(mte-203) thermodynamics heat...

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Thermodynamics&

Heat transferLecture 03

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Instructor: Dr. Qari Khalid Waheed

Institute of Mechatronics Engineering,

UET Peshawar.

BY: Engr. Muhammad Usman Khan

(MtE-203)

Fall 2015

Reversible & Irreversible Processes

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Reversible process - a process that, once having take place it can be reversed. In doing so, it leaves no change in the system or boundary.

Irreversible process - a process that cannot return both the system and surrounding to their original conditions.

Fig 1: (a) Reversible Process (b) Irreversible Process

(a) (b)

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Reversible non-flow processes

Constant Volume processWorking substance is contained in a rigid vessel.

Boundaries of the system immovableWork done = zero, i.e. W = 0All heat supplied increase the internal energy of the system

From non-flow energy equation, for unit mass;Q + W = u2 – u1

But, W = 0Q = u2 – u1

For mass, m, of the working substance;Q = U2 – U1

And for a perfect gas, we haveQ = mcv (T2 – T1)

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Constant Volume process

(a) (b)

Fig 2: Constant volume process for (a) a vapour and (b) a perfect gas

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Reversible non-flow processes

Constant Pressure processFor a constant pressure process the boundaries must move against an external

resistance, as heat is suppliedFluid in a cylinder behind a piston.Work done is not zero, i.e. W =/= 0

For any reversible process, we know that;

W = – 𝒗1

𝒗2𝒑 𝒅𝒗Since 𝒑 is constant;

W = – 𝒑 v1

v2 𝒅𝒗 = – 𝒑 (𝒗2 – 𝒗1)

As we know that from non-flow energy equation;

Q + W = u2 – u1 => Q = (u2 – u1 ) – WHence for a reversible constant pressure process;

Q = (u2 – u1 ) + 𝒑 (𝒗2 – 𝒗1)

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Q = (u2 + 𝒑𝒗2) – (u1 + 𝒑𝒗1)

From enthalpy equation, we have;

h = u + p𝒗, hence;

Q = h2 – h1

Or for mass, m, of a fluid;

Q = H2 – H1

And for a perfect gas, we have

Q = mcp (T2 – T1)

Example 3.1

Constant Pressure process (cont’d)

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(a) (b)

Fig 3: Constant pressure process for (a) a vapour and (b) a perfect gas

Constant Pressure process

The shaded area represent the work done by the fluid.

Isothermal Process

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Process at constant temperature.

In an isothermal expansion heat must be added continuously in order to keep the temperature at initial value.

In an isothermal compression heat must be removed from the fluid continuously during the process.

After calculation of the heat flow, the work done can be obtained using the non-flow equation for unit mass.

Q + W = u2 – u1

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Fig 4: Isothermal process for a vapour on a p-v diagram

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Isothermal Process (cont’d)

For perfect gas;

As we know that from ideal-gas equation of state;

pv = RT

Since T is constant (isothermal process) and R is also constant, Hence;

pv = RT = Constant

Therefore for isothermal process, pv =Constant, OR;

p1v1 = p2v2

Work done for unit mass is given by;

W = - 𝟏𝟐𝒑 𝒅𝒗 ----------(1)

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Fig 4: Isothermal process for a perfect gas on a p-v diagram

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Since; pv = Constant, => p = c/v, where c = constant.

Therefore

W = – 𝒗1

𝒗2 𝒄𝒅𝒗

𝒗= - 𝒄 =

i.e. for unit mass of gas;

W = p1v1 ln (𝒗𝟏

𝒗𝟐)---------- (2)

Or

W = p2v2 ln (𝒗𝟏

𝒗𝟐)

For mass, m, of the gas;

W = p1V1 ln (𝒗𝟏

𝒗𝟐) ----------- (3)


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p1v1 = p2v2 => 𝒗𝟏

𝒗𝟐= 𝒑𝟐

𝒑𝟏, putting into Eq. (2);

W = p1v1 ln (𝒑𝟐

𝒑𝟏) per unit mass of gas, &

W = p1V1 ln(𝒑𝟐

𝒑𝟏) for mass, m, of the gas.

Also; p1v1 = p2v2 = RT =constant. Hence;

W = RT ln (𝒑𝟐

𝒑𝟏) per unit mass of gas, and

W = mRT ln (𝒑𝟐

𝒑𝟏) for mass, m, of the gas.

