Module 4. Second law of thermodynamics

Lesson 9

GENERAL EXPRESSION FOR CHANGE IN ENTROPY, NUMERICAL PROBLEMS

9.1 General Expression for Change in Entropy of a Perfect Gas during a Non Flow process

We know that first law of thermodynamics applied to a non-flow reversible process is:

δQ = dU + δW

Or δQ = mC_v∆T + P.dV

Incremental change in Entropy during a reversible process is given as:  

Total change in entropy during process 1-2 will be

……… (Eq. 9.1)

 By equation 9.1, we can calculate change in entropy in terms of change in temperature and change in volume during a non flow reversible process 1-2.

 From the perfect gas equation, we know that

Putting it in eq 9.1

……… (Eq. 9.2)

By equation 9.2, we can calculate the change in entropy in terms of change in temperature and change in pressure during a non flow reversible process 1-2

……… (Eq. 9.3)

With the help of equation 9.3, we can calculate change in Entropy in terms of change in pressure and change in volume during a non-flow reversible process 1-2.

9.1.1  Change of entropy during a polytropic process

We know that a small heat exchange, δQ during a polytropic process is given as

……… (Eq. 9.4)

 This is the expression for change in entropy in terms of change in volume only

We know that during a polytropc process,

Putting this value in eq 9.4

;

……… (Eq. 9.5)

This is the expression for change in entropy in terms of change in temperature only.

  Putting these values

……… (Eq. 9.6)

This is the expression for change in entropy in terms of change in pressure only.

Putting these values

……… (Eq. 9.7)

9.2  Numerical Problems

9.2.1 A cyclic heat engine operates between a heat source temperature of 900^o C and heat sink temperature of 30^oC. Find the least rate of heat rejection per kW of net power output produced by the engine.

Solution

Given that

                Absolute temperature of source

                  T_h = 900 + 273 = 1173 K

                 Absolute temperature of heat sink,

                   T_c = 30 + 273 = 303 K

Now the least rate of heat rejection will only come in case of reversible heat engine, because it has the highest efficiency to convert heat from heat source into work.

Efficiency of reversible heat engine between the given source & sink temperatures is calculated as:

Thus heat absorbed from heat source for producing 1 kW of power

Hence the least heat rejected will be equal to

Q_c = Q_h – W

                = 1.35 – 1

                = 0.35 kW = 350 W (Ans)

9.2.2 Air at 15^oC and 1.05 bar occupies 0.02 m³. The air is heated at constant volume until the pressure is 4.2 bar, and then cooled at constant pressure back to its original temperature. Calculate the net heat flow to or from the air and net entropy change.

Solution

The two processes are ‘constant volume heating’, and then ‘constant pressure cooling’ to the same initial temperature 15⁰C

Initial conditions of air are given as

                             Pressure, P₁ = 1.05 x 10⁵ Pa

                        Temperature T₁ = 15+273 = 288 K

                              Volume, V₁ = 0.02 m³

Conditions after constant volume heating are given as

                                             P₂ = 4.2 x 10⁵ Pa

                                            V₂ = V₁ = 0.02 m³

Final temperature,

T₂ can be calculated from gas law 

= 1152 K

Then the air undergoes the constant pressure cooling process 2-3 and conditions after that will be

P₃ = P₂ = 4.2 x 10⁵ Pa (Constant Pressure Process)

T₃ = T₁ = 288 K          (Given)

V₃ can be calculated from gas law

Mass of air can be calculated by the given conditions at any state as.

= 0.0254 kg

Net Heat Exchanged by air

Heat absorbed during constant volume heating

= mCv (T₂−T₁)

= .0254 x 717 x (1152-288)

= 15735 J

 Heat rejected during constant pressure cooling

= mCp (T₂−T₃)

= .0254 x 1005 x (1152-288)

= 22055 J

 Net heat exchange = Heat Rejected – Heat absorbed

= 22055 – 15735

= 6320 J

i.e.          Net heat rejected by air = 6320 J (Ans)

 Net Entropy Change of Air:-

Entropy change during Constant Volume process 1-2 is given as

Entropy change during constant pressure process 2-3 is given as

Net Entropy Change = S₃ – S₁

= .0254 (-0.399) = -0.0101 kJ/K= (Ans)

Negative sign means the entropy of air is decreasing.

9.2.3 Equal masses of a certain gas at same pressures but at different temperatures T₁ and T₂ (T₁> T₂) are mixed in a thermally insulated system. Find the net change in entropy of the Universe due to change in entropy of the system. Show that change is necessarily positive.

Solution

Let after mixing, the temperature of mixture is T_c

Now change in entropy of gas mass whose temperature has decreased from T₁ to T_c

……… [Sub Eq. (i)]

Change in entropy of gas mass whose temperature has increased from T₂ to T_c

……… [Sub Eq. (ii)]

So Net Change in Entropy of System

= ds₁ +ds₂

Also, we know that A.M (Arithmetic mean) is always greater then G.M (Geometric mean). So change is always +ve. Hence it is proved that change is necessarily positive.

