ENGINEERING THERMODYNAMICS STUDY MATERIAL

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 1 THERMODYNAMIC CONCEPTS: 1. Define: (i) Reversible Process (iv) Intensive and Extensive properties (ii) System and Surroundings (v) Work done by a system. (iii) State of a System [DEC 2005] 10 Marks (i) REVERSIBLE PROCESS: The process is said to be reversible when the surrounding and system both are restored to their initial states at the end of the process. • The condition for reversible process is a system must be in thermodynamic equilibrium at all states. • The change in state of system should be finite means there should not be any spontaneous change in state from the equilibrium. • Example – Frictionless adiabatic expansion or compression of gas. Frictionless isothermal expansion or compression of gas. V O G C E (ii) SYSTEM AND SURROUNDINGS: System The system is defined as a prescribed region of space or finite quantity of matter surrounded by an envelope called boundary. • The boundary may be real physical surface or it may be some imaginary surface. The boundary may be fixed or it may be moving. • In a broad sense, the system is defined as a specified region where changes due to transfer of mass or energy (i.e. heat and work) or both to be studied. • In a system it is not necessary that the volume or shape should remain fixed. Prof.S Venkatesh Rao 1

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Surrounding The space and matter external to the thermodynamic system and outside boundary is called surrounding. Universe When system and surrounding are put together, it is called universe. (iii) STATE OF A SYSTEM: The state is exact condition of the system. State is condition of system at particular instant of time, which can be described by properties such as pressure, temperature, volume, etc. • To define the state at least two properties are necessary. • Even if one property is changed, the state will change V O G C E (iv) INTENSIVE AND EXTENSIVE PROPERTIES: (a) Intensive property The properties, which do not depend upon the mass of the system, are known as intensive property. • Example –Pressure, Temperature, Density, etc. (b) Extensive property The properties, which depend on the mass of the system, are known as extensive property. • Example – Enthalpy, Entropy, Total Volume etc. (v) WORK DONE BY A SYSTEM. In order to include all thermodynamic processes, work is said to be done by a system when the sole effect external to the system can be reduced to the raising of a mass against gravity. OR Prof.S Venkatesh Rao 2

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Work may be defined as energy transferred, without transfer of mass, across the boundary of a system because of an intensive property difference other than temperature, existing between the system and the surroundings. 2. Define Point function and Path function with example. [MAY 2006, MAY 2007, MAY 2008] 3 Marks Point function Point function is defined as a quantity, whose valve does not depends on the particular path followed during a process, but depend only on end states • The each thermodynamic property has single value at each state i.e. properties of system depends upon state of system. • Therefore, all thermodynamic properties are called as point function V O G C E Path function Path function is defined as a quantity, whose valve depends on the particular path followed during a process • Consider process 1.2 as shown in figure • System is undergoing a change from state 1 to state 2. • This change depends on intermediate states through which system changes its condition from point 1 to 2. • System can change its state by two paths. i.e. Path A and Path B as shown in figure • The work transfer can be represented by area under P.V. diagram. • Therefore work transfer will be different for path A and path B. • Thus work transfer depends upon the path of system and is called path function Prof.S Venkatesh Rao 3

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 3. State the Zeroth law of thermodynamics. What is its significance? [DEC 2007] 4 marks Zeroth Law of thermodynamics states “If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.” • Let A, B and C A B AB are three bodies at different C C temperature. • If A and C are brought into good contact, energy in the form of heat will V O G C E transfer, from a body at higher temperature to a body at lower temperature. • After some time they will be in thermal equilibrium. • Now if B and C are brought in good contact, after some time they will also be in thermal equilibrium. • Then A and B are also in thermal equilibrium. • It helps to measure the temperature of any body 4. Explain with diagram quasi—static process. [DEC 2008] 4 marks Let us consider a system of gas contained in a cylinder (Figure 1). The system initially is in equilibrium state, represented by the properties p1, v1, t1. The weight on the piston just balances the upward force exerted by the gas. If the weight is removed, there will be an unbalanced force between the system and the surroundings, and under gas pressure, the piston will move up till it hits Figure 1:non quasi static process the stops. The system again comes to an equilibrium state, being described by the properties p2, v2, t2. But the intermediate states passed through by the system are none equilibrium states, which cannot be described by thermodynamic coordinates. Prof.S Venkatesh Rao 4

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Figure 2 shows points 1 and 2 as the initial and final equilibrium states jointed by a dotted line, which has got no meaning otherwise. Figure 2:p-v diagram of non quasi-V static processO G Now if the single weight on the piston is made up ofC many very small pieces of weights (Figure 3) andE these weights are removed one by one very slowly from the top of the piston, at any instant of the upward travel of the piston. If the gas system is isolated, the departure of the state of the system from the thermodynamic equilibrium state will be infinitesimally small. So every state passed through by the system will be an equilibrium state. Such a process, which is but a locus of all the equilibrium points passed through by the system, is known as quasi-static process (Figure 4) • ‘Quasi’ meaning ‘almost’. • Infinite slowness is the characteristic feature of a quasi-static process. • A quasi-static process is thus a succession of equilibrium states. • A quasi-static process is also called a reversible process. Prof.S Venkatesh Rao 5

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 5. Represent the following processes on P-V and T-S diagram starting from the same point. (i) Isentropic process (ii) Isobaric process (iii) lsochoric process (iv) Isothermal process (v) Polytropic process. [DEC 2011] 5 Marks V O G C E 6. Define the following terms: (i) System [DEC 2011] 10 Marks (ii) Quasi-static process (iii) Reversible process (iv) Thermodynamic work (v) Thermodynamic equilibrium. (i) System The system is defined as a prescribed region of space or finite quantity of matter surrounded by an envelope called boundary (ii) Quasi-static process A process, which is but a locus of all the equilibrium points passed through by the system, is known as quasi-static process Prof.S Venkatesh Rao 6

Vishwatmak Om Gurudev College Of Engineering Thermodynamics (iii) Reversible process A reversible process (also sometimes known as quasi-static process) is one which can be stopped at any stage and reversed so that the system and surroundings are exactly restored to their initial states. (iv) Thermodynamic work In order to include all thermodynamic processes, work is said to be done by a system when the sole effect external to the system can be reduced to the raising of a mass against gravity. (v) Thermodynamic equilibrium. If no spontaneous change in the properties of the system can occur even after a catalytic or other such small disturbance, then the system is said to be in thermodynamic equilibrium. V O G C E Prof.S Venkatesh Rao 7

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. Thermodynamics is an axiomatic science which deals with the relations among heat, work and properties of systems which are in equilibrium. It basically entails four laws or axioms known as Zeroth, First, Second and Third law of thermodynamics. 2. A system is a finite quantity of matter or a prescribed region of space. A system may be a closed, open or isolated system. 3. A phase is a quantity of matter which is homogeneous throughout in chemical composition and physical structure. V O G C E 4. A homogeneous system is one which consists of a single phase. 5. A heterogeneous system is one which consists of two or more phases. 6. A pure substance is one that has a homogeneous and invariable chemical composition even though there is a change of phase. 7. A system is in thermodynamic equilibrium if temperature and pressure at all points are same ; there should be no velocity gradient. 8. A property of a system is a characteristic of the system which depends upon its state, but not upon how the state is reached. Intensive properties do not depend on the mass of the system. Extensive properties depend on the mass of the system. 9. State is the condition of the system at an instant of time as described or measured by its properties. Or each unique condition of a system is called a state. 10. A process occurs when the system undergoes a change in state or an energy transfer takes place at a steady state. Prof.S Venkatesh Rao 8

