Thermodynamics Solution

1. a) An insulated insulated rigid rigid vessel vessel contains contains some some powdered powdered coal and and air at a pressur pressuree of 10 bar and and a 0 temperature of 20 C. The coal is ignited, there results a rise in the pressure and temperature 0 of the contents of the vessel, the final temperature is 538 C. Taking the vessel and the contents to be the system under consideration evaluate the increase in the energy of the system. b) The The insulation is now removed. A heat transfer. of 50 kJ from the system system causes 0 the temperature to fall to the initial value, 20 C. Evaluate the increase in the energy of the system during this process. c) Taking the initial energy of the system 32 kJ, write down the energy values after after process (a) and after process (b) Solution: a) The vessel is rigid i.e., dv = 0 W = 0 System boundary is insulated, Q = 0 st By 1 law Q = ∆E + W ∆E = 0 (Though there is a change in temperature and pressure of the system, there is no change in energy. All that happens is chemical energy energ y of the system is converted into internal energy). b) W = 0 ; Insulation removed: Q = - 50 kJ -50 = 0 + ∆E ∆E = -50 kJ or decrease decrease in energy energy = 50 kJ c) During process (a) ∆E = 0 Initial energy = energy after process = 32 kJ For process (b) initial energy = Final energy in process (a) = 32 k J ∆E for process (b) = -50 kJ  – Einitial = - 50 kJ i.e., Efinal – E Efinal = -50 -50 + 32 32 = 18 kJ 2. On a warm warm summer summer day, a housewife housewife decides to beat the the heat by closing closing the the windows windows and doors in the kitchen and opening the refrigerator door. At first she feels cool and refreshed, but after a while the effect begins to to wear off. Evaluate the situation as it relates to first law, law, considering the room including the refrigerator as the system. Solution: Room

..

+ -

We

System

Ref 

At first the temperature of air in the room falls since it communicates with the cool refrigerator. This makes the housewife feel cool. Considering the room and its contents as a system, and assuming walls, windows and doors nonconducting, Q = 0. For the operation of refrigerator, electricity is supplied from outside and hence electrical work We = is done on the system. From first law of TD Q = ∆E + W e 0 = ∆E - We ∆E = We Positive sign of energy indicates the increase in energy of the system with with time. As the energy is increasing, the temperature of air increases and hence effect of coolness gradually begins to wear off.

Dr. T.N. Shridhar, Professor, NIE, Mysore

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3. The average heat transfer from a person to the surroundings when he is not actively working is about 950 kJ/hr. suppose that in the auditorium containing 1000 people the ventilation system fails. a) How much does the internal energy of air in the auditorium increase during the first 15 minutes after the ventilation fails? b) Considering the auditorium and all the people as system and assuming no heat transfer. to surroundings, how much does the int. energy of the system change? How do you account for the fact that the temperature of air increases? Solution: a) Average heat transfer per person = 960 kJ/hr = 960 / 60 = 15.83 kJ /min Average heat transfer / person for 15 min = 237.5 kJ Average heat transfer for 15 min in the auditorium containing 1000 people Q = 237.5 x 1000 = 237500 kJ/min From first law of TD, we have Q = ∆E + W 237500 = ∆E + 0 ∆E = 237.5 MJ b) Considering the auditorium and all the people as system, Q = 0; W = 0 Q = ∆E + W ∆E = 0 0 = ∆E + 0 Increase in internal energy of the air due to increase in its temperature is compensated by the decrease in internal energy of the people. 4. A household refrigerator is loaded with fresh food and closed. Consider the whole refrigerator and the contents as a system. The machine uses 1 kWhr of electrical energy in cooling the food and the internal energy of the food (system) decreases by 5250 kJ, as the temperature drops. Find the magnitude and direction of heat transfer during the process. Solution: Ref  Power + Supply Contents

Given: Electrical work, = 1 kWhr We = 860 Kcal = 860 (4.187) = - 3600.8 kJ Given, ∆E = - 5250 kJ From first law of TD Q = ∆E + We = - 5250 – 3600.8 = - 8850 kJ Negative sign indicates heat flows from the refrigerator to the surroundings 5. A closed system undergoes a constant volume process in which 85 kJ of heat is supplied to it. The system then undergoes a constant pressure process in which 90 kJ of heat is rejected by

