressors, as well as fans, are devices used to increase the

pressure of a fluid. Work is supplied to these devices from an external source

through a rotating shaft.

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

( )net e iW m h h− = −

( )net i eW m h h= −

in outm m=

Pumps The work required when pumping an incompressible liquid in

an adiabatic steady-state, steady-flow process is given by

The enthalpy difference can be written as

The pumping process for an incompressible liquid isessentially isothermal, and the internal energy change isapproximately zero. Since v2 = v1 = v the work input to thepump becomes

2 22 1

2 1 2 1( )2

V VQ W m h h g z z −

− = − + + −

( ) ( ) ( )2 1 2 1 2 1h h u u Pv Pv − = − + −

( )2 2

2 12 1 2 1( )

2V VW m v P P g z z

−− = − + + −

( )2 1W m v P P− = −

( ), 2 1in pumpW m v P P= −

Throttling Valves Throttling valves are any kind of flow-restricting devices that

cause a significant pressure drop in the fluid. The pressure drop in the fluid is often accompanied by a large

drop in temperature, and for that reason throttling devices arecommonly used in refrigeration and air-conditioningapplications.

in outm m=

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

i i e em h m h=

i eh h=

Mixing Chambers The mixing of two fluids occurs frequently in engineering

applications. The section where the mixing process takes placeis called a mixing chamber.

The ordinary shower is an example of a mixing chamber.

in outm m=∑ ∑

1 2 3m m m+ =

2 3 1m m m= −

in outE E=

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

1 1 2 2 3 3m h m h m h+ =

1 1 3 1 2 3 3m h m m h m h + − =

1 2 3 21 3( ) ( )m h h m h h− = −

231 3

1 2

( )( )h hm mh h−

=−

Heat Exchangers Heat exchangers are normally well-insulated devices that

allow energy exchange between hot and cold fluids withoutmixing the fluids.

in outm m=

1 2 wm m m= =

3 4 Rm m m= =

in outE E=

2 2

2 2i e

neti i i e e enetinlet exit

V VQ m h gz W m h gz

+ + + = + + +

∑ ∑

1 1 3 3 2 2 4 4m h m h m h m h+ = +

2 31 4( ) ( )w Rm h h m h h− = −

Example

1.Steam at 0.4 MPa, 300 oC, enters an adiabaticnozzle with a low velocity and leaves at 0.2MPawith a quality of 90%. Find the exit velocity, inm/s.

2. The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in Figure below.

a) Compare the magnitudes of h, ke, and pe.b) Determine the work done per unit mass of

the steam flowing through the turbine.c) Calculate the mass flow rate of the steam.

3. Nitrogen gas is compressed in a steady-state,steady-flow, adiabatic process from 0,1 Mpa, 25 oC.During the compression process the temperaturebecome 125 oC. If the mass flow rate is 0.2kg/s,determine the work done on the nitrogen, in kW.(use cp=1.039kJ/kg.K)

4. Saturated steam at 0.4 MPa is throttled to0.1MPa, 100 oC. Determine the quality of thesteam at 0.4MPa.

5. Steam at 0.2MPa, 300oC, enters a mixingchamber and is mixed with cold water at 20oC,0.2MPa, to produce 20kg/s of saturated liquid at0.2MPa. What are the required steam and coldwater flow rates?

6. Two tanks (Tank A and Tank B) are separated by a partition. Initially TankA contains 2-kg steam at 1 MPa and 300°C while Tank B contains 3-kgsaturated liquid–vapor mixture with a vapor mass fraction of 50 percent.Now the partition is removed and the two sides are allowed to mix until themechanical and thermal equilibrium are established. If the pressure at thefinal state is 300 kPa, determine (a) the temperature and quality of thesteam (if mixture) at the final state and (b) the amount of heat lost from thetanks.

7. Air enters an adiabatic nozzle steadily at 300 kPa,200°C, and 30 m/s and leaves at 100 kPa and 180m/s. The inlet area of the nozzle is 80 cm2.Determine (a) the mass flow rate through thenozzle, (b) the exit temperature of the air, and (c)the exit area of the nozzle.

8. Steam flows steadily through an adiabatic turbine.The inlet conditions of the steam are 10 MPa,450°C, and 80 m/s, and the exit conditions are 10kPa, 92 percent quality, and 50 m/s. The mass flowrate of the steam is 12 kg/s. Determine (a) thechange in kinetic energy, (b) the power output, and(c) the turbine inlet area.

9. Refrigerant-134a is throttled from the saturatedliquid state at 700 kPa to a pressure of 160 kPa.Determine the temperature drop during thisprocess and the final specific volume of therefrigerant.

10. In steam power plants, open feedwater heaters are frequentlyutilized to heat the feedwater by mixing it with steam bled off theturbine at some intermediate stage. Consider an open feedwaterheater that operates at a pressure of 1000 kPa. Feedwater at 50°Cand 1000 kPa is to be heated with superheated steam at 200°Cand 1000 kPa. In an ideal feedwater heater, the mixture leavesthe heater as saturated liquid at the feedwater pressure.Determine the ratio of the mass flow rates of the feedwater andthe superheated vapor for this case.

12. An adiabatic air compressor is to be powered by adirect-coupled adiabatic steam turbine that is alsodriving a generator. Steam enters the turbine at 12.5MPa and 500°C at a rate of 25 kg/s and exits at 10kPa and a quality of 0.92. Air enters the compressorat 98 kPa and 295 K at a rate of 10 kg/s and exits at 1MPa and 620 K. Determine the net power deliveredto the generator by the turbine.