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 11. Any process or series of processes whose end states are identical is termed a cycle. 12. The pressure of a system is the force exerted by the system on unit area of boundaries. Vacuum is defined as the absence of pressure. 13. A reversible process is one which can be stopped at any stage and reversed so that the system and surroundings are exactly restored to their initial states. An irreversible process is one in which heat is transferred through a finite temperature. 14. Zeroth law of thermodynamics states that if two systems are each equal in temperature to a third, they are equal in temperature to each other. V O G C E 15. Infinite slowness is the characteristic feature of a quasi-static process. A quasi-static process is a succession of equilibrium states. It is also called a reversible process. Prof.S Venkatesh Rao 9

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 NIL DEC 2005 Q1 (a) Fill in the blanks: (i) To describe a state at least two ______________ properties are required. 1 Q 1(c) Define: (i) Reversible Process (iv) Intensive and Extensive properties (ii) System and Surroundings (v) Work done by a system. V(iii) State of a System 10 O G C E MAY 2006 Q 1 (a) Define Point function and Path function with example. 3 DEC 2006 Q 1 (a)Explain the term: Irreversibility. Name the factors that render processes irreversible. 4 Q1 (b) Explain the terms Path and point functions with suitable examples. 4 MAY 2007 Q 1(a) Explain the terms Path and point functions with suitable examples. 4 DEC 2007 Q 1(a) State the Zeroth law of thermodynamics. What is its significance? 4 MAY 2008 Q 1(a) Define point function and path function with example. 3 Q 7 Write short notes on (d) Classical and statistical approach of thermodynamics. 5 DEC 2008 (RC) 10 Q 1(d) Explain with diagram quasi—static process 4 Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2009 (RC) NIL DEC 2009 (RC) NIL MAY 2010 (RC) Q1(a) Explain intensive and extensive properties with examples. 4 DEC 2010 (RC) NIL V O G C E MAY 2011(RC) NIL DEC 2011 (RC) Q 1 (b) Represant the following processes on P-V and T-S diagram starting from the same point. (i) Isentropic process (ii) Isobaric process (iii) lsochoric process (iv) Isothermal process (v) Polytropic process. 5 Q 1 (d) Define thermodynamic property. List the types of property giving three examples of each. 5 Q 2 (b) Define the following terms : (i) System (ii) Quasi-static process (iii) Reversible process (iv) Thermodynamic work (v) Thermodynamic equilibrium. 10 MAY 2012(RC) Q 1 (a) Define a thermodynamic system. Differentiate between open system, closed system and isolated system. 4 Prof.S Venkatesh Rao 11

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 2 FIRST LAW OF THERMODYNAMICS 1. State and explain first law of thermodynamic for closed system. [DEC 2009] 4 Marks OR State the first law of thermodynamics and prove that for non-flow process it leads to Q=W+∆U [DEC 2008] 4 Marks OR State the first law of thermodynamics for closed system undergoing a cycle. [MAY 2007] 4 Marks OR State and explain first law of thermodynamics for a closed system undergoing a process. [MAY 2009] 4 Marks First law of thermodynamics for closed system undergoing a cycle “When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the surroundings is equal to net work done by the system on its surroundings” Mathematically, V O G C E The symbol (integral symbol with a circle in the middle) is used to indicate that the integration is to be performed over the entire cycle. or ∮ (������������������������ − ������������������������) = ������������ … . ������������������������������������ ������������ ������������������������������������������������������������������������ ������������������������������������������������������������������������ Prof.S Venkatesh Rao 12

Vishwatmak Om Gurudev College Of Engineering Thermodynamics First law of thermodynamics for a closed system undergoing a process. When a process is executed by a system, the change in stored energy of the system is numerically equal to the net heat interactions minus the net work interaction during the process. ∴ E2 – E1 = Q – W ∴ ΔE = Q – W or [ Q = Δ E + W ] Where E represents the total stored energy. If the electric, magnetic and chemical energies are absent and changes in potential and kinetic energy for a closed system are neglected, the above equation can be written as ������������ − ������������ = ∆ ������������ = ������������������������ − ������������������������ Where U represents the total Internal energy. Or ������������ = ������������ + ∆������������ V O G C E 2. Show that internal energy is property of system [DEC 2009, MAY 2011] 4 Marks • The internal energy (U) of the system is a property as it is dependent only on its and state and not on the path followed by the process. • This can be proved considering the process shown in figure. • Consider the initial condition of the system a represented by the point 1. A cycle may be executed either following the path 1-A-2- C-1 or 1-A-2-B-1. equation • Applying the ∫ δQ = ∫ δW to the above system in steps, we can write for the cycle 1-A-2-B-1 followed by the system. Prof.S Venkatesh Rao 13

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ∫ ∫2viaA (δQ − δW ) + IviaB (δQ - δW ) = 0 .........(1) 12 And for the path followed by the system 1 - A - 2 - C - 1 ∫ ∫2viaA (δQ − δW ) + IviaC (δQ - δW ) = 0 .........(2) 12 From equation 1 and 2 we get ∫ ∫2viaA (δQ − dW ) = IviaC (δQ - δW ) .........(3) 12 • It is obvious from the equation 11, that the value of the integral ∫ ( )1 δQ - δW 2 remains the same irrespective of the path followed by the system to reach from state 2 to state 1. • But the change of (δQ – δW) during the execution of the process represents ∆U as per first law of thermodynamics , therefore the internal energy is a point function and not a path function and it represents a property of the system. V O G C E 3. Derive expression for work done in a reversible polytropic process executed by a closed system. [MAY 2011] 4 marks We know that for any reversible process, For a process in pvn = constant, we have (Since the constant C, can be written as ������������1������������1������������ ������������������������ ������������2������������2������������) 14 or Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 4. Prove that heat absorbed or rejected during a polytropic process for ideal gas is given by [MAY 2006] 6 Marks OR Develop the following expression for the heat transfer from a mass of gas undergoing reversible expansion process obeying the polytropic law, pVn = constant V O G C E [DEC 2009] 8 Marks OR Prove that heat absorbed or rejected during a polytropic process for ideal gas is given by — [DEC 2007, MAY 2008] 6 Marks Using non-flow energy equation, the heat flow/transfer during the process can be found, Prof.S Venkatesh Rao 15

Vishwatmak Om Gurudev College Of Engineering Thermodynamics ∴ ������������ = ������������−������������ × ������������������������ (������������1 − ������������2) ������������������������������������������������������������ ������������������������ = ������������ ������������−1 ������������−1 V O G C E 5. Write a generalized form of the first law of thermodynamics for an open system. Explain the terms. Reduce this equation for SSSF process [MAY 2005] 5 Marks OR State Steady Flow Energy Equation. How is it applied to nozzle? [DEC 2007] 4 Marks OR Explain steady flow energy equation. Apply it to nozzle, turbine and compressor. [DEC 2007] 5 Marks The first law of thermodynamics applied to steady flow processes gives us steady flow energy equation (S.F.E.E) which is mathematical statement of conservation of energy that includes all energy terms associated with mass flow As per law of conservation of energy Total energy at inlet + Heat interaction = Total energy at outlet + Work interaction Prof.S Venkatesh Rao 16

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Using suffix ‘1’ for the energies of the entering section 1-1 and suffix ‘2’ for energies at the leaving section 2-2. We have Mathematically ( )• • • Q− W = m ∆ pe + ∆ ke + ∆ u + ∆ w f kJ/s ……….on time basis (1) V OWhere G C E ( )q − w = ∆ pe + ∆ ke + ∆ u + ∆ w f kJ/kg ……….on mass basis (2) ������������, ������������ = ℎ������������������������������������ ������������������������������������ ������������������������������������������������ ������������������������������������������������������������������������������������������������ ������������������������ ������������������������/������������������������ Above two equations (1) & (2) are known as ‘Steady Flow Energy Equation’ (S.F.E.E.) Prof.S Venkatesh Rao 17

