16.1. TWO IMPORTANT ‘Tds’ THERMODYNAMIC RELATIONS
We have, 
or 
or 
In differential form, it can be written as 
or TdS = dU + pdV …………………..(16.1)
or Tds = du + pdv ( for unit mass )
The above equation is very important since it establishes a relationship between all thermodynamic properties and does not involve path functions like work and heat. It is interesting, therefore, to note that equation δQ = dU + pdV and δQ = TdS are true for reversible process but equation TdS = dU + pdV is true for all processes reversible as well as irreversible. This equation is true for any two equilibrium states of a system.
Since H = U + pV
It follows that dH = dU + pdV + Vdp
Substituting this relation in Equation (16.1)
Thus TdS = dH - Vdp ………………….(16.2)
or Tds = dh – vdp (for unit mass)
Equation (16.1) and (16.2) are two important ‘TdS’ thermodynamic relations.
16.2. ENTROPY CHANGE IN A PROCESS FOR IDEAL GAS:
Various processes are shown on p-v and T-s diagrams in Fig. 16.1. The entropy change in these processes is determined as follows.
Fig. 16.1. Polytropic processes on P-v and T-s diagrams.
16.2.1. Constant volume process (Isochoric process) (Fig. 16.2):
V1 = V2 = V From (16.1), we have TdS = dU + pdV But dV = 0 and dU = m Cv dT; therefore TdS = m Cv dT + 0 = m Cv dT
or S2 – S1 = m Cv ln For unit mass, s2 – s1 = Cv ln |
Fig. 16.2. Constant volume process |
[ If Cv is constant for the gas]
16.2.2. Constant pressure process (Isobaric process) (Fig. 16.3):
P1 = P2 = P From Equation (16.2), we have TdS = dH - Vdp But dP = 0 and dH = mCpdT; therefore TdS = m Cp dT + 0 = m CpdT or S2 – S1 = m Cp ln For unit mass, s2 – s1 = Cp ln |
Fig. 16.3 Constant pressure process |
[ If Cp is constant for the gas]
16.2.3. Constant temperature process (Isothermal process) (Fig. 16.4):
PV = C = mRT or T1 = T2 = T From Equation (16.1), we have TdS = dU + pdV But dU = m Cv dT =0 as dT = 0; therefore TdS = 0 + pdV = pdV
or Using perfect gas relation
or S2 – S1 = m R ln For unit mass, or s2 – s1 = R ln |
Fig. 16.4. Constant temperature process |
, we can write
16.2.4. Reversible and an irreversible adiabatic process (isentropic process)(Fig. 16.5):
For reversible adiabatic process, we have,
But δQ = 0 (for adiabatic process) Therefore, S2 – S1 = 0 or S1 = S2 Therefore a reversible adiabatic process is an isentropic process For an irreversible adiabatic process, for unit mass, we have,
But δQ = 0 (for adiabatic process) Therefore, S2 – S1 > 0 or S2 > S1 |
Fig. 16.5. Adiabatic process |
16.2.5. Polytropic process:
we have, 
But for a gas, 
. Using this in the above equation, we get
S2 – S1 = 
For unit mass,
s2 – s1 = 
16.2.6. Equal-internal energy process:
U1 = U2 = U
From Equation (16.1), we have TdS = dU + pdV
But dU = 0; therefore TdS = 0 + pdV = pdV
dS = 
or 
or S2 – S1 = 
For unit mass,
or s2 – s1 = 
16.2.7. Change in Entropy for an Equal Enthalpy Process
H1 = H2 = H
From Equation (16.2), we have TdS = dH - Vdp
But dH = 0 ; therefore Tds = 0 - Vdp = - Vdp
dS = 
or 
or S2 – S1 = 
= 
For unit mass,
s2 – s1 = 
16.3. CARNOT CYCLE ON T-s DIAGRAM Let a heat engine operate on a Carnot cycle as discussed in ‘Fig. 14.11 of Lesson 14’. It works on four reversible processes as shown on T-S diagram in Fig. 16.6. It is clear from the T-S diagram that, The amount of heat transferred by the reservoir at temperature ‘TH’ to the working substance, 1Q2 = T1(S2 – S1) The amount of heat transferred by the working substance to the reservoir at temperature ‘TL’, 3Q4 = T3 (S4 – S3) = T3 (S1 – S2). or 3Q4 = - T3 (S2 – S1). |
|
Since the system undergoes a cycle, the net heat transfer should be equal to the net work done by the system.
or Net work done by the system = Net heat transfer by the system = 1Q2 + 3Q4
= T1 (S2 – S1) + [- T3 (S2 – S1)]
= [T1 (S2 – S1) - T3 (S2 – S1)]
Thermal efficiency of the engine working on cycle = 
16.4. SECOND LAW OF THERMODYNAMICS FOR A CONTROL VOLUME (C.V.)
The second law of thermodynamics can be applied to an open system (C.V.) by a procedure similar to the application of First law for open system. Figure 16.6 shows an open system.