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For perfect gas from joule’s law;

U2 - U1 = mcv (T2 – T1)

Since process is isothermal, T2 = T1

U2 - U1 = 0

And similarly from non-flow energy equation;

u2 - u1 = Q + W, => u2 = u1

Hence, for an isothermal process of perfect gas:

Q + W = 0

Example 3.3

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Adiabatic Process: The process in which no heat is transferred to or from the fluid during the process.

Can be reversible or irreversible.

Here considering only reversible adiabatic process.

From non-flow energy equation, we have;

u2 - u1 = Q + W

Q = 0, (process is adiabatic);

u2 - u1 = W, for any adiabatic non-flow process

For perfect gas, consider the non flow energy equation in differential form;

dQ + dW = du => dQ = du – dW ------------------- (*)

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Reversible adiabatic non-flow processes

Since process is irreversible; => dW = - pdv, putting into Eq. (*);

dQ = du + pdv = 0 ----------- (*1)

From enthalpy, h = u + pv;

dh = du + pdv + vdp , OR

du + pdv = dh – pdv, using Eq.(*1)

dh – pdv = 0 => dh = pdv

Using pv = RT => p = RT/v, substituting into Eq.(*1);

du + 𝑹𝑻𝒅𝒗

𝒗= 0 -----------------(*2)

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Reversible adiabatic non-flow processes (cont’d)

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From specific internal energy equation;

u = cvT OR du = cvdT

Therefore Eq.(*2) become:

cvdT + 𝑹𝑻𝒅𝒗

𝒗= 0

Divitind by T, and then integrating; we have

cv𝒅𝑻

𝑻= 𝑹𝒅𝒗

𝒗= 0 => cvln T + R ln v = constant

Since, RT = pv => T = (pv)/R, Therefore;

cv𝒍𝒏𝒑𝒗

𝑹= 𝑹 𝒍𝒏 𝒗 = Constant

Dividing by Cv

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𝒍𝒏𝒑𝒗

𝑹=

𝑹

Cv𝒍𝒏 𝒗 = Constant-------(*3)

As, Cv = 𝑹

(𝜸−𝟏)OR

𝑹

Cv= (𝜸 − 𝟏) (Ratio of Specific Heat Capacities )

Substituting into Eq.(*3);

𝒍𝒏𝒑𝒗

𝑹= (𝜸 − 𝟏) ln v = Constant, OR

𝒍𝒏𝒑𝒗

𝑹= ln (vꝩ- 1) = Constant, {As, ln ab = b ln a}

OR

𝒍𝒏 (𝒑𝒗𝒗ꝩ− 1

𝑹) = constant

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𝒍𝒏 (𝒑𝒗ꝩ

𝑹) = constant, By eliminating log (ln);

𝒑𝒗ꝩ

𝑹= 𝐞constant

OR

𝒑𝒗ꝩ = Constant.Each perfect gas has its own value of 𝜸.

See the proof for Work done in the book.

Example 3.4

Example 3.5

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Fig 5: Reversible adiabatic processes a perfect gas on a p-v diagram

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Polytropic Processes

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The Polytropic process is one in which the pressure-volume relation is given as

pvn = constant

The exponent n may have any value from minus infinity to plus infinity depending on the process.

Some of the more common values are given below;

Process Exponent n

Constant pressure 0

Constant volume ∞

Isothermal & ideal gas 1

Adiabatic & ideal gas k = CP/CV

We know that, for any reversible process:

W = – 𝒑 𝒅𝒗

For the process in which, pvn = constant, we have p = c/vn , c is constant.

W = – 𝒄 𝒗1

𝒗2 𝒅𝒗

vn= – 𝒄 [

v – n+1

– 𝒏+𝟏] 𝑣2𝑣1

= – 𝒄 [V2

–n +1–V1–n +1

– 𝒏+𝟏]

=

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Polytropic Processes (cont’d)

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Since the constant c can be written as p1 𝒗 1n = p2 𝒗 2

n :

Work input = p2 𝒗 2 – p1 𝒗 1

𝒏 – 𝟏

The above equation is true for any working substance undergoing a reversible polytropic process.

We can also write as𝑷𝟏

𝑷𝟐

= (𝒗𝟐

𝒗𝟏) 𝒏

Example 3.6

Example 3.7

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For an ideal gas under going a polytropic process the boundary work is


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Questions ?

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