9.2.4 0.2 kg of air at pressure = 1.5 bar and temperature = 300 K is compressed to a pressure of 15 bar according to the law PV^1.25 = Constant, Determine:

(i) Initial and final parameters of air, (ii) Work done on or by air (iii) Heat flow to or from the air, (iv) Change of entropy

Solution

m= 0.2 kg;  P₁ = 1.5 x 10⁵  Pa;  T₁= 300^o K;   P₂ = 15x10⁵ Pa;   n =1.25

(i) For a polytrophic process with value of n = 1.25,  

= 0.115 m³ (Ans)

Also

= 0.018 m³ Ans

(ii) Work exchange during a polytropic process is given as

= - 40285.6 J

= - 40.28 kJ (Ans)

 Negative value of work exchange shows that work is done on the gas.

(iii) Heat Flow during polytropic process

= -15 kJ (Ans)

Negative value of heat exchange shows that heat is rejected by the gas

(iv) Change in Entropy for a polytropic process

= -39.9 ≈ - 40J/^o K or -.04 kJ/^o K   (Decrease in Entropy)

                                                                           (Ans)

9.3  Unsolved Problems

9.3.1 A Carnot cycle operates between source and sink temperatures of 260⁰ C and – 17.8°C. If the system receives 100 kJ from the source, find (i) efficiency of the system, (ii) the net work transfer, (iii) heat rejected to the sink.        

[Ans. 52.2%; 52.2 kJ; 47.8 kJ]

9.3.2 Source A can supply energy at a rate of 11,000 kJ/min at 320⁰C. A second source B can supply energy at a rate 110,000 kJ/min at 68⁰C. Which source A or B, would you choose to supply energy to an ideal reversible engine that is to produce large amount of power if the temperature of the surroundings is 40⁰C?

[Ans. Source B]  

9.3.3 A heat engine is supplied with 278 kJ/s of heat at a constant fixed temperature of 283⁰C and the heat rejection takes place at 5⁰C. The following results were reported: (i) 208 kJ/s are rejected, (ii) 139 kJ/s are rejected, (ii) 70 kJ/s are rejected.

Classify which of the results report a reversible cycle or irreversible cycle or impossible results.

[Ans. (i) Irreversible (ii) Reversible (iii) Impossible]   

9.3.4 0.03m³ of nitrogen contained in a cylinder behind a piston is initially at 1.05 bar and 15⁰C. The gas is compressed isothermally and reversibly until the pressure is 4.2 bar. Calculate the change of entropy, the heat flow, and the work done, and sketch the process on P-v and T-s diagrams. Assume nitrogen to act as a perfect gas. Molecular weight of nitrogen = 28.

[Ans. 0.01516 kJ/K (decrease) ; 4.37 kJ (heat rejected) ; 4.37 kJ]

9.3.5 0.05 kg of carbon dioxide (molecular weight = 44) is compressed from 1 bar and 15⁰ C, until the pressure is 8.3 bar, and the volume is then 0.004 m³. Calculate the change of entropy. Take c_p for carbon dioxide as 0.88 kJ/kg K, and assume carbon dioxide to be a perfect gas.

[Ans. 0.0113 kJ/K (decrease)]

9.3.6 A rigid cylinder containing 0.006 m³ of nitrogen (molecular weight 28) at 1.04 bar, 15⁰C, is heated reversibly until the temperature is 90⁰ C. Calculate the change of entropy and the heat supplied. Sketch the process on T-s diagram. Take the isentropic index, γ, for nitrogen as 1.4, and assume that nitrogen is a perfect gas.

[Ans. 0.00125 kJ/K; 0.407 kJ]

9.3.7 1 kg of air is allowed to expand reversibly in a cylinder behind a piston in such a way that the temperature remains constant at 260⁰ C while the volume is doubled. The piston is then moved in, and heat is rejected by the air reversibly at constant pressure until the volume is the same as it was initially. Calculate the net heat flow and the overall change of entropy. Sketch the processes on T-s diagram.

[Ans. -161.9 kJ/kg; -0.497 kJ/kg K]

9.3.8 A quantity of gas (mean molecular weight 36.2) is compressed according to the law pvⁿ = constant, the initial pressure and volume being 1.03 bar and 0.98 m³ respectively. The temperature at the start of compression is 17⁰C and at the end it is 115⁰C. The amount of heat rejected during compression is 3.78 kJ, c_p = 0.92. Calculate :

(i)Value of n,          (ii) Final pressure,           (iii) Change in entropy.

[Ans. (i) 1.33; (ii) 1.107 bar; (iii) 0.228 kJ/kg K]