Vishwatmak Om Gurudev College Of Engineering Thermodynamics (S.F.E.E.) to Turbine In a steam or gas turbine steam or gas is passed through the turbine and part of its energy is converted into work in the turbine. This output of the turbine runs a generator to produce electricity as shown in Fig. The steam or gas leaves the turbine at lower pressure or temperature. Applying energy equation to the system. Here, Z1 = Z2 (i.e., Δ Z = 0) V O •• • ( ∆ ke + ∆ h) kJ/s G C − Q−W =m E The sign of Q is negative because heat is rejected (or comes out of the boundary). The sign of W is positive because work is done by the system (or work comes out of the boundary). (S.F.E.E.) to Compressor A centrifugal compressor compresses air and supplies the same at moderate pressure and in large quantity. Applying energy equation to the system. Here, Z1 = Z2 (i.e., Δ Z = 0) •  − •  = • ( ∆ ke + ∆ h) kJ/s − Q− W m  •• = • ( ∆ ke + ∆ h) kJ/s − Q+W m Prof.S Venkatesh Rao 18

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The Q is taken as negative as heat is lost from the system and W is taken as negative as work is supplied to the system. (S.F.E.E.) to Nozzle In case of a nozzle as the enthalpy of the fluid decreases and pressure drops simultaneously the flow of fluid is accelerated. This is generally used to convert the part of the energy of steam into kinetic energy of steam supplied to the turbine. For this system, V Δ PE = 0O G W=0 C E Q=0 Applying the energy equation to the system, ∆ ������������������������ + ∆ℎ = 0 ∆ ������������������������ = −∆ℎ = (ℎ1 − ℎ2) ������������21 − ������������12 = (ℎ1 − ℎ2) 2000 ������������2 = �������������12 + 2000(ℎ1 − ℎ2) 6. What is Joule-Thomson coefficient? What conclusion can be drawn from a given value of this coefficient? [MAY 2006, DEC 2006, MAY 2008, DEC 2009, MAY 2010] OR Explain in brief Joule Thompson Porous Plug Experiment. Draw an inversion curve and explain the significance of Joule Thompson Coefficient. [MAY 2011] 6 Marks OR Prof.S Venkatesh Rao 19

Vishwatmak Om Gurudev College Of Engineering Thermodynamics What is throttling process? Define Joules Thomson coefficient, inversion point and inversion curve. [MAY 2012] 6 Marks When a fluid passes through a restriction such as a porous plug, a capillary tube, or an ordinary valve, its pressure decreases and the process is called the throttling process. The porous plug is shown in Fig. 1. V O G C E Figure 1. The Joule-Thomson porous plug experiment. In this system, Q = 0 ( system is isolated) W = 0 ( there is no work interaction) ΔPE= 0 ( inlet and outlet are at the same level) ΔKE= 0 ( kinetic energy does not change significantly) Applying the energy equation to the system h1 = h2 This shows that enthalpy remains constant during adiabatic throttling process. Throttling process was investigated by Joule and Thompson (Lord Kelvin) in their famous porous plug experiment (Fig.1). A fluid at a fixed temperature and pressure T1 and P1 (thus fixed enthalpy) is forced to flow through a porous plug, and its temperature and pressure downstream (T2 and P2) are measured. Prof.S Venkatesh Rao 20

Vishwatmak Om Gurudev College Of Engineering Thermodynamics The experiment is repeated for different sizes of porous plugs, each giving a different set of T2 and P2. Plotting the temperatures against the pressures gives us an h= constant line on a T-P diagram, as shown in Fig 2. V O G C E Repeating the experiment for different sets of inlet pressure and temperature and plotting the results, we can construct a T-P diagram for a substance with several h= constant lines, as shown in Fig 3. JOULE CO-EFFICIENT The slope of a constant enthalpy line or a p-T diagram of a fluid during a throttling (h= constant) process at a particular state is called Joule-Thompson co-efficient, µ and is given by. Prof.S Venkatesh Rao 21

Vishwatmak Om Gurudev College Of Engineering Thermodynamics µ =  ∂T  h ∂p • The temperature behavior of a fluid during a throttling (h= constant) process is described by the Joule-Thomson coefficient, and may be positive, zero or negative value • Thus the Joule-Thomson coefficient is a measure of the change in temperature with pressure during a constant-enthalpy process. Notice that if V o O• The states where µ = 0 are called 'inversion states' and locus of these states is called the G Cinversion curve. E • The region inside the inversion curve is the cooling region since µ is positive, and temperature falls with fall in pressure. • The region outside the inversion curve is the heating region since µ is negative and temperature rises with fall in pressure. • The temperature at the intersection of the P= 0 line (ordinate) and the upper part of the inversion line is called the maximum inversion temperature. • Cooling can take place only if the initial temperature before throttling is below the maxi-mum inversion temperature. • This temperature is about 5 Tc • It is clear from this diagram that a cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion temperature. This presents a problem for substances whose maximum inversion temperature is well below room temperature. • For hydrogen, for example, the maximum inversion temperature is = - 68°C. Thus hydrogen must be cooled below this temperature if any further cooling is to be achieved by throttling. Prof.S Venkatesh Rao 22

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 7. Show that the work done in a steady flow process is given by - ∫vdp. [DEC 2000] 6 marks OR Show that the reversible adiabatic work transfer for a steady flow system is given by the relation -∫v dp [MAY 2005] 8 Marks In steady flow processes − ∫ vdp , neglecting change in potential energy and kinetic energy represents the shaft work. Mathematically − ∫ vdp = W shaft V  O−∫ 2 vdp the work done during the steady flow process 1-2 and G 1 represents C E graphically represents the area behind the curve representing process 1-2 on P- v diagram as shown in fig. Considering the change in potential and kinetic energy the − ∫ vdp in steady shaft flow process represents the sum of shaft work, change in potential energy and change in kinetic energy. Prof.S Venkatesh Rao 23

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Mathematically − ∫ vdp = W + ∆PE + ∆KE Proof For the steady flow processes energy equation can be written as q − w = ∆PE + ∆KE + ∆h for mass unit In differential form δq = δw + dPE + dKE + dh (1) (2) But from first law of thermodynamics (3) V O δq = δw + du G C E and as δw = Pdv for reversible process δq = Pdv + du from defination of enthalpy h = u + pv dh = du + Pdv + v.dp Substituting the values of δq & dh from equation (2) & (3) in equation (1) − ∫ vdp = ∫ δw + ∆PE + ∆KE − ∫ vdp = W + ∆PE + ∆KE Neglecting∆ ������������������������ & ∆ ������������������������ − � ������������ ������������������������ = ������������ Prof.S Venkatesh Rao 24

