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Engineering › Thermodynamics ›

First Law

Describes the first Law and Introduces Enthalpy and Entropy

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Contents

Definitions
Forms Of Energy
The General Energy Equation
The Application Of The First Law To Particular Cases
Reversible Processes
Heat Entropy
An Isentropic Process - Entropy Remains Constant
Changes In Entropy
Page Comments

Definitions

S.T.P. ( Standard Temperature and Pressure ) $60^0\;F.\;and\;14.7\;psi$
N.T.P. ( Normal temperature and Pressure ) $32^0\;F.|;and|;14.7\;psi$
HEAT ENGINE This is any machine that can convert heat into work and visa versa.
WORKING SUBSTANCE ( W.S.) The W.S. is used as the carrier for the heat energy. The Heat Engine carries out the conversion process by a series of changes of state of the working substance.
THE STATE OF THE WORKING SUBSTANCE. The state of the W.S. is defined by the values of it's properties i.e. Pressure (P); Volume (V); Temperature (T); Internal Energy (E); Enthalpy (H)and Entropy ( $\phi$ .
Note. These Properties are sometimes called " Functions of State"
THE FIRST LAW OF THERMODYNAMICS. (Sometimes called the Law of the conservation of Energy.) Energy can neither be created nor destroyed but only transforms from one form to another.
HEAT AND WORK UNITS ( Imperial ). One BTU is the heat required to raise one pound of water through $\frac{1}{180}$ of the range $32^0 F\;-\;212^0F$ . One CHU is the heat required to raise one pound of water through $\frac{1}{100}$ of the range $0^0C\;-\;100^0C$

One Foot Pound is the work done when a force of one pound moves one foot in the direction of it's application.

THE MECHANICAL EQUIVALENT OF HEAT ( J ).
From experimental work it has been established that 778 ft.lbs give a temperature rise of

one degree F. to one lb. of water. 778 ft.lb = 1 BTU
1400 ft.lb. = 1 CHU

THE THERMAL EFFICIENCY OF AN ENGINE. This is the means of comparing the ability of an engine to convert heat energy into work. Work output = Heat Energy supplied - Heat Energy rejected in the WS and in losses.
$W\;=\;Q_s\;-\;Q_R$
(1)

$\therefore\;\;\;\;\;Thermal\;\eta \;=\;\frac{Mechanical\;Work\;Out put}{Heat\;Energy\;Supplied}$

(2)

$=\;\frac{Q_S\;-\;Q_R}{Q_S}\;=\;1\;-\;\frac{Q_R}{Q_S}$

(3)

Examples of $\eta$

$Steam\;Locomotive\;\approx 4\%$

(4)

$I.C.\;Engine\;\approx \;18\%\;-\;30\%$

(5)

$Modern\;Power\;Station\;\approx 30\%$

(6)

From this it can be seen that the conversion of heat into work involves considerable losses in AVAILABLE ENERGY whilst work to heat can achieve 100%.

Forms Of Energy

Work and heat
Potential Energy (PE) The energy possessed by a quantity of matter by virtue of it's position above some specified datum.
Kinetic Energy (KE) Energy possessed by virtue of the translational Velocity.
Internal Energy This is the Sum of the Molecular KE ( Translational and Rotational )and the PE of the individual molecules.

The General Energy Equation

Imagine a heat engine taking in a WS in a State 1. The WS passes through the machine having $Q_S$ units of heat supplied to it and giving a work output of W units. The WS then leaves the machine with energies $\;E_2\;(KE)_2\;(PE)_2$

Applying The first Law

Energy Entering the System = Energy Leaving the System

The work done to drive one unit mass of air through the system is given by:-

$Work\;done\;=\;W_2L_2\;-\;W_1L_1\;=\;P_2a_2L_2\;-\;P_1a_1L_1\;=\;P_2V_2\;-\;P_1V_1$

(7)

$Thus\;\;\;\;E_1\;+\;(KE)_1\;+\;(PE)_1\;+\;Q_S\;=\;E_2\;+\;(KE)_2\;+\;(PE)_2\;+\;W\;+\;(P_2V_2\;-\;P_1V_1)$

(8)

________________________________________________________________________________________

$\therefore\;\;\;\;(E_1\;+\;P_1)\;+\;(KE)_1\;+\;(PE)_1\;+\;Q_S\;=\;(E_2\;+\;P_2)\;+\;(KE)_2\;+\;(PE)_2\;+\;W$

(9)

The Equation E;+;PV is called ENTHALPY ( H )

In a Petrol; Engine we can say that KE and PE are negligible.

$Thus \;\;\;\;\;H_1\;+\;Q_S\;=\;H_2\;+\;W$

(10)

Therefore the Heat supplied to the WS = The Work done by the WS + The Gain in Enthalpy

In a Gas Turbine we can ignore the PE Term.

