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Thermodynamics&

Heat transferLecture 02

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Instructor: Dr. Qari Khalid Waheed

Institute of Mechatronics Engineering,

UET Peshawar.

BY: Engr. Muhammad Usman Khan

(MtE-203)

Fall 2015

The Working Fluid

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The matter contained within the boundaries of a system is defined as the working fluid.

When two independent properties of the fluid are known then the thermodynamic state of the fluid is defined.

The working fluid can be liquid, vapour, or gaseous phase.

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Liquid, Vapour and Gas

When a liquid is heated at any one constant pressure there is one fixed temperature at which bubble of vapour form in the liquid; this phenomena is called boiling.

Consider a P-v diagram:

The points P,Q, and R represents

The boiling-points of a liquid

At pressure Pp , PQ & PR

respectively.

Fig:1: Boiling-points plotted on a P-v graph

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Liquid, Vapour and Gas (cont’d)

Enthalpy of Vaporization:

Further heating of a liquid at its boiling-point at constantpressure resulting vapour (change in phase); during thischange of phase the temperature & pressure remain constant.The heat supplied is known as enthalpy of vaporization.

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Fig:2. Points of complete vaporization Plotted on P-V graph.

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Liquid, Vapour and Gas

When a dry saturated vapour is heated at constant pressureits temperature rises & it becomes superheated.

The difference between the actual temperature of thesuperheated vapour and the saturation temperature at thevapour is called the degree of superheat. For example, In theFig:3, the vapour is superheated at PQ and T3, and the degreeof superheat is T3 – T2.

There is a corresponding saturation temperature for eachsaturation pressure.

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7Fig:3. Isothermal for a vapour plotted on a P-V graph

Use of Vapour Tables

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Extract from tables of properties of wet stream

Fig:4. Points identified on a P-V diagram of wet Stream

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Properties of wet vapour

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For a wet vapour the total volume of the mixture is given by the volume of the

liquid present plus the volume of dry vapour present.

Therefore the specific volume is given by:

𝒗 = 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒍𝒊𝒒𝒖𝒊𝒅+𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒅𝒓𝒚 𝒗𝒂𝒑𝒐𝒖𝒓

𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒆𝒕 𝒗𝒂𝒑𝒐𝒖𝒓

• Now for 1 kg of wet vapour there are 𝒙 kg of dry vapour and (1- 𝒙) kg of liquid,

where 𝒙 is the dryness fraction as defined, hence;

𝒗 = 𝒗 f (1- 𝒙) + 𝒗 g 𝒙

• The volume of the liquid is usually negligibly small compared to the volume of dry

saturated vapour, hence for most practical problems;

𝒗 = 𝒙 𝒗g

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Properties of wet vapour (cont’d)

The enthalpy of a wet vapour is given by the sum of the enthalpy of theliquid plus the enthalpy of the dry vapour, i.e.

h = (1- x) hf + xhg

Therefore

h = hf + x (hg - hf)

i.e.

h = hf + x hfg

Similarly, the internal energy of a wet vapour is given by the sum of theinternal energy of the liquid and the internal energy of the dry vapour,

i.e. u = (1- x) uf + xug

or u = hf + x (ug - uf)

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Example 2.1

Example 2.2

Read properties of superheated vapour (Reading assignment)

Example 2.3 --- Example 2.7

The Characteristic equation of state

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Any relation among the pressure, temperature, and specific volume of asubstance is called an equation of state. The simplest and best-knownequation of state is the ideal-gas equation of state, given as

PV = RT

where R is the gas constant.

• Caution should be exercised in using this relation since an ideal gas is a

fictitious substance

• Real gases exhibit ideal-gas behavior at relatively low pressures and high

temperatures.For a mass, m, occupying a volume, V,

pV = mRT --------------> (A)

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The Characteristic equation of state (cont’d)

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The mass of any substance per amount of substance is known as the molar mass, ḿ, i.e.

ḿ = 𝒎

𝒏-----------------> (B)

• 𝒎 is the mass and 𝒏 is the amount of substance.

• Normal unit for ḿ is kg/kmol

Substituting Eq. (B) into Eq. (A)

pV = n ḿRT oR ḿR = 𝒑𝑽

𝒏𝑻-----------> (C)

The quantity 𝒑𝑽

𝒏𝑻is a constant for all gases & is known as molar gas constant.

Represented by Ȓ.