Vishwatmak Om Gurudev College Of Engineering Thermodynamics FOR LAST MINUTE REVISION 1. Internal energy is the heat energy stored in a gas. The internal energy of a perfect gas is a function of temperature only. 2. First law of thermodynamics states: — Heat and work are mutually convertible but since energy can neither be created nor destroyed, the total energy associated with an energy conversion remains constant. Or — No machine can produce energy without corresponding expenditure of energy, i.e., it is impossible to construct a perpetual motion machine of first kind. First law can be expressed as follows : Q = ΔE + W Q = ΔU + W ... if electric, magnetic, chemical energies are absent and changes in potential and kinetic energies are neglected. V O G C E 3. There can be no machine which would continuously supply mechanical work without some form of energy disappearing simultaneously. Such a fictitious machine is called a perpetual motion machine of the first kind, or in brief, PMM1. A PMM1 is thus impossible. 4. The energy of an isolated system is always constant. 25 5. W & Q & ∆������������ in case of (i) Reversible constant volume process (v = constant) W = 0 ; Q = m cv (T2 – T1); ∆������������ = ������������ ������������������������ ������������������������������������������������ �������������������������������������������������� Prof.S Venkatesh Rao

Vishwatmak Om Gurudev College Of Engineering Thermodynamics (ii) Reversible constant pressure process (p = constant) W = p(V2 – V1) =m R (T2 –T1 ) ; Q = m cp (T2 – T1) ; ∆������������ = ������������ ������������������������ ������������������������������������������������ �������������������������������������������������� (iii) Reversible temperature or isothermal process (pv = constant) W = p1V1 loge r ; Q = W ; ∆������������ = ������������ ������������ where r = expansion or compression ratio.=������������������������������������������������ = ������������������������ ������������������������ (iv) Reversible adiabatic process (pvγ = constant) ������������ = ������������������������������������������������−������������������������������������������������ = ������������ ������������ (������������������������−������������������������) = ������������������������������������(������������������������ − ������������������������) = −∆������������ ; ������������ = 0 ; ∆������������ = 0 ������������−������������ ������������−������������ V (v) Polytropic reversible process (pvn = constant)O G ������������ = ������������������������������������������������ C−������������������������������������������������= ������������ ������������ (������������������������ − ������������������������) ; ������������ = ������������ − ������������ × ������������ ; ∆������������ = ������������ ������������������������ ������������������������������������������������ �������������������������������������������������� ������������ E−������������ ������������ − ������������ ������������ − ������������ ������������������������������������������������������������ ������������������������ = ������������ − ������������ × ������������������������ ������������ − ������������ 6. Steady flow equation can be expressed as follows: In terms of I.E. ( )• • • Q− W = m ∆ pe + ∆ ke + ∆ u + ∆ w f kJ/s on time basis ( )q − w = ∆ pe + ∆ ke + ∆ u + ∆ w f kJ/kg on mass basis In terms of enthalapy •• • (∆ pe + ∆ ke + ∆ h) kJ/s on time basis Q− W =m Prof.S Venkatesh Rao 26

Vishwatmak Om Gurudev College Of Engineering Thermodynamics q − w = (∆ pe + ∆ ke + ∆ h) kJ/kg on mass basis ∆ pe = change in potential energy , in kJ/kg = g(Z2 − Z1 ) kJ/kg 1000 ∆ke = Change in kinettic energy, in kJ/kg = C 2 − C12 kJ/kg 2 2000 ∆u = Change in specific inernal energy of fluid, in kJ/kg = (u2 − u1 ) = Cv (T2 − T1 ) (For gases) ∆w f = Change in flow work, in kJ/kg = ( p2v2 − p1v1 ) 7. The law of conservation of mass V O ���������̇ ��� = ������������������������������������������������ = ������������������������������������������������ G ������������������������ ������������������������ C E 8. During adiabatic throttling process enthalpy remains constant. The slope of a constant enthalpy line on a p-T diagram is called Joule-Thompson co-efficient, μ. Prof.S Venkatesh Rao 27

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION QUESTIONS MAY 2005 Q 1 Fill in the blanks: An engine producing continuous work and no other effect is called as _________ 1 Q 2 (a) Write a generalized form of the first law of thermodynamics for an open system. Explain the terms. Reduce this equation for SSSF process 5 Q 3 (c) Show that the reversible adiabatic work transfer for a steady flow system is given by the relation -∫v dp 8 DEC 2005 Q 1 Fill in the blanks : (iv) Joule-Thomson coefficient is zero at ______________ point. 1 V O G C E MAY 2006 Q 2 (a) Prove that heat absorbed or rejected during a polytropic process for ideal gas is given by 6 Q 2 (c) What is Joule-Thomson coefficient ? What conclusion can be drawn from a given value of this coefficient? 4 DEC 2006 Q 2 (a) What is Joule-Thompson coefficient. State its significance. 4 Q 3 (a) State Steady Flow Energy Equation. How is it applied to nozzle? 4 MAY 2007 Q 2 (a) State the first law of thermodynamics for closed, system undergoing a cycle. 4 Q 3 (a) Explain steady flow energy equation. Apply it to nozzle turbine and compressor. 5 DEC 2007 Q 2 (a) Prove that heat absorbed or rejected during a polytropic process for ideal gas is given by — Prof.S Venkatesh Rao 28

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 6V Q 3 (a) Explain steady flow energy equation. Apply it to nozzle, turbine and compressor. 6O G MAY 2008C Q 2 (a) Prove that heat absorbed or rejected during a polytropic process for ideal gas is givenE by 6 Q 2 (c) What is Joule-Thomson coefficient ? What conclusion can be drawn form a given value of this coefficient? 4 DEC 2008 (RC) Q 1 (b) State the first law of thermodynamics and prove that for non-flow process it leads to Q=W+∆U 4 Q 3 (b) Why specific heat at constant pressure is greater than specific heat at constant volume Explain with neat figure 8 MAY 2009 (RC) Q 2 (b) Write a note on Joule’s experiment. 6 Q 2 (c) State and explain first law of thermodynamics for a closed system undergoing a process. 4 DEC 2009 (RC) Q1( a) Show that internal energy is property of system. 4 Q1(c) State and explain first low of thermodynamic for closed system. 4 Q2(b) Develop the following expression for the heat transfer from a mass of gas undergoing reversible expansion process obeying the polytropic law, pVn = constant 8 4 Q3(a) Define Joule -Thomson coefficient, inversion point and inversion curve. Q(c) Show that the work done in a steady flow process is given by - ∫pdv. 6 Prof.S Venkatesh Rao 29

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2010 (RC) Q3(c) What is Joule-Thompson coefficient? What conclusions can be drawn from a given values of this coefficient ? 4 V O G C E Prof.S Venkatesh Rao 30

Vishwatmak Om Gurudev College Of Engineering Thermodynamics GRADED PROBLEMS TYPE I: FIRST LAW APPLIED TO NON CYCLIC PROCESS Type I (A): General Processes 1. A fluid is confined in a cylinder by a spring-loaded, frictionless piston, so that the Pressure in the fluid is a linear function of the volume. (p = a + bV). The internal energy of the fluid is given by the following equation U=34+3.15pV, Where U is in kJ, p= kPa and V in cubic metre. If the fluid changes from an initial state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06 m3, with no work other than that done on the piston, find the direction and magnitude of the V Owork and heat transfer. 10 [MAY 2007] G C E 2. A 10kg copper block (specific heats 0.4 kJ/kg K) at 900C is Immersed into an insulated tank containing 0.36 m3 of water at 30 0C. Applying first law, calculate equilibrium temperature. 2 [MAY 2004] 3. If the internal energy of a perfect gas is zero at absolute zero temperature, show that when its volume is v and its pressure P. its internal energy is given by u= Pv/ (γ-1) where γ=CP/CV 8 [DEC 2004] 4. Calculate the internal energy and enthalpy of 1kg air occupying 0.03m3 at 3MPa. 3 MAY 2003] TYPE I (B): SPECIFIED PROCESSES (i): Isothermal Process 5. 0.04 m3 of nitrogen, initially at pressure 1.05 bar and temperature I5°C is compressed in a cylinder isothermally and reversible until the pressure is 4.2 bar. Calculate i. mass of the gas ii. work done Prof.S Venkatesh Rao 31