$Thus \;\;\;\;\;H_1\;+(KE_1)\;+\;Q_S\;=\;H_2\;+\;(KE_2)\;+\;W$

(11)

Therefore the Heat supplied to the WS = Work done by the WS + Gain in KE + Gain in Enthalpy

For A non Flow Process There are no PV; KE; or PE terms

$\therefore\;\;\;\;\;E_!\;+\;Q_1\;=\;E_2\;+\;W$

(12)

i.e. The Heat Supplied TO the WS = Work done BY hte WS + GAIN in IE

EXAMPLE.

Twenty lbs.of gas per second flow through a Gas Turbine. The Inlet pressure is 100 psi and the Specific Volume is 3 cu ft/lb. The Inlet Velocity is 500 ft/sec.

In passing through the Engine, the IE drops by 40 BTU/lb and 100 BTU/sec.are lost through Radiation. If the gas is discharged at 1000 ft/sec. and at 50 psi with a Specific Volume of 4 cu ft/lb, find the HP developed by the Turbine.

$E_1\;+\;P_1V_1\;+\;(KE)_1\;+\;Q_S\;=\;E_2\;+\;P_2V_2\;=\;(KE)_2\;+\;W$

(13)

$\therefore\;\;\;\;\;W\;=\;(E_1\;-\;E_2)\;+\;(P_1V_1\;-\;P_2V_2)\;+(KE_1\;-\;KE_2)\;+\;Q_S$

(14)

$But\;\;\;\;\;E_1\;-\;E_2\;=\;40\;BTU/lb.$

(15)

$And\;\;\;\;\;\;P_1V_1\;-\;P_2V_2\;=\;\frac{100\times 144\times 3}{778}\;-\;\frac{50\times 144\times 4}{778}\;=\;18.5\;BTU/lb.$

(16)

$And\;also\;\;\;\;\;\;KE_1\;-\;KE_2\;=\;\frac{500^2}{2g\times 778}\;-\;\frac{1000^2}{2g\times 778}\;=\;15\;BTU/lb.$

(17)

$Q_S\;=\;-\;\frac{100}{20}\;BTU/lb.$

(18)

$\therefore\;\;\;\;\;W\;=\;40\;-\;18.5\;-\;15\;-\;5\;=\;38.5\;BTU/lb$

(19)

$\therefore\;\;\;\;\;H. P.\; Out put\;=\;\frac{38.5\times 20\times 778}{550}\;=\;1090\;H.P.$

(20)

The Application Of The First Law To Particular Cases

$(E_1\;+\;P_1V_1)\;+\;PE_1\;+\;KE_1\;+\;Q_S\;=\;(E_2\;+\;P_2V_2)\;+\;PE_2\;+\;KE_2\;+\;W$

(21)

1) A Constant Volume Process.

The Work output = 0 and Constant Volume Processes are invariably Non-Flow. Basically nearly all the terms in the above equation are eliminated with the exception of:-

$E_1\;+\;Q_S\;=\;E_2$

(22)

i.e. all the heat supplied goes to increasing the Internal Energy of the WS

2) Constant Pressure

a) Non-Flow

$Force\;on\;the\;Piston\;=\;P\;A$

(23)

$Work\;Done\;=\;P\;A\;l\;=\;P(V_2\;-\;V_1)$

(24)

In the First Law equation most terms are eliminated and we are left with:-

$E_1\;+\;Q_S\;=\;E_2\;+\;W$

(25)

$\therefore\;\;\;\;\;Q_S\;=\;(E_2\;-\;E_1)\;+\;P(V_2\;-\;V_1)\;=\;(E_2\;+\;P_2V_2)\;-\;(E_1\;+\;P_1V_1)\;=\;H_2\;-\;H_1$

(26)

i.e. The Heat Supplied = The Gain in ENTHALPY

b) Flow

Considering that all changes to PE and KE are negligible. The First Law becomes:-

$H_1\;+\;Q_S\;=\;H_2\;+\;W$

(27)

It is impossible to devise a machine which operates a Constant Pressure flow Process giving a Constant Work Output. Thus in all practical cases the heat supplied equals in gain in Enthalpy.

Adiabatic Process. This is a Process during which there is NO heat transfer of heat between the WS and the surroundings.

a) Non Flow

$E_1\;=\;E_2\;+\;W$

(28)

$\thjerefore\;\;\;\;\;Work\;Done\;=\;(E_1\;-\;E_2)\;\;\;\;\;\;\;i.e. Loss\; of \;Internal\; Energy$

(29)

b) Flow ( All Rotary Turbines and Compressors)

$H_1\;=\;H_2\;+\;W$

(30)

Therefore the Work Done = The loss in Enthalpy.

3) Throttling.

Consider two points 1 and 2 which are far enough from the orifice so that the KE ters are negligible. No heat is supplied and there is no work output.

$\therefore\;\;\;\;\;H_1\;=\;H_2$

(31)

Hence there is no change in Enthalpy.

EXAMPLE

Two identical Pistons and cylinders are filled with 2 cu ft. of a WS at 50 psi. Enough heat is added to each to raise the Internal Energy by 50 BTU. In the first cylinder the piston is locked solid and in the second the piston is allowed to move so that the pressure remains constant. Find the heat added in each case if the piston are is one sq. ft and in the second case the piston moves by six inches.

