**9.4.1 Finding out of Governing Equation for Reversible Adiabatic Process**

We know dq= Tds = du + Pdv = c_{v} dT + Pdv for adiabatic process dq = 0 i.e., ds = 0 thus

c_{u}dT = – Pdv … (9.21)

Again dq = Tds = udh -vdp = c_{p}dt – vdP

0 = c_{p}dT – vdP

c_{p}dT= vdP

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

cp dt / cu d = – udp / p du

or dp / p + c_{p} / c_{u} 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

P_{1} u_{1} = p_{2} u_{2}

P_{2} / p_{1} = ( u_{1}/u_{2})^{γ}

For ideal gas pu = RT

or p_{1} u_{1} / T_{1} = P_{2 }U_{2} / T_{2}

P_{2} / P_{1} = ( U_{1} / U_{2}) .T_{2} / T_{1}

(9.24) ( u_{2} / u_{1}) . T_{2} / T_{1}) = ( u_{1} / u_{2}) γ

or T_{2} / T_{1} = ( u_{1} / u_{2}) γ-1

again p_{1} u_{1} / T_{1} = P_{2}u_{2} / T_{2}

v_{1} / u_{2} = ( p_{2} / p_{1}) x ( T_{1} / T_{2})

thus ( 9.25) => T_{2} / T_{1}= ( P_{2} / P_{1}) γ-1 (T_{1} / T_{2})^{γ-1}

( T_{2} / T_{1}) 1+γ-1 = ( P_{2} / P_{1} )^{γ-1}

T_{2} / T_{1} = ( 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[ u_{2} – u_{1}]

= 1 / γ-1 [ Cu_{2} – Cu_{1}]

= 1 / γ-1 [ p_{2} u_{2} u^{γ}-1 – p_{1} u_{1} u^{1-γ}]

= 1 / γ-1 [ p_{2} u_{2 }– p_{1} u_{1}]

= R ( T_{2} –T_{1}) / γ-1 = RT_{1} / γ-1 [ T_{2} / T_{1}-1]

= RT_{1} / γ-1 [(P_{2}/P_{1})^{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

**9.4.5 Reversible Adiabatic Steady Flow Process**

From equation (5.9) in Chapter 5, we have

= c_{p }( T_{1} – T_{2}) = γ R / γ -1 ( T_{1} – T_{2})

= γ / γ -1 ( P_{1} U_{1} – P_{2} u_{2}) = γ / γ-1 p_{1 }u_{1} [ 1 – ( p_{2} / p_{1}) γ-1 /γ]

If kinetic and potential energy neglected, then

w _{1-2 }= γ p_{1} u_{1} / γ-1 [ 1 –( p_{2} / p_{1}) ^{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 V_{2} / V_{1}

W_{1-2} = mRT In V_{2} / V_{1 }= mRT In P_{1} / P_{2}

The heat transfer Q_{1- 2} is given by 1st law,

O_{1-2} = U_{2}-U_{l}+W_{1-2}

= O+W_{1-2}

=mRT In V_{2} /V_{1} = mRT In P_{1} / P_{2}

Q_{1-2} = T ( S_{2 }–S_{1})

_{2 }–S

_{1}= 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, P_{1} v]’ = P_{2} v_{2′}

Or In P_{1} + n In v_{1} = In P_{2} + n In v_{2}

n = ln p_{1} – ln p_{2} / ln u_{2} – ln u_{1}

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

Again P_{1} v_{1}^{n }= P_{2} v^{n}_{2}

T_{2} / T_{1} = ( u_{1} / u_{2}) ^{n-1}

T_{2} / T_{1}= ( P_{2} /P_{1}) ^{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

s_{2} – s_{1} = c_{u} in T_{2} / T_{1} + R In u_{2} /u_{1}

= r / γ-1 in T_{2} / T_{1} + R / n-1 in T_{1} / T_{2}

= (n-γ) r / (γ-1) (n-1) in T_{2} / T_{1}

Similarly s_{2} – s_{1} = ( n – γ / (γ-1) (n-1) R In P_{1}/ P_{2})

And s_{2} –s_{1} = n – γ / (γ-1) (n-1) R In u_{1} / u_{2}

_{ }

From equations (9.36), (9.37) and (9.38), if n = y is put then s_{2 }– s_{1} = 0 i.e., the specific

entropy change becomes zero.

For n < γ, s_{1} > s_{2}, i.e., specific entropy of the gas decreases .

and for n > γ, s_{2} > s_{1}, 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

q_{1-2} = u_{2} – u_{1} +W_{1 -2}

q_{l-2} – W_{1-}2 = u_{2} – u_{1} = c_{v} (T_{2}– T_{1})

= r / γ -1 ( t_{2} – t_{1}) = p_{2} u_{2} – p_{1} u_{1} / γ -1

For steady flow process, we have

q_{1- 2} – W_{1 – 2 }= h_{2}– h_{1}

= cp(T_{2}-T_{1})

γR / γ -1 ( T_{1}-t_{2})

= γ / γ -1 ( p_{2} u_{2} – p_{1} u_{1})

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.