Vishwatmak Om Gurudev College Of Engineering Thermodynamics [DEC-95-7] iii. heat flow. iv. Also determine: ∆U,∆H,∆S (ii): Isentropic Process 6. Air at 1.02 bar, 22°C initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate (i) Final temperature (ii) Final Volume (iii) Work done. 8 [MAY 2009 (RC)] Also determine: ∆U,∆H,∆S 7. A perfect gas is compressed isentropically from 1 bar and 17°C to the final pressure 6 bar The work done on the gas is 60 kJ/kg. Calculate Cp, Cv, R and molecular weight of the gas. [MAY-97-5]: V O G C E (iii): Polytropic Process 8. A Cylinder fitted with a piston has an initial volume of 0.1 m3 and contains nitrogen at 150 KPa, 25°C. The piston is moved compressing nitrogen until the pressure is one MPa and the temperature is 150°C. During this compression process, heat is transferred from the nitrogen, and work done on the nitrogen is 20 kJ. Determine the amount of this heat transfer. For nitrogen, assume R= 0.2968 kJ/kg-K and CV=0.1695 kJ/kg.K. Also determine: W,∆U,∆H,∆S 5 [MAY 2005] 9. 0.124 m3 of Argon gas at 1.25 bar and 30°C is contained in a cylinder and piston assembly. The gas is compressed in a polytropic process to 8 bar and 200°C. Calculate (i) Change in internal energy (iii) Work interaction (ii) Change in enthalpy (iv) Heat transfer, during the process. 8 [DEC 2008 (OC)] 10. 3 kg of air at a pressure of 5 bar and 250°C expands reversibly in a polytropic process. The final pressure is 1.5 bar and index of expansion is 1.25. Find the change in entropy during a process. Take Cv = 0.652 kJ/kgK and γ= 1.4. 10 [MAY 2009 (RC)] 11. 0.2 kg of air with initial pressure 1.5 bar and temperature 300 K is compressed to a pressure of 15 bar according to PV1.25 = C. Determine :— (i) Work transfer Prof.S Venkatesh Rao 32

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 8 [MAY 2009 (OC)] (ii) Heat transfer (iii) Change in entropy (iv) Change in internal energy. 12. The internal energy and equation of state of a closed gas system are given by U = (189 + 1.25t) kJ/kg and Pv=0.61 T , where t is the temperature in °C, and T in °K, P is the pressure on kN/m2 and v is the specific volume in m3/kg of the system. If the temperature of 5 Kg of gas is raised T from 200°C to 400°C at 1. constant pressure, 2. according to the law. Pv1.3 = constant, calculate the work and heat transfer in each case 8 [DEC-92]: V O G C E TYPE II: TWO OR MORE PROCESSES NOT FORMING CYCLE 13. Three moles of an ideal gas being initially at a temperature T0 =273 K, were isothermally expanded 5 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total amount of heat transfer to the gas during the (isothermal and isochoric process equals 80 kJ. Find specific heat ratio γ=CP/CV for this gas 10 [DEC 2005] 14. 1.5 m3 of air at 1 bar and 25°C is compressed isothermally to a pressure of 5 bar. Calculate the work done, heat interacted and the change of entropy during the process. If this air is now expanded back to original volume according to the law PV1.3 = constant, calculate change of entropy during expansion Represent the processes on P-V. and T-S diagrams. Take CP = 1.0 kJ/kg °K and γ = 1.4. 10 [MAY 2006, MAY 2008] Also determine: W,∆U,∆H,∆S for each process & total 15. A fluid contained in a cylinder receives 150 kJ mechanical energy by means of a paddle wheel together with 50 kJ in the form of heat. At the same time a piston in the cylinder moves in such a way that the pressure remains constant at 200 kN/m2 during the fluid expansion from 2m3 to 5 m3. What is the change in internal energy and in enthalpy? 10 [DEC 2006] Prof.S Venkatesh Rao 33

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 16. Air at 20°C and 1.05 bar occupies 0.025 m3. The air is heated at constant volume until the pressure is 4.5 bar and then cooled at constant pressure back to original temperature. Sketch the processes on P-V and T-S diagram and calculate— (i) The heat flow from the air. (ii) The net entropy change. 8 [DEC 2008 (OC)] TYPE III: TWO OR MORE PROCESSES FORMING CYCLE [FIRST LAW APPLIED TO CYCLIC PROCESS] 17. A system contains 0.15 m3 of air at 3.8 bar and 150°C. A reversible adiabatic expansion takes place till pressure falls to 1.03 bar. The gas is then heated at constant pressure till enthalpy increases by 60.7 kJ. Determine total work done. If these processes are replaced by a single reversible polytropic process giving same work between initial and final states, determine index of expansion. 10 [DEC 2008 (RC)] V O G C E 18. A closed system executes a quasi-static process absorbing 80 kJ of heat and expanding from 2 m3 to 2.25 m3 against constant pressure of 2 bar. A non-quasi-static process is used to bring the system to its initial state during which it absorbs 120 kJ of heat. Calculate the work done in the second process. 8 [DEC 2004] 19. One kg of air initially at a pressure of 1 bar and temperature 20°C undergoes following processes to complete the cycle – i. Isothermal compression to a pressure of 6 bar. ii. Constant pressure heat addition process to such temperature that subsequent adiabatic expansion completes the cycle. Calculate:- i. Net heat addition ii. Net work obtained [MAY-95-8] Also determine : ∆U,∆H,∆S for each process &for cycle 20. A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four processes. The sum of all heat transfer during the cycle is 340 kJ. The system completes Prof.S Venkatesh Rao 34

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 200 cycles/min. Complete the following table showing the method for each process and compute the net rate of work output in kW. 10 [DEC 2005] TYPE V: FIRST LAW APPLIED TO FLOW PROCESSES V O G C E 21. A steam turbine operating under steady flow conditions receives steam at the following conditions :- Pressure =12 bar; Temperature = 190°C; Specific volume =0.16 m3/kg Specific internal energy = 2,400 kJ/kg Velocity = 20 m/sec Elevation=5m the steam leaves the turbine at the following rate Pressure =0.3 bar Temperature = 70° C Specific volume =5.25 m3/kg Specific internal energy =2,120kJ/kg Velocity = 60 m/sec Elevation= 2m The turbine loses heat to the surrounding at the rate of 18 kJ/min, the mass flow rate of steam is 0.5 kg/sec Calculate the power output of the turbine. 7 [DEC-93] 22. In a steam power station, steam flows steadily through a 0.2 m diameter pipeline from the boiler to the turbine. At the boiler end, the steam conditions are found to be : p = 4 MPa, t =400° C, h=3213.6 kJ/kg and v = 0.073 m3/kg. At the turbine end, the conditions are found to Prof.S Venkatesh Rao 35