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the system and 15 kJ of work is done on it. Finally the system is brought back to its original state by a reversible adiabatic process. Determine i) The magnitude and direction of work  transfer during the adiabatic process. ii) The energy of the system at all end states if the energy at the initial state is 100 kJ. Solution: 3

2

p 1 V Process 1-2: Constant volume process i.e., dv = 0 i.e., W1-2 = 0 Q1-2 = (E2 – E1) + W1-2 85 = (E2 – E1) + 0 (E2 – E1) = 85 kJ E2 -100 = 85 E2 = 185 kJ But E1 = 100 kJ Process 2-3: (Constant pressure process) Q2-3 = (E3 – E2) + W2-3 -90 = (E3 – E2) – 15 (E3 – E2) = -75 kJ E3 = -75 + 185 = 110 kJ But E2 = 185 kJ



In a cyclic process δQ

  δW 

Q1-2 + Q2-3 + Q3-1 = W1-2 + W2-3 + W3-1 85 – 90 + 0 = 0 -15 + W3-1 W3-1 = 10 kJ 6. A system undergoes a constant pressure process which is followed by a constant volume process. during the constant pressure process, 125 kJ of heat is transferred to the system and 50 kJ of work is done by the system. during a constant volume process, 125 kJ of heat is rejected from the system. find the work interaction if a rev. adiabatic process restores the system to the initial state. Solution: Process 1-2: Constant pressure process We have Q12 = 125 W12 = + 50 Q12 = E2 – E1 + W12 (E2 – E1) = 75 kJ 125 = (E2 – E1) + 50 Process 2-3: Constant volume process, i.e., dv = 0 W23 = 0 We have Q23 = -125 Q2-3 = (E3 – E2) + W2-3 -125 = (E3 – E2) + 0 (E3 – E2) = - 125 kJ Q3-1 = 0



For a cyclic process δQ

  δW 

125 – 125 + 0 = 50 + 0 + W3-1

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W3-1 = - 50 kJ or Process 3-1: (Rev. Adiabatic process) dQ = 0 Q3-1 = E1 – E3 + W3-1 0 = (E1 – E2) + (E2 – E3) + W3-1 W3-1 = -50 kJ = -75 + 125 + W3-1 7. A system executes a cyclic process which includes four processes: 1-2, 2-3, 3-4 and 4-1. The magnitudes of the energy transfer are shown in the following table. Process heat transfer Work transfer Change in internal Q (kJ) W(N-m) energy ∆U (kJ) 1-2 10 0 10 2-3 -25 15 x 10 -40 kJ 3-4 60 23.5 kJ 36.5 kJ 4-1 -15 -8.5 x 10 -6.5 kJ



Find the magnitude of the unknown quantities in kJ (Hint: For the process 3-4, use dU   0 8. A system composed of a stone having a mass of 10 kg and a bucket containing 100 kg of  water are at the same temperature the stone being at a height of 10.2 m above the water level. Stone falls into the water. Determine ∆U, ∆kE, ∆PE, Q and W for the following cases. a) The stone is above to enter the water. b) The stone has just come to rest in the bucket and c) Heat has been transferred to the surroundings in such an amount that the stone and water are at the same temperature they were initially. st Solution: The 1 law of THERMODYNAMICS. is m 2 2 Q1-2 = U2 – U1 + (v2  – v1 ) + mg (h2 – h1) + W1-2 --- (1) 2 a) The stone is about to enter the water Assuming no heat transfer. to or from the stone a sit falls, ∆U = 0 Q1-2 = 0 W1-2 = 0 Equation: 1) reduces to 0 = ∆kE + ∆pE or - ∆kE = ∆pE = mg (h2 – h1) = 10 (9.81) (-10.2) = - 1000 J = - 1kJ i.e., ∆kE = 1 kJ & ∆pE = - 1kJ b) Just after the stone comes to rest in the bucket, Q1-2 = 0 ; W1-2 = 0 ∆kE = 0 Equation 1) reduce to ∆ p E = -∆U = mg (h 2 – h1) = - 1kJ ∆U = 1kJ & ∆pE = - 1kJ c) After heat has been transferred so that stone & water are at the same temperature they were initially, ∆U = 0  In this case ∆U = 0 ∆kE= 0 W1-2 = 0 Equation 1) reduces to Q1-2 = ∆PE = mg (h2 – h1) = -1kJ

Dr. T.N. Shridhar, Professor, NIE, Mysore

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The Pure Substance The system encountered in thermodynamics is often quite less complex and consists of fluids that don not change chemically, or exhibit significant electrical, magnetic or capillary effects. These relatively simple systems are given the generic name th e Pure Substance. Definition A system is set to be a pure substance if it is (i) homogeneous in chemical composition, (ii) homogeneous in chemical aggregation and (iii) invariable in chemical aggregation.

