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
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
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
9.4.5 Reversible Adiabatic Steady Flow Process
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
=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.
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
220.127.116.11 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.
18.104.22.168 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
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.
22.214.171.124 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.