Case 1

$Heat\; Supplied\; =\; Gain\;in\;IE\;=\;50\;BTU$

(32)

Case 2

$Heat\; Supplied\; =\;Work\;Done\;+ Gain\;in\;IE\;$

(33)

$=\;P(V_2\;-\;V_1)\;+\; Gain\;in\;IE\;$

(34)

$=\;\frac{50\times 144}{2}\;+\; Gain\;in\;IE\;$

(35)

$=\;\frac{50\times 144}{2\times 778}\;+\; 50\;=\;54.6 \;BTU$

(36)

Reversible Processes

The Principle of Reversibility. A Thermodynamic Process is reversible if it may be reversed in sense by some infinitely small change in the conditions changing it. When reversed it will pass back through each stage it assumed in the direct sense and will have the same State at each point in the direct and reverse process.

Implications of the Principle. An equal amount of work will be absorbed in the reversed process as is given out in the direct process. In addition an equal amount of heat will be rejected in the reversed process as was supplied during the direct process.

Conditions of Reversibility.

If there is heat transfer between bodies of different temperature, the process can no be reversible. Thus all heat transfer process must be Isothermal.
If there is friction between moving parts the Process can not be reversible. Friction ALWAYS represents a loss.
There can be no Internal Friction due to inter molecular forces,
During the process there must be no Eddies or Turbulence in the WS since these are irreversible.

From the Conditions it can be seen that NO practical Process ca be Thermodynamically Reversible>

Work Done In a Reversible Process.

Imagine the WS enclosed in a frictionless Cylinder and Piston at State 1. Now some Process is performed and the WS changes to State 2

Consider the piston at some point where the Pressure is "P". Let it move a small distance $\delta\;l$ and let the corresponding drop in pressure be $\delta\;P.$

$Initial\;Pressure\;=\;P$

(37)

$Final\;Pressure\;=\;P\;-\;\delta \,P$

(38)

$Mean\;Pressure\;=\;P\;-\;\frac{\delta \,P}{2}$

(39)

$Force\;on\;Piston\;=\;\left( P\;-\;\frac{\delta \,P}{2} \right)A$

(40)

$\therefore\;\;\;\;\;\delta\;W\;=\;\left( P\;-\;\frac{\delta \,P}{2} \right)A\,\delta\,l\;=\;\left(P\;-\;\frac{\delta\,P}{2} \right)\delta\,V$

(41)

$Thus \;\;\;\;\;\delta\,W\;=\;P\delta\,V\;-\;\frac{\delta\,P}{2}\delta\,V$

(42)

$\therefore\;\;\;\;\;Total\;Work\;Done\;=\;\lim_{\delta\,V\rightarrow 0}\sum_{V_1}^{V_2}P\,\delta\,V$

(43)

$=\;\int_{V_1}^{V_2}P\;dv$

(44)

$If\;\;\;\;\;V_2\gg \;V_1\;\;\;\;\;W+ve\;\;\;\;(Expansion)$

(45)

$If\;\;\;\;\;V_2\ll \;V_1\;\;\;\;\;W+ve\;\;\;\;(Compression)$

(46)

Curve or Line of State

$Area\;of\;Strip\;=\;P\;dv$

(47)

If we take an infinite number of such strips then $\delta\,v\rightarrow 0$ and the sum of their Areas equals the area under the Curve

$\therefore\;\;\;\;\;Area\;under\;the\;Curve\;=\;V_1\;-\;V_2\;=\;\lim_{\delta\,v\rightarrow 0}\sum_{v_1}^{v_2}}p\;\delta\,v$

(48)

$\therefore\;\;\;\;\;Area\;under\;the\;Curve\;=\;\int_{V_1}^{V_2}P\,dv\;=\;Work\;Done$

(49)

N.B. This only applies to a Reversible Process.

Heat Entropy

Construct a Graph with Absolute Temperature as the Ordinate such that the Area under the Curve of State represents the HEAT supplied or Rejected.

The quantity along the abscissa is a Property or FUNCTION OF STATE and is called ENTROPY . This is given the letter $\Phi$

$\delta\,Q\;=\;T\;\delta\,\Phi$

(50)

An Isentropic Process - Entropy Remains Constant

$Reversible\;Adiabatic\;=\;Isentropic$

(51)

N.B. It is possible to have Adiabatic Processes which are NOT reversible and Isentropic Processes which are not Adiabatic.

e.g. A Rotary Compressor. It is Adiabatic since no heat goes in or out of the machine. However Entropy increases due to heat produced internally by friction. If , however just enough heat is removed by a cooling system to counter act the heat produced by friction, then Entropy will remain constant but the process is no longer Adiabatic. These Processes are not Reversible.

Changes In Entropy

Entropy is a function of State so it's value depends upon the State. Changes in it's value depend only on the initial and final state and not on the process which cause the Change.

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