Vishwatmak Om Gurudev College Of Engineering Thermodynamics be p = 3.5 MPa, t = 392 0 C, h = 3202.6 kJ .1kg and v = 0.084 m3 /kg. There is a heat loss of 8.5kJ/kg from the pipeline. Calculate the steam flow rate. 10 [MAY 2007] 23. In a gas turbine gases flow through the turbine at the rate of 25 kg/sec and power developed is 18000 kW. The enthalpies of gas at inlet and outlet are 1350 kJ / kg and 470 kJ / kg respectively. The velocities of the gases at the inlet and outlet are 60 m/s and 15.0 m/s respectively. Calculate the rate at which the heat is rejected from the turbine. Also, find the area of inlet pipe if the specific volume of gases at inlet is 0.5 m3/kg 8 [DEC 2007, DEC 2006] 24. A large SSSF expansion engine has two low velocity flows of water entering. High pressure steam enters at point 1 with 2kg/s 2 MPa, 5000 C and 0.5 kg/s cooling water at 120 KPa, 30°C enters at point 2. A single flow exits at point3, with 150 KPa, 80% quality, through at 0.15 m diameter exhaust pipe: There is a heat toss of 300 KW Find the. exhaust velocity and the power output of the engine. 10 [MAY 2005] V O G C E 25. In a steam plant 1 kg of water per second is supplied to the boiler. The enthalpy and velocity of water entering the boiler are 800 kJ/kg and 5 m/s. The water receives 2200 kJ/kg of heat in the boiler at constant pressure. The steam after passing through the turbine comes out with a velocity of 50 m/s and its enthalpy is 2520 kJ/kg. The inlet is 4 m above the turbine exit. Assuming the heat losses from the boiler and turbine to the surroundings are 20 kJ/s, calculate the power developed by the turbine. Consider the boiler and turbine as single system. 12 [DEC 2008 (RC)] 26. At inlet to a certain nozzle the enthalpy of fluid passing is 2800 kJ/kg and the velocity is 50 m/s. At discharge end the enthalpy is 2600 kJ/kg. The nozzle is horizontal and there is negligible heat loss from it. (i) Find the velocity at the exit of the nozzle. (ii) If the inlet area is 900 cm2 and specific volume at inlet is 0.187 m3/kg. Find the mass flow rate.(iii) If the specific volume at the nozzle exit is 0.498 m3/kg, find the exit area of nozzle. 10 [MAY 2009 (OC)] 27. Air at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800OC . It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity Prof.S Venkatesh Rao 36

Vishwatmak Om Gurudev College Of Engineering Thermodynamics of 60 m/s to a nozzle where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/sec, Calculate . (i) The rate of heat transfer to, the air in the heat exchanger. (ii) The power output from the turbine assuming no heat loss. (iii) The velocity of exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h=CP t where cp is the specific heat equal to 1.005 kJ/kgK and t is the temperature. 12 [DEC 2004] V O G C E Prof.S Venkatesh Rao 37

Vishwatmak Om Gurudev College Of Engineering Thermodynamics EXAMINATION PROBLEMS (SOLVE AS HOME WORK) MAY 2005V Q 2 (b) A large SSSF expansion engine has two low velocity flows of water entering. HighO pressure steam enters at point 1 with 2kg/s 2 MPa, 5000 C and 0.5 kg/s cooling water at 120G KPa, 30°C enters at point 2. A single flow exits at point3, with 150 KPa, 80% quality, throughC at 0.15 m diameter exhaust pipe: There is a heat toss of 300 KW Find the exhaust velocity andE the power output of the engine. 10 Q 2 (c) A Cylinder fitted with a piston has an initial volume of 0.1 m3 and contains nitrogen at 150 KPa, 25°C. The piston is moved compressing nitrogen until the pressure is one MPa and the temperature is 150°C. During this compression process, heat is transferred from the nitrogen, and work done on the nitrogen is 20 kJ. Determine the amount of this heat transfer. For nitrogen, assume R= 0.2968 kJ/kg-K and CV=0.1695 kJ/kg.K. 5 DEC 2005 Q 2 (a) Three moles of an ideal gas being initially at a temperature T0 =273 K, were isothermally expanded 5 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total amount of heat transfer to the gas during the (isothermal and isochoric process )equals 80 kJ. Find specific heat ratio γ=CP/CV for this gas 10 Q 2 (b) A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four processes. The sum of all heat transfer during the cycle is 340 kJ. The system completes 200 cycles/min. Complete the following table showing the method for each process and compute the net rate of work output in kW. Prof.S Venkatesh Rao 10 38

Vishwatmak Om Gurudev College Of Engineering Thermodynamics MAY 2006 Q 1 (a) Calculate the internal energy and enthalpy of 1 kg air occupying 0.03 m3 at 3 MPa 3 Q 2 (b) 1.5 m3 of air at 1 bar and 25°C is compressed isothermally to a pressure of 5 bar. Calculate the work done, heat interacted and the change of entropy during the process. If this air is now expanded back to original volume according to the law PV1.3 = constant, calculate change of entropy during expansion Represent the processes on P-V. and T-S diagrams. Take CP = 1.0 kJ/kg °K and γ = 1.4. 10 DEC 2006 Q 2 (c) A fluid contained in a cylinder receives 150 kJ mechanical energy by means of a paddle wheel together with 50 kJ In the form of heat. At the same time a piston in the cylinder moves in such a way that the pressure remains constant at 200 kN/m2 during the fluid expansion from 2m3 to 5m3. What is the change in internal energy and in enthalpy? 10 Q 3 (c) In a gas turbine gases flow through the turbine at the rate of 25kg/sec and power developed is 18 MW. The enthalpies of gas at inlet and outlet are l350kJ/kg and 470kJ/kg respectively. The velocities of the gases at the inlet and outlet are 60m/s and 150m/s respectively. Calculate the rate at which the heat is rejected from the turbine. Also, find the area of inlet pipe if the specific volume of gases at inlet is 0.5m3 /kg. 10 V O G C E MAY 2007 Q 2 (c) A fluid is confined in a cylinder by a spring-loaded, frictionless piston, so that the Pressure in the fluid is a linear function of the volume. (p = a + bV). The internal energy of the fluid is given by the following equation U=34+3.15pV, Where U is in kJ, p= kPa and V in cubic metre. If the fluid changes from an initial state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06 m3, with no work other than that done on the piston, find the direction and magnitude of the work and heat transfer. 10 Q 3 (b) In a steam power station, steam flows steadily through a 0.2 m diameter pipeline from the boiler to the turbine. At the boiler end, the steam conditions are found to be : p = 4 MPa, t =400° C, h=3213.6 kJ/kg and v = 0.073 m3/kg. At the turbine end, the conditions are found to Prof.S Venkatesh Rao 39

Vishwatmak Om Gurudev College Of Engineering Thermodynamics be p = 3.5 MPa, t = 392 0 C, h = 3202.6 kJ .1kg and v = 0.084 m3 /kg. There is a heat loss of 8.5 kJ/kg from the pipeline. Calculate the steam flow rate. 10 DEC 2007 Q 2 (c) A closed system executes a quasi-static process absorbing 80kJ of heat and expanding from 2 m3 to 2.25 m3 against a constant pressure of 2 bar. A, non-quasi-static process is used to bring the system to its initial state during which it absorbs 120 kJ of heat. Calculate the work done in the second process. 8 Q 3 (c) In a gas turbine gases flow through the turbine at the rate of 25 kg/sec and power developed is 18000 kW. The enthalpies of gas at inlet and outlet are 1350 kJ / kg and 470 kJ / kg respectively. The velocities of the gases at the inlet and outlet are 60 m/s and 15.0 m/s respectively. Calculate the rate at which the heat is rejected from the turbine. Also, find the area of inlet pipe if the specific volume of gases at inlet is 0.5 m3/kg. 8 V O G C E MAY 2008 Q 1 (b) Calculate the pressure of 1 kg air occupying 0.03 m3 at 200°C. 2 Q 2 (b) 1.5 m3 of air at 1 bar and 27°C is compressed isothermally to a pressure of 5 bar. Calculate the, work done, heat interacted and the change of entropy during the process. If this air is now expanded back to original volume according to the law PV1.3 = const. Calculate change of entropy during expansion. Represent the processes on P. V. and T-S diagrams. Take CP= 1.0 kJ/kg °K and γ = 1.4. 10 DEC 2008 (RC) Q 2 (a) In a steam plant 1 kg of water per second is supplied to the boiler. The enthalpy and velocity of water entering the boiler are 800 kJ/kg and 5 m/s. The water receives 2200 kJ/kg of heat in the boiler at constant pressure. The steam after passing through the turbine comes out with a velocity of 50 m/s and its enthalpy is 2520 kJ/kg. The inlet is 4 m above the turbine exit. Assuming the heat losses from the boiler and turbine to the surroundings are 20 kJ/s, calculate the power developed by the turbine. Consider the boiler and turbine as single system. 12 Prof.S Venkatesh Rao 40