Homogeneous in chemical composition means that the composition of each part of the system is same as the composition of any other part. Homogeneous in chemical aggregation implies that the chemical elements must be chemically combined in the same way in all parts of the system. Invariable in chemical aggregation means that the chemical aggregation should not vary with respect to time. Steam

Water

H2 + ½ O2 (Gas)

H 2 + O2 (Gas)

Water

Water

(i) (ii) (iii) Satisfies condition (i) Satisfies condition (i) Does not satisfies condition (i) Satisfies condition (ii) Does not satisfies condition (ii) Satisfies condition (iii) Figure Illustration of the definition of pure substance In figure three systems are shown. The system (i) shown in the figure is a mixture of steam and water. It is homogeneous in chemical composition because in every part of the system we have, for every atom of oxygen we have two atoms of hydrogen, whether the sample is taken from steam or water. The same is through for system (ii) consisting of water and uncombined mixture of hydrogen and oxygen. System (iii) however is not homogeneous in chemical composition because in the upper part of the system hydrogen and oxygen are present in the ratio 1:1 where as in the bottom portion they are present in the ratio 2:1. System (i) also satisfies condition (ii), because both hydrogen and oxygen have combined chemically in every part of the system. System (ii) on the other hand does not satisfies condition (ii) because the bottom part of the system has two elements namely hydrogen and oxygen have chemically combined where as in the upper part of the system the (ii) elements appear as a mixture of two individual gases. Invariable in chemical aggregation means that the state of chemical combination of the system should not change with time. Thus the mixture of hydrogen and oxygen, if it is changing into steam during the time the system was under consideration, then the systems chemical

Dr. T.N. Shridhar, Professor, NIE, Mysore

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aggregation is varying with time and hence this system is not a pure substance. Thus the system (i) is a pure substance where as the systems (ii) and (iii) are not pure substances. The Two Property Rule for a Pure Substance The thermodynamics state of a pure substance of a given mass can be fixed by specifying two independent properties provided (i) the substance is in equilibrium and (ii) the effects of gravity, motion, capillarity, electricity and magnetism are negligible.

The above rule indicates that if the values of two properties of a pure substance are fixed then the values for all other properties are fixed. This means that there is a definite relation between the two independent properties and each of the other properties. Each of these relations is called “Equation of state” for a pure substance. The equation of state for a pure substance can be in any one of the following forms: (i) Algebraic equation (example: perfect gas equation), (ii) Tables (example: steam tables) and (iii) Charts (example: Mollier chart for steam). Specific heat, C When interaction of heat takes place between a closed system and its surroundings, the internal energy of the system changes. If δQ is the amount of heat transferred to raise the temperature of 1 kg of substance by dT, then, specific heat C = δQ/dT

As we know, the specific heat of gas depends not only on the temperature but also upon the type of the heating process. i.e., specific heat of a gas depends on whether the gas is heated under constant volume or under constant pressure process.  We have dQ = m CV. dT for a rev. non-flow process at constant volume and dQ = m Cp. dT for a rev. non-flow process at constant pressure For a perfect gas, Cp & C V are constant for any one gas at all pressure and temperatures. Hence, integrating above equations. Flow of heat in a rev. constant pressure process = m Cp (T2 – T1) Flow of heat in a rev. constant volume process = m CV (T2 – T1) The internal energy of a perfect gas is a function of temperature only. i.e, u = f (T), to evaluate this function, let 1 kg of gas be heated at constant volume From non-flow energy equation, δQ = dU + δW δW = 0 since volume remains constant δQ = dU = CV. dT Int. U = CVT + k where k is a constant For mass m, Int. energy = m CVT Any process between state 1 to state 2, Change in int. energy = m CV (T2 – T1) (U2 – U1) = m CV (T2 – T1) We can also find the relationship between Cp & CV & shown that C  p  R rR  γ ; Cp – Cv = R ; C v  & C P  C V  r  1  r  1

Dr. T.N. Shridhar, Professor, NIE, Mysore

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Enthalpy: Consider a system undergoing a quasi equilibrium constant pressure process. We st have from 1 law of thermodynamics for a non-flow process, Q1-2 = U2 – U1 + W1-2 W1-2 =  12 pdv Since pressure is constant W1-2 = p ( V2 – V1) Q1-2 = U2 – U1 + p ( V2 – V1) = (U2 + p2V2) – (U1 + p1V1) i.e., heat transfer during the process is given in terms of the change in the quantity (U + pV) between initial and final states. Therefore, it find more convenient in thermodynamics to define this sum as a property called Enthalpy (H) i.e., H = U + pV In a constant pressure quasi equilibrium process, the heat transfer is equal to the change in enthalpy which includes both the change in internal energy and the work for this particular process.

