# AUS: +61-280-07-5697

9.4.1 Finding out of Governing Equation for Reversible Adiabatic Process

We know dq= Tds = du + Pdv = cv dT + Pdv for adiabatic process dq = 0 i.e., ds = 0 thus

cudT = – Pdv …                                                         (9.21)

Again                                            dq = Tds = udh -vdp = cpdt – vdP

0 = cpdT – vdP

cpdT= vdP

Dividing equation (9.21) by equation (9.22), we get

cp dt / cu d = – udp / p du

or                     dp / p + cp / cu du / v = 0

or                            dp / p + γ du / u = 0

ʃ dp / p + ʃ du / v = in c where c is constant

in p + γ in u = in c

or                                  In ( pu γ) = In C

or                                         pu γ = C where    C = constant

For process 1 – 2 we can have

P1 u1 = p2 u2

P2 / p1 = ( u1/u2)γ

For ideal gas   pu = RT

or                                 p1 u1 / T1  = P2 U2 / T2

P2 / P1 = ( U1 / U2) .T2 / T1

(9.24)                 ( u2 / u1) . T2 / T1) =  ( u1 / u2) γ

or                                        T2 / T1 = ( u1 / u2) γ-1

again                             p1 u1 / T1 = P2u2 / T2

v1 / u2 = ( p2 / p1) x ( T1 / T2)

thus ( 9.25) =>                     T2 / T1= ( P2 / P1) γ-1 (T1 / T2)γ-1

( T2 / T1) 1+γ-1 = ( P2 / P1 )γ-1

T2 / T1 = ( P2 /P1) γ-1/γ

Equations (9.24), (9.25) and (9.26) will give the relation among P, v, and T in a reversible adiabatic process for an ideal gas.

9.4.2 The Internal Energy Change of an Ideal Gas for Reversible Adiabatic

Process

We have                      Tds = du + Pdv = 0

Or                               du  = – Pdv

= C / γ -1[ u2 – u1]

= 1 / γ-1 [ Cu2 – Cu1]

= 1 / γ-1 [ p2 u2 uγ-1 – p1 u1 u1-γ]

= 1 / γ-1 [ p2 u2 – p1 u1]

= R ( T2 –T1) / γ-1 = RT1 / γ-1 [ T2 / T1-1]

= RT1 / γ-1 [(P2/P1)γ-1/γ -1]

9.4.3 The Enthalpy Change of an Ideal Gas for a Reversible Adiabatic Process

Tds = dh –udp =0

dh = udp

9.4.4 The Work done by an Ideal Gas in a Reversible Adiabatic Process

We know                     dQ = dU +dW = 0

Or                                dW = -dU

i.e., work is done at the expense of the internal energy. If m is the mass of the gas.

Then :.             W1-2=U1-U2=m(u1-u2)·

= m( p1 u1 – p2 u2) / γ -1 = mr ( T1 –T2) / γ -1

= m R T1 / γ – 1 [ 1 – ( P2 /P1) γ-1 / γ

From equation (5.9) in Chapter 5, we have

= cp ( T1 – T2) = γ R / γ -1 ( T1 – T2)

= γ / γ -1 ( P1 U1 – P2 u2) =  γ / γ-1 p1 u1 [ 1 – ( p2 / p1) γ-1 /γ]

If kinetic and potential energy neglected, then

w 1-2 = γ p1 u1 / γ-1 [ 1 –( p2 / p1) γ -1 / γ

9.4.6 Reversible Isothermal Process

If ideal gas changes its state from 1 to 2 isothermally, the work done is given by

= mRT In V2 / V1

W1-2 = mRT In V2 / V1 = mRT In P1 / P2

The heat transfer Q1- 2 is given by 1st law,

O1-2 = U2-Ul+W1-2

= O+W1-2

=mRT In V2 /V1 = mRT  In P1 / P2

Q1-2 = T ( S2 –S1)

[ S2 –S1 = entropy change (not specific entropy change)].

9.4.7 Polytropic Process

In practice Pv” = constant is an equation of polytropic process. Here n is the index of a

process. This equation is an equation of many processes depending upon the value of n.

γ is a property of the gas but n is not.

Pv” =constant

For two states 1 – 2, P1 v]’ = P2 v2′

Or                          In P1 + n In v1 = In P2 + n In v2

n = ln p1 – ln p2 / ln u2 – ln u1

Thus n can be found out from P1, P2, v1 and v2 values.

Again              P1 v1n = P2 vn2

T2 / T1 = ( u1 / u2) n-1

T2 / T1= ( P2 /P1) n-1 /n

9.4.7.1 Entropy Change in a Polytropic Process

In a reversible adiabatic process dS = 0. But in a polytropic process the entropy changes.

From equation (9.18) we can write

s2 – s1 = cu in T2 / T1 + R In u2 /u1

= r / γ-1 in T2 / T1 + R / n-1 in T1 / T2

= (n-γ) r / (γ-1) (n-1) in T2 / T1

Similarly         s2 – s1 = ( n – γ / (γ-1) (n-1)  R In P1/ P2)

And                      s2 –s1 =  n – γ / (γ-1) (n-1)  R In u1 / u2

From equations (9.36), (9.37) and (9.38), if n = y is put then s2 – s1 = 0 i.e., the specific

entropy change becomes zero.

For                  n < γ, s1 > s2, i.e., specific entropy of the gas decreases .

and for                        n > γ, s2 > s1, i.e., specific entropy of the gas increases.

If the gas is cooled, the entropy decrease is possible, and if reversible heat transfer

takes place, the increase of entropy is possible.

9.4.7.2 Heat and Work in a Polytropic Process

From 1st law

q1-2 = u2 – u1 +W1 -2

ql-2 – W1-2 = u2 – u1 = cv (T2– T1)

= r / γ -1 ( t2 – t1) = p2 u2 – p1 u1 / γ -1

For steady flow process, we have

q1- 2 – W1 – 2 = h2– h1

= cp(T2-T1)

γR / γ -1 ( T1-t2)

= γ / γ -1 ( p2 u2 – p1 u1)

For a polytropic process (K.E. and P.E. neglected)

Equations (9.39) and (9.41) are used to determine heat and work quantities for a

closed and a steady flow system respectively.

9.4.7.3 Polytropic Specific Heat

For                                          pv n = constant

Now                            Tds = du + pdu

Generally heat and work both are involved in a polytropic process. To calculate heat transfer during the polytropic process, it is first to evaluate the work via either ʃ Pdv or ʃ v d p for closed or open steady flow process respectively. By applying 1st law, we will then get heat transfer.