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Q 6 (a) A system contains 0.15 m3 of air at 3.8 bar and 150°C. A reversible adiabatic expansion takes place till pressure falls to 1.03 bar. The gas is then heated at constant pressure till enthalpy increases by 60.7 kJ. Determine total work done. If these processes are replaced by a single reversible polytropic process giving same work between initial and final states, determine index of expansion. 10 MAY 2009 (RC) Q 2 (a) 3 kg of air at a pressure of 5 bar and 250°C expands reversibly in a polytropic process. The final pressure is 1.5 bar and index of expansion is 1.25. Find the change in entropy during a process. Take Cv = 0.652 kJ/kgK and γ= 1.4. 10 Q 7(b) Air at 1.02 bar, 22°C initially occupying a cylinder volume of 0.015 m3, is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate (i) Final temperature (ii) Final Volume (iii) Work done. 8 V O G C E DEC 2009 (RC) Q1(d)A gas undergoes a reversible non-flow process according to the relation p= (-3 +15) where V is the volume in m3and p is the pressure in bar. Determine work done when the volume changes from 3 to 6 m3. 5 Q2 (a) 60 liters of an ideal gas at 290 K and 1 bar is compressed adiabatically to 10 bar. It is then cooled at constant volume and further expanded isothermally so as to reach the condition from where it started. Evaluate 1) pressure at the end of constant volume cooling, 2) change in internal energy during constant volume process, 3) net work done and heat transfer during the cycle. Assume Cp =14.25kJ/kgK and Cv= 10.l5kJ/kgK 12 Q3(b) In a steady flow process, the fluid flows through a machine at the rate of 15 kg/min. The entrance and exit parameters of the machine are Velocity 5 m/s and Sm/s ,Pressure 100 kPa and 700 kPa, Specific volume 0.45 m3/kg and 0.125 m3/kg resp. The working fluid leaves the machine with internal energy 160 kJ/kg greater than that at entrance and during the process 7200kJ/min of heat is lost to the surrounding. Assuming entrance and exit pipe to be at the Prof.S Venkatesh Rao 41

Vishwatmak Om Gurudev College Of Engineering Thermodynamics same level, calculate the shaft work and the ratio of inlet pipe diameter to outlet pipe diameter. 10 MAY 2010 (RC) Q2 (a) Air at a temperature of 20 °C passes through a heat exchanger at a velocity of 40 m/s where it’s temperature is raised to 8200C. It then enters a turbine with same velocity of 40 m/s and expands till the temperature falls to 620 °C. On leaving the turbine, the air is taken at a velocity of 55 m/s to a nozzle where it expands untill the temperature has fallen to 510 °C. If the air flow rate is 2.5 kg/s, calculate (i) rate of heat transfer to the air in the heat exchanger. (ii) the power output from the turbine assuming no heat loss. (iii) the velocity at exit from the nozzle, assuming no heat loss. 10 Q5 (a) Air at 20 °C and 1.05 bar occupies 0.025 m3. The air is heated at constant volume untill the pressure is 4.5 bar and then cooled at constant pressure back to original temperature. Sketch the processes on T-S and P-V diagram and calculate (i) The heat flow from air (ii) The net entropy change. 7 V O G C E DEC 2010 (RC) Q2(b) A system containing 0.2 m3 of air at a pressure of 4 bar and 160°C expands isentropically to pressure of 1.06 bar and after this the gas is heated at the constant pressure till the enthalpy increases by 65 kJ. Calculate the work done. Now imagine that these processes are replaced by a single eversible polytropic process producing same work between initial and final state; find the index of .expansion in this case. Take Cp of air = 1.005 kJ/kg K. 12 MAY 2011 (RC) Q 2 (c) A turbocompressQr delivers 2.33 m3 is of air at 0.276 MPa, 43° C which is heated at this pressure to 4300 C and finally expanded in a turbine which delivers 860 kW.Dur ing expansion there is a heat transfer of 0.09 MJ/s to the surroundings. Calculate the turbine exhaust temperature if changes in kinetic and potential energy are negligible 10 Prof.S Venkatesh Rao 42

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Q 4 (b) 0.5 kg of air initially at 25 o C is heated reversibly at constant pressure till the volume is doubled and is then heated reversibly at constant volume till the pressure is doubled. For the total path find the change in entropy 8 Q 6 (c) A gas initially at 14.4 bar and 360°C is expanded isothermally to a pressure of 2.24 bar .It is-then cooled at constant volume till the pressure falls to 1.02 bar. Finally an adiabatic compression brings the gas back to the initial state. The mass of the gas is 0.23 kg and Cp= 1 KJ/kg K. Draw the p-v diagram and determine: i) The value of the· adiabatic index of compression.(ii)The change in internal energy of the gas during the adiabatic process. 10 MAY 2011 (RC) Q 2. (a) An air compressor takes in air at 1 bar and 20° C and discharges into a pipe having inlet diameter 20 mm. The average velocity of air at a point in the pipe close to the discharge is 7·7 m/sec and the discharge pressure is 3 bar. Neglecting the inlet air velocity and assuming the compression of air as adiabatic. Calculate the power input to the compressor. Take γ=1-4 and R = 286·7 J/kg0 K 10 V O DEC 2011 (RC) G NIL C E MAY 2012 (RC) Q 3 (b) A closed system contains air at pressure 1 bar, temperature 290 K and volume 0·02 m3 this system undergoes a thermodynamic cycle consisting of the following three process in series :- Process-1-2 : Constant volume heat addition till pressure becomes 4 bar. Process 2-3 : Constant pressure cooling. Process 3-1 : Isothermal heating to initial state. Represent the cycle on T-s and p-v plots and evaluate the change in entropy for each process. 10 Prof.S Venkatesh Rao 43

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Q 4 (b) In a steady flow system, fluid_ flows at the rate of 15 kg/min. It enters at a pressure of 100 kPa, a velocity of 5 m/s, Internal energy 50 kJ/kg and specific volume 0.45 m3/kg. It leaves the system at a pressure of 700 kPa, a velocity of 8m/s, internal energy·160 kJ/kg and specific volume 0·125 m3/kg. During its flow through system there is a heat loss of 7200kJ/kg. Assuming entrance and exit pipes to be at the same level. Calculate the shaft work and ratio of inlet pipe diameter to outlet pipe diameter. 10 V O G C E Prof.S Venkatesh Rao 44