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The Characteristic equation of state (cont’d)

ḿR = Ȓ = 𝒑𝑽

𝒏𝑻

Or pV = n Ȓ TSince

ḿR = Ȓ

R = Ȓ

ḿ

• The value of Ȓ has been shown to be 8314.5 Nm/kmol K.

Example 2.8

Example 2.9

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The heat required to raise unit mass through one degree rise;

dQ = mc dT

• Where m is the mass, dT is the increase (change) in temperature, & cis the specific heat capacity.

Specific Heat Capacity at Constant Volume, Cv is:

Cv =(𝝏𝒖

𝝏𝑻)v

Specific Heat Capacity at Constant pressure, Cp is:

Cp =(𝝏𝒉

𝝏𝑻)p

Specific Heat Capacity

We can write:

dQ = mcp dT for a reversible non-flow process at constant pressure.

and

dQ = mcv dT for a reversible non-flow process at constant volume.

For a perfect gas the values of Cv and Cp are constant, Hence byintegrating the above equations;

Q = mcp (T2 – T1), for a reversible constant pressure process.

Q = mcv (T2 – T1), for a reversible constant volume process.

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Specific Heat Capacity (cont’d)

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Joule’s law states that the internal energy of a perfect gas is a function of the absolute temperature only, i.e. u = f(T).

• Let unit mass of a perfect gas be heated at constant volume. From the non-flow energy equation;

dQ + dW = dusince the volume remains constant, then no work-done, i.e. dW = 0, therefore;

dQ = duAt constant volume for a perfect gas, for unit mass;

dQ = Cv dTTherefore, dQ = du = Cv dTBy integration

u = Cv T + k. where k is constant

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Joule’s Law

For a perfect gas it can be assumed that, u = 0 when T = 0, hence the constant k is zero,

i.e.

Specific internal energy, u = Cv T for a perfect gas,

internal energy, U = mcv T for mass, m, of a perfect gas.

Gain in internal energy is given by;

U2 – U1 = mcv (T2 – T1)

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Joule’s Law (cont’d)

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Relationship between the Specific Heat

Capacities

Let a perfect gas be heated at constant pressure from T1 to T2. From the non-flow equation. i.e.

Q + W = U2 – U1

And for a perfect gas, from equation;

U2 – U1 = mcv (T2 – T1). Hence;

Q + W = mcv (T2 – T1) ----------------------> (*)During a constant pressure process, the work done is given by the pressure times the change in volume, i.e. W = - p(V2 – V1). Then using the equations,

pV2 = mRT2 and pV1 = mRT1

we haveW = - mR(T2 – T1)

Substituting value of W into equation (*), we get;

Q - mR(T2 – T1) = mcv (T2 – T1)-------------> (**)

Therefore

Q = m(Cv + R)(T2 – T1)

But for a constant pressure process

Q = mCp(T2 – T1)

Hence by equating the two equations for the heat flow, Q, we have;

m(Cv + R)(T2 – T1) = mCp(T2 – T1)

therefore

Cv + R = Cp

Or

Cp - Cv = R

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Relationship between the Specific Heat

Capacities (cont’d)

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As specific enthalpy, h = u + pv ------------> (*1)

For a perfect gas, from equation, pv = RT

From joule’s law equation, u = CvT. hence by putting in the equation (*1)

h = CvT +RT = (Cv+ R)T

But

Cp - Cv = R or Cv + R = Cp

Therefore h is given by;

h = CpT

for mass, m, of a perfect gas

H = m CpT

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Specific enthalpy of a perfect gas

The ratio of the specific heat capacity at constant pressure to the specific heat capacity at constant volume is given by the symbol 𝛾(gamma).

i.e. 𝜸 = Cp/ Cv ---------------------------> (*2)

As, Cp – Cv = R

Dividing both sides by Cv

Cp/ Cv - 1 = R/ Cv

Therefore, using equation (*2), we have;

𝜸 - 1 = R/ Cv

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Ratio of Specific Heat Capacities

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Therefore

Cv = 𝑹

(𝜸−𝟏)-------------> (*3)

Again from equation (*2), Cp= 𝜸 Cv

Substituting into equation (*3);

Cp= 𝜸 Cv = Cv = 𝜸𝑹

(𝜸−𝟏)

Example 2.10

Example 2.11

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Ratio of Specific Heat Capacities (cont’d)

Questions ?

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