Fig. 16.6. The general case for an open system
Case 1: For more than one inlet and one exit of control volume
For a flow process as shown in Figure 16.6, the entropy rate balance gives
………. (16.3)
In the tabove equation the equality holds for a reversible process and the inequality for an irreversible process.
and 
is rate of entropy change within the system (C.V).
and 
rate of entropy transfer in and out of the system (C.V) accompanying mass flow.
is rate of heat transfer during time dt at the location of boundary where the instantaneous surface temperature is T
represents the accompanying rate of entropy transfer.
sign for more than one inlet and one exit of control volume
For the steady state steady flow (SSSF) process, which is defined as a process in which the conditions within the control volume do not vary with time.
i.e. 
and 
For steady state steady flow process, the change in entropy from Equation (16.3) can be given by the following equation
Case2: For one inlet and one exit of control volume
Many applications in engineering problems involve one inlet and one exit of control volume. For such cases the steady state and steady flow entropy rate balance is given by
Example 16.1. 5 kg of oxygen is heated in a reversible non-flow constant volume process from temperature of 60°C until the pressure is doubled. If CP =0.913 kJ/kgK, determine, (i) Final temperature. (ii) work done, (iii) change in internal energy. (iv) heat transferred, (v) change in enthalpy and (vi) change in entropy. Assume CV= 0.653 kJ/kgK,
Solution:
Given: m = 5 kg; T1= 60 °C = 60 + 273 = 333 K; p2 = 2p1 CP =0.913 kJ/kgK; dV=0
(i) Determine the final temperature, T2;
Formula: For constant volume process, V1 = V2, we have
T2 = 
T1
Answer: T2 = 
x T1= 2 x 333 = 666 K or 393°C
(ii) Determine the work done, 1W2;
Formula: work done, 1W2 = ∫pdV
Answer: 1W2 = ∫pdV= 0 K or 0°C
(iii) Determine the change in internal energy, (U2 - U1);
Formula: U2- U1= m CV (T2 -T1)
Answer: U2 - U1= m CV (T2 -T1) = 5 x 0.653 x (666-333) = 1087 kJ
(iv) Determine the heat transfer, 1Q2;
Formula: 1Q2= (U2- U1)+1W2
Answer: 1Q2= (U2- U1)+1W2= 1087 + 0 = 1087 kJ
(v) Determine the change in enthalpy, (H2 - H1);
Formula: H2-H1= m CP(T2 - T1)
Answer: H2-H1= m CP(T2 - T1) = 5 x0.913 x (666 - 333) = 1520 kJ
(vi) Determine the change in entropy, (S2 - S1);
Formula: S2-S1 = 
Answer: 
= 2.263 kJ/K
Example 16.2: An insulated cylinder contains 25 kg of nitrogen. Its volume capacity is 5 m3. Paddle work is done on the gas by stirring it till the pressure in the vessel gets increased from 5 bar to 10 bar. Determine (i) change in internal energy. (ii) work done (iii) heat transferred and (iv) change in entropy. Assume Cp =1.04 kJ/kgK and CV = 0.7432 kJ/kgK for nitrogen.
Solution:
Given: m = 25 kg;p1= 5 bar = 5 x102kN/m2 ; p2= 10 bar = 10 x102 kN/m2
V1 = V2 = 5m3and it is constant for both end states.