Vishwatmak Om Gurudev College Of Engineering Thermodynamics CHAPTER 3 SECOND LAW OF THERMODYNAMICS 1. Explain with examples the limitations of First Law of Thermodynamics. [MAY 2011] 5 Marks OR What are the limitation of first law of thermodynamics? [MAY 2012] 4 Marks If a system executes a closed cycle, then according to the first law of thermodynamics, we can write. � ������������������������ = � ������������������������ V O G C E The first law of thermodynamics merely states that the work transfer equals the heat transfer and does not place any restriction on the direction of heat and work transfer. According to this law it is assumed that any change of thermodynamic state can take place in either direction. But it has been found that this is not the case, particularly in the inter-conversion of heat and work. It would be obvious from the following examples that the processes naturally proceed in certain directions and not in the opposite directions, even though the reversal of the processes does not violate the first law. 1. If two metal blocks at temperature T1 and T2 (T1 > T2) are brought into contact with each other and are isolated from the surroundings, heat flows from the high temperature block to the low temperature block till the temperature of both blocks are equal. From experience we know that the reverse process by which heat flows from the low temperature block to the high temperature block is impossible, i.e., the original temperature T1 and T2 cannot be restored, spontaneously. Prof.S Venkatesh Rao 45

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 2. If work is supplied to a thermally insulated closed system with the help of paddle wheel, the energy of the system will increase. But the reverse process is impossible i.e. with the cooling of the system to its original energy state, obtaining back the entire paddle work is not possible. 3. A running automobile can be stopped by using the brake where the kinetic energy of automobile is converted into heat which increases the energy of the brake drum. But is not possible to accelerate the automobile by the cooling of the hot brake drum. 4. There is no change in the temperature of a gas during the free expansion through a valve into an evacuated vessel, and no work is done, but the volume increases and the pressure decreases. It the above process takes place in opposite direction then the gas at low pressure should rush back through the valve and compress itself to its original state. Such a reversal of the process is found to be impossible. V O G C E 5. As the car drives up the hill, the level of the fuel in the tank drops. If this process of driving up the hill is reversed i.e., the car is coasted down the hill, the fuel consumed cannot be reproduced. From the above examples, it may be concluded that the first law of thermodynamics provides a necessary but not sufficient condition for process to take place.  From the above examples, it may be noted it that the limitations of First Law of thermodynamics are  First law of thermodynamics does not help to predict whether the system will or will not undergo a particular change.  No restriction has been imposed by the first law on the possibility of conversion of energy from one form into another.  The first law does not throw any light on what portion of heat may be converted into useful form of energy, i.e. work. Prof.S Venkatesh Rao 46

Vishwatmak Om Gurudev College Of Engineering Thermodynamics 2. Define the terms – 1. Thermal reservoir 2. Source 3. Sink 4. Heat engine [DEC 2007, MAY 2012] 5. Thermal Efficiency of Heat Engine 6. Refrigerator [MAY 2012] 7. Heat pump [DEC 2007, MAY 2012] 8. C.O.P. of refrigerator and C.O.P. of heat pump [DEC 2007, MAY 2011] V O G C E ENERGY RESERVOIRS, SOURCE & SINK • A thermal reservoir is defined as a large body of infinite heat capacity, which is capable of absorbing or rejecting an unlimited quantity of heat without suffering appreciable changes in its thermodynamic coordinates. • The changes that do take place in the large body as heat enters or leaves are so very slow and so very minute that all processes within it are quasi-static. • The thermal reservoir FROM which heat Q1 is transferred to the system operating in a heat engine cycle is called the source or High temperature reservoir (HTR) • The thermal energy reservoir TO which heat Q2 is rejected from the system during a cycle is the sink or Low temperature reservoir (LTR). HEAT ENGINE AND EFFICIENCY • Heat Engine is a device used to convert heat energy into useful work when operating in a cyclic process. • It is a cyclically operating system in which only heat and work cross the boundary of the system. • Consider a system shown in Figure 1 which represents a steam power plant. Prof.S Venkatesh Rao 47

Vishwatmak Om Gurudev College Of Engineering Thermodynamics Steam is generated in the boiler at high pressure and supplied to the steam turbine. • The steam turbine develops the work using part of the energy in the steam and the steam coming out of the turbine is condensed in the condenser and fed back to the boiler, with the help of a pump to complete the cycle. • The boiler, turbine, condenser or pump individually, do not constitute a heat engine, as only a part of the cycle takes place in each of these components. • All the components are necessary to complete the cycle. • Diagrammatically Heat Engine is represented as V O G C E • Referring to Figure 1, the net heat entering into the system is (Q1 – Q2) and it must be equal to the sum of the turbine and pump work, as per first law of thermodynamics � ������������������������ = � ������������������������ ∴ ������������������������ − ������������������������ = ������������������������ − ������������������������ = ������������������������������������������������ = ������������ Prof.S Venkatesh Rao 48

Vishwatmak Om Gurudev College Of Engineering Thermodynamics • The efficiency of the heat engine is defined as ������������ = ������������������������������������ ������������������������������������������������ ������������������������������������������������������������������������ = ������������ = ������������������������ − ������������������������ = ������������ − ������������������������ ������������������������������������������������ ������������������������������������������������������������������������������������������������ ������������������������ ������������������������ ������������������������ HEAT PUMP, REFRIGERATOR & THEIR C.O.P. • A Reversed Heat Engine is a device which working in a cycle delivers heat from low temperature to a high temperature. • The heat engine and Reversed Heat Engine are represented diagrammatically in Figure 2. This model will be adopted in further discussion. • The efficiency of heat pump is defined as per the objective in mind. • The efficiency term is replaced by coefficient of performance (C.O.P.) for Reversed Heat Engine just to avoid confusion between heat engine and heat pump. • Sometimes a term Energy performance ratio is also used. V SOURCE, O SOURCE, AT T1>T2 G AT T1>T2 C Q1 E Q1 H. W H. W E E Q2 Q2 SINK, AT T2 SINK, AT T2 Figure 2 Diagrammatic representations of heat engine and heat pump • If the purpose of Reversed Heat Engine is to extract the heat from the sink which is at low temperature (T2) and supply it to a room at temperature T1 for heating, then C.O.P. of the system is given by (C .O.P )HP = Q1 = Q1 W Q1 − Q2 Prof.S Venkatesh Rao 49

Vishwatmak Om Gurudev College Of Engineering Thermodynamics • If the purpose of Reversed Heat Engine is to extract the heat from the sink which is at low temperature (T2) to maintain it at this low temperature compared with source temperature T1 and reject it to the source at temperature T1 then C.O.P. of the system is given by (C .O.P ). Re f = Q2 = Q2 W Q1 − Q2 • • This form of heat pump is known as REFRIGERATOR • The whole object of the study of thermodynamics is to design a more efficient engine or heat pump. • That means, maximum percentage of heat supplied to the engine should be converted into useful work in case of heat engine and minimum work should be V O G C E supplied for a given amount of heat to be extracted from the low temperature source in case of the heat pump. 3. Show that COP of heat pump is equal to one plus COP of refrigerator when working between same temperatures limits. [MAY 2007] 4 Marks OR Prove that (COP)HP = 1 + (COP)Ref [DEC 2010] 4 Marks OR Define C.O.P of Heat Pump and Refrigerator and derive the relationship between the two.[MAY 2011] 4 Marks We know that C.O.P. of a refrigerator is given by (C .O.P ). Re f = Q2 = Q2 and W Q1 − Q2 C.O.P. of a Heat pump is given by (C .O.P )HP = Q1 = Q1 W Q1 − Q2 ������������������������ (������������. ������������. ������������)������������������������������������ + 1 = ������������2 + 1 = ������������2 − (������������1 − ������������2) = ������������1 = (������������. ������������. ������������)������������������������ ������������1 − ������������2 ������������1 − ������������2 ������������1 − ������������2 ∴ (������������������������������������)������������������������ = ������������ + (������������������������������������)������������������������������������ Prof.S Venkatesh Rao 50