The enthalpy of a fluid is the property of the fluid, since it consists of the sum of a property and the product of the two properties. Since enthalpy is a property, like internal energy, pressure, specific volume and temperature, it can be introduced into any problem whether the process is a flow or a non-flow process. For a perfect gas, we have h = u + pV = CV T + RT = (CV + R) T = CpT i.e., h = CpT & H = mCpT For any process, δQ = dH = mCpdT For a process between states 1 & 2 Change in enthalpy = (H2 – H1) = mCp (T2 – T1) Specific heat at Constant Volume: st When heat interaction takes place at constant volume, δW = 0 and from 1 law of  thermodynamics, for unit mass, (δq)V = dU

The amount of heat supplied or removed per degree change in temperature, when the system is kept under constant volume, is called as the specific heat at constant volume, Or

CV =

 δQ    dU   dT   dT   V  V 

Or dU = CV dT Specific heat at Constant pressure When heat interaction is at constant pressure, (δq)p = dh

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The amount of heat added or removed per degree change in temperature, when the system is kept under constant pressure, is called as the specific heat at constant pressure. Or Cp =

 δQ    dh   dT   dT   p  p

Or dh = Cp. dT st

Application of 1 law of thermodynamics to non-flow or closed system: a) Constant volume process (V = constant) st Applying 1 law of thermodynamics to the process, Q1-2 = U2 – U1 + W1-2 = U2 – U1 + 0 i.e., Q1-2 = CV (T2 – T1) For mass ‘m’ of a substance, Q = mCV (T2 – T1) b) Constant pressure (p = Constant) st Applying 1 law of thermodynamics to the process, Q1-2 = u2 – u1 + W1-2 The work done, W1-2 =  12 p d V = p ( V2 – V1) i.e., Q1-2 = u2 – u1 + p ( V2 – V1) = (u2 + pV2) – (u1 + pV1) = h2 – h1 i.e., Q = Cp (T2 – T1) For mass ‘m’ of a substance, Q = mCp (T2 – T1) c) Constant temperature process (Isothermal process, T = constant) st Applying 1 law of thermodynamics to the process, Q1-2 = U2 – U1 + W1-2 = CV (T2 – T2) + W1-2 i.e., Q1-2 = W1-2  T  1  T  2 Q1-2 = p1V1 lnV2 /V1 = p1 V1 ln p1 /p2 γ 

d) Reversible adiabatic process (pV ) = constant Applying 1st law of thermodynamics to the process, Q1-2 = U2 – U1 + W1-2 O = u2 – U1 + W1-2 --- (1)  p V    p 2V 2 Or (U1 – U2) = 1 1 γ   1

(U1 – U2) =

 R T 1  T 2 

γ   1

The above equation is true for an adiabatic process whether the process is reversible or not. In an adiabatic experiment, the work done W1-2 by the fluid is at the expense of a reduction

Dr. T.N. Shridhar, Professor, NIE, Mysore

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in the internal energy of the fluid. Similarly in an adiabatic composition process, all the work done on the fluid goes to increase the internal energy of the fluid. 

To derive pV = C: For a reversible adiabatic process We have δq = du + δu For a reversible process, δw = p d V δq = du+ p dV = O  For an adiabatic process δq = 0  RT  Also for a perfect gas, pV = RT or p = V  dV  dU + RT V  Also, u = CV T or du = CV dT dV  CV dT + RT V  dT  dV   R 0 or CV T  V  Int., CV ln T + R ln V = constant Sub. T = pV/R Pv Cv ln  R ln v = constant  R  pV   R  Or ln ln V  = constant  R C V   R