R = Cp - CV = 1.04 - 0.7432 = 0.2968 kJ/kgK. Insulated cylinder, 1Q2= 0
(i) Determine the change in internal energy, (U2 - U1);
Formula: (U2 -U1) = m CV (T2-T1) = CV (mT2 - mT1)
Finding unknown, mT1 and mT2 ;
The mass of the gas in the cylinder is given by
Therefore, mT1 = 
= 8.423 x103 kgK
and mT2 = 
= 16.846 x103 kgK
Answer: U2 - U1= CV(mT2 − mT1) = 0.7432 x (16.346 − 8.423)x103= 6260 kJ
(ii) Determine work done, 1W2 ;
Formula: 1Q2 = U2 − U1+ 1W2
therefore 1W2= − (U2 − U1) = − 6260 kJ
Answer: 1W2= − (U2 − U1) = − 6260 kJ
(iii) Determine heat transfer, 1Q2 ;
Answer: 1Q2 = 0 kJ
(iv) Determine the change in entropy, (S2 - S1);
Formula: For constant volume process, 
Finding unknown, 
;
For constant volume, 
= 2
Answer: S2-S1 = 
= 12.878 kJ/kg K
Example 16.3: Nitrogen is heated in a steady flow process from 700 kPa and 27°Cto 620 kPa and927oC. Derive an expression for the change in entropy and calculate the change of entropy during the process if CP = 39.65 – 7.65x103/T + 1.5x106/T2 kJ/kg-mole-K and the value of R is 0.3026 kJ/kgK. Assume molecular weight of nitrogen as 28.016.
Solution:
Given: p1= 700 kPa; p2= 620 kPa; T1 = 27+273 = 300 K; T2 = 927 + 273 = 1200 K
CP = 39.65 – 7.65x103/T + 1.5x106/T2 ; R= 0.3026 kJ/kgK; Mol. wt. = 28.016 kg/mol
Determine the change in entropy:
Formula: We know Tds = du + pdv = d(h – pv) + pdv = dh – vdp
or 
Answer: (S2-S1) 
= 1.5215 kJ/kg K
Example 16.4: 5 kg of air is compressed in a reversible steady flow polytropic process from 100 kpa and 40°Cto 1000 kpa and during this process the law followed by the gas is pV1.25= C. Determine the change in entropy CV = 0.717 kJ/kgK , R = 0.287 kJ/kgK.
Solution:
Given: m = 5 kg; Ti =40 °C= 40 + 273 = 313 K; pV1.25= C
Pi = 100 kPa = 1 x 105 N/m2 ;Pe= 1000 kPa = 1 x 106 N/m2;
R =CP - CV or CP = R + CV or CP = 0.287 + 0.717 = 1.005 kJ/kgK
Determine the change in entropy:
Formula: For unit mass, Tds = dh – vdp
TdS = dH – Vdp or 
or 
Finding unknown, Te ;
Equation of state, 
Therefore, 
or Te = 
= 496 K
Answer: (Se-Si) = 
= − 0.9887 kJ/K
Example 16.5: Air is throttled from 1000 kPa at 27°Cto 100 kPa in a steady flow adiabatic process. Changes in kinetic and potential energies are negligible. Determine the change in entropy per kg of air flow. CP= l.004 kJ/kgK, R =0.287 kJ/kgK.
Solution:
Given: Ti=27 °C = 27 + 273 = 300 K; Pi = 1000 kPa; Pe= 100 kPa
R =0.287 kJ/kgK CP= l.004 kJ/kgK ΔPE = 0 ΔKE = 0
Determine the change in entropy per kg of air flow:
Formula: In steady flow energy equation for this process (throttling), we have
hi = he or he − hi = 0 ;
or CP(Te-Ti) = 0 or Te = Ti
Now we have, Tds = dh – vdp = − vdp ;
or 
Answer: (Se-Si) = 
= 0.6608 kJ/kgK
Example 16.6: A gas undergoes a non-flow process according to the law p = (0.16/V + 2.1) bar, where V is the volume in cubic meters. Initial volume is 0.6 m3 and the final volume is 0.2 m3. Determine the change in enthalpy during the process if 25 kJ of heat is rejected from the system.
Solution:
Given: V1= 0.6 m3; V2 = 0.2 m3; 1Q2 = - 25 KJ; p = (0.16/V + 2.1) bar
Determine the change in enthalpy(H2 – H1):
Formula: Equation for energy balance for closed system is given by
δQ = dU + pdV
As H =U+pV; Therefore δQ = d(H – pV) + pdV
or δQ =dH – pdV - Vdp + pdV
or δQ =dH − Vdp
Integrate, 
1Q2 = (H2 – H1) − 
or (H2 – H1) = 1Q2 + 
Finding unknown,
;
Now, p = 
bar; p = 
x 105 N/m2
dp = − 
= 17550 J = 17.55 kJ
Answer: (H2 – H1) = 1Q2 + 
= −25 + 17.55 = − 7.45 kJ
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