 R or  γ   1 C V 

Also, CV =

ln

 pV 

ln

 (γ   1) ln V  = constant

 R  pV   R

 γ   1

 ln V γ 1 = constant γ  1

or ln

 pVxV   R

= constant

r 

i.e., ln or

 pV 

 pV r 

= constant

 R

constant

=e

= constant  R i.e., pV = constant we have pV = RT  RT  orp= V   sub. This value of p in pV = C

Dr. T.N. Shridhar, Professor, NIE, Mysore

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 RT   V =C V 

Also, V =

 RT  P

-1

or TV

= constant

--- (a)



sub. This in equation pressure = C γ 

  RT    = constant p     p   

2

T   p

γ 1

= constant

T 

or  p

 γ 1   γ      

= constant --- (b)

For a reversible adiabatic process for a perfect gas between states 1 & 2, we can write 



p1V1 = p2V2 -1

T1V1 T 1 r 1

 p1

= T2V2

 r 

T 2 r 1

 p 2

or

 p 2  p1

-1

 V      1   V 2   or

or r 

T 2 T 1

T 2 T 1

r 

--- (c)

 V      1   V 2  

  p     2    p1  

γ  1

--- (d)

r 1 r 

--- (e)

The work done in an adiabatic process is W = u1 – u2 The gain in I.E. of a perfect gas, is u2 – u1 = CV (T2 – T1) W = CV (T1 – T2)  R But CV = γ   1

W =

 R(T 1  T 2 )

Using pV = RT,

γ   1 W=

  p 2V 2 γ   1

 p1V 1

n

e) Poly tropic process (pV = constant) st Applying 1 law of thermodynamics, Q1-2 = u2 – u1 + W1-2  R T 1  T 2  = (u2 – u1) + n 1  R T 1  T 2  i.e., Q = - CV (T1 – T2) n 1  R  γ   n W  Also CV = sub. & simplifying Q =   γ   1   n  1  In a poly tropic process, the index n depends on the heat and work quantities during the process.

Dr. T.N. Shridhar, Professor, NIE, Mysore

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3

0

9. A cylinder contains 0.45 m of a gas at 1 bar & 80 C. The gas is compressed to a volume of  3 0.13 m , the final pressure being 5 bar. Determine i) the mass of the gas, ii) the value of  index ‘n’ for composition, iii) the increase in internal energy of the gas and iv) the heat received or rejected by the gas during compression. (Take  = 1.4, R = 294.2 J/kg-K). 3 5 3 Solution: V1 = 0.45 m p1 = 1 x 10 Pa V2 = 0.13 m T1 = 353 K 5 p2 = 5 x 10 Pa 1 x10 5 x 0.45 i) We have p1V1 = mRT1 m = = 0.433 kg 294.2 x353 n

 V     p n n ii) p1V1 = p2V2 i.e.,  1   2  V 2    p1 n 0.45     5  Or      n = 1.296  0.13   1  n 1 0.296 T 2  V 1   0.45       T2 = 509.7 K iii)  T 1  V 2    0.13  Increase in int. energy, U = mCv (T2 – T1)  R

(T2 – T1) r  1 294.2 = 0.433 x  509.7  353 1.4  1 = 49.9 kJ = 0.433 x

iv) We have Q = U + W  p V    p 2V 2 mR T 1  T 2  0.433 294.27  353  509.7  W= 1 1 = = n 1 n 1 0.296 = - 67.44 kJ Q = 49.9 – 67.44 = - 17.54 kJ Heat rejected = - 17.54 kJ 10. The properties of a certain fluid are related as follows U = 196 + 0.718 t 0 pv = 0.287 (t + 273) where u is the sp. Internal energy (kJ/kg), t is in C, 2 3 p is pressure (kN/m ) and v is sp. Volume (m  /kg). For this fluid, find Cv & Cp  du   du Solution: By definition sp. Heat at constant volume Cv =     dt   v dt 

CV =

d 

(196 + 0.718 t) dt  0 = 0.718 kJ/kg C  dh   d  u   pv  Also, Cp =     dt   p dt 

Dr. T.N. Shridhar, Professor, NIE, Mysore

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= =



du dt  d  dt  d 



d  dt 

 pv 

196  0.718t  

d  dt 

 0.287t  0.287 x 273

 0.718t  0.287t  dt  0 = 1.005 kJ/kg C 11. A fluid system consisting of 4.17 kg of a pure substance has an energy E of 85 kJ. The kinetic energy of the system is 17 kJ and its gravitational potential energy is 5 kJ. The system undergoes an adiabatic process in which the final sp. i.e., is 150 kJ/kg, the final kinetic energy is 1.9 kJ and the final gravitational potential energy is 1.1 kJ. The effects due to electricity, capillary and magnetism are assumed to be absent. a) Evaluate the initial value of  the sp. i.e., of the fluid. b) Determine the magnitude and sign of the work done during the process. Solution: Total initial energy E1 =KE1 + PE1 +U1 85 = 17 + 5 + U1 U1 = 63 kJ 63 Initial sp. i.e., =  15.108kJ  / kg 4.17 Final state: E2 = kε2 + Pε2 + U2 = 1.9 + 1.1 + 4.17 (150) = 628.5 kJ st From 1 law, δQ = E2 – E1 + W 0 = 628.5 – 63 + W = - 565.5 kJ 12. A mass of 0.2 kg of a pure substance at a pressure of 1 bar and a temperature of 313 k  3 occupies a volume of 0.15 m . Given that the int. energy of the substance is 31.5 kJ, evaluate the sp. Enthalpy of the substance. 5 2 Solution: m = 0.2 kg P = 1 x 10 N/m T = 313 k v = 0.15 v = 31.5 kJ We have, enthalpy = U + Pv 3 5 = 31.5 x 10 + 1 x 10 x 0.15 = 46.5 kJ sp. Enthalpy = 46.5/0.2 = 232.5 kJ/kg 3

13. A gas enters a system at an initial pressure of 0.45 MPa and flow rate of 0.25 m /s and leaves 3 at a pressure of 0.9 MPa and 0.09 m  /s. During its passage through the system the increase in i.e., is 20 kJ/s. Find the change of enthalpy of the medium. 6 3 Solution: p1 = 0.45 x 10 Pa V1 = 0.25 m  /s 6 3 p2 = 0.9 x 10 Pa V2 = 0.09 m  /s 3 (u2 – u1) = 20 x 10 J/s st We have from 1 law for a constant pressure quasi static process Q1-2 = (u2 + p2V2) – (u1 + p1V1) = (H2 – H1) = Change in enthalpy = (u2 – u1) + p2V2 – p1V1 3 6 6 = 20 x 10 + 0.9 x 10 x 0.09 – 0.45 x 10 x 0.25

Dr. T.N. Shridhar, Professor, NIE, Mysore

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(H2 – H1) = - 11.5 kJ/s There is a decrease in enthalpy during the process 0

14. A closed system of constant volume experiences a temperature rise of 20 C when a certain process occurs. The heat transferred in the process is 18 kJ. The specific heat at constant 0 volume for the pure substance comprising the system is 1.2 kJ/kg C, and the system contains 2 kg of this substance. Determine the change in the internal energy and the work done. 0 0 U= ? Solution: T = 20 C Q = + 18 kJ Cv = 1.2 kJ/kg C m = 2kg W=? Change in int. energy, v = mCv T = 2 (1.2) (20) = 48 kJ st From 1 law of thermodynamics Q = U + W + 18 = 48 + W W = - 30 kJ 3

15. The stationary mass of gas is compressed without friction from an initial state of 2 m and 2 x 5 2 3 5 2 10 N/m to a final state of 1 m and 2 x 10 N/m , the pressure remaining the same. There is a transfer of 360 kJ, of heat from the gas during the process. How much does the internal energy of the gas change? 3 3 Solution: p1 = p2 = 2 bar V1 = 2m V2 = 1m Q = -360 kJ U = ? 5 5 W = pdV = 2 x 10 (1-2) = - 2 x 10 J st From 1 law of thermodynamics, Q = U + W 5 - 360000 = U – 2 x 10 U = - 160 kJ 16. The internal energy of a certain substance is given by the following equation 3 u = 3.56 pv + 84 where u is given in kJ/kg, p is in KPa and v in m /kg. A system composed of 3 kg of this substance expands from an initial pressure of 500 KPa and a volume of 0.22 3 1.2 m to a final pressure of 100 KPa in a process pv = constant. i) If the expansion is quasistatic, find Q, U, and W for the process. ii) In another process the same system expands according to the same pressure volume relationship as in part (i) and from the same initial state to the same final state as in part (i) but the heat transfer in this case is 30 kJ. Find the work transfer for this process. iii) Explain the difference in work transfer in parts (i) and (ii). 3 Solution: internal energy equation is, u = pv + 84, V1 = 0.22 m p1 = 500 kPa p2 = 100 kPa, 1.2 Process is pv = C i) u = 3.56 pv + 84 u = u2 – u1 = 3.56 (p2 v2 – p1 v1) per kg U = U2 – U1 = 3.56 (p2 V2 – p1 V1) for 3 kg 1.2 1.2 V2 = (p1 /p2)1/1.2 V1 = 0.8412 m3 We have p1 V1 = p2 V2 U = 3.56 (100 x 103 x 0.8412 – 500 x 103 x 0.22) = - 92.134 kJ  p V    p 2V 2 For a quasi static process, W = pdv = 1 1 n 1  100 x10 3 x0.8412  500 x10 3 x0.22



0.2

= + 129.4 kJ

Dr. T.N. Shridhar, Professor, NIE, Mysore

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From 1st law of thermodynamics, Q = U + W = - 92.134 + 129.4 = 37.27 kJ ii) Here Q = 30 kJ Q = U + W W = 122.134 kJ 30 = -92.134 + W iii) The work in (ii) is not equal to pdv since the process is not quasi-static. 17. A fluid is contained in a cylinder by a spring-loaded, frictionless piston so that the pressure in the fluid is a linear friction of the volume (p = a + bv). The internal energy of the fluid is 3 given by the following equation, U = 34 + 3.15 pV where U is in kJ, p is in kPa and V in m . 3 If the fluid changes from an initial state of 170 kPa, 0.03 m to a final state of 400 kPa, 0.06 3 m , with no work other than that on the piston, find the direction and magnitude of the work  and heat transfer. Solution: Change in internal energy of the fluid during the process,  U U2 – U1 = 3.15 (p2V2 – p1V1) = 3.15 (400 x 0.06 – 170 x 0.03) = 59.54 kJ Now p = a + bV Or 170 = a + b (0.03) 400 = a + b (0.06) Solve above two equations 2 230 = b (0.03) b = 7666.67 kN/m a = - 60 kN/m2 Work transfer involved during the process W1-2 =  12 pdv =  12 (a + bV) dV = a (V2 – V1) +



2

b. V 2

 V 12 

2

 0.06 = - 60 (0.06-0.03) + 7666.67

2

 0.03 2  2

= 8.55 kJ From 1 law of thermodynamics Q1-2 = (U2 – U1) + W1-2 = 59.54 + 8.55 = 68.09 kJ i.e., heat flow into the system during the process. st

18. A piston cylinder arrangement has a gas in the cylinder space. During a constant pressure expansion to a larger volume the work effect for the gas are 1.6 kJ, the heat added to the gas and cylinder arrangement is 3.2 kJ and the friction between the piston and cylinder wall amounts to 0.24 kJ. Determine the change in internal energy of the entire apparatus. (Gas, cylinder, piston). Solution: W1-2 = 1.6 kJ Q1-2 = 3.2 kJ (Q)f  = - 0.24 U = ? Q1-2 = U + W - Qf  3.2 = U + 1.6 – 0.24 U = 1.84 kJ 3

19. A system receives 42 kJ of heat white expanding with volume change of 0.123 m against an 2 atmosphere of 12 N/m . A mass of 80 kg in the surroundings is also lifted a distance of 6 mts.

Dr. T.N. Shridhar, Professor, NIE, Mysore

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i) Find the change in energy of the system. ii) The system is returned to its initial volume by an adiabatic process which require 100 kJ of work. Find the change in energy of the system. iii) Determine the total change in energy of the s ystem. 3 4 2 Solution: Q = 42 kJ, V = 0.123 m , p = 12 x 10 N/m m = 80 kg d = 6 mt W during adiabatic process = - 100 kJ i) Q = E + W Now, W = pV + W 4 = 1.2 x 10 x 0.123 + 80 (9.81) (6) = 19.469 kJ E = Q – W = 42 – 19.469 = 22.531 kJ ii) Q = 0, W = - 100 kJ Q = E + W 0 = E – 100 E = 100 kJ iii) Total change in energy of the system, E = Q – W (E)i + (E)ii = 122.531 kJ = 42 – [(-100) + 22.531] = 119.47 kJ 20. A thermally insulated battery is being discharges at atmosphere pressure and constant volume. During a 1 hr test it is found that a current of 50A and 2v flows while the 0 0 temperature increases from 20 C to 32.5 C. Find the change in internal energy of the cell during the period of operation. Solution: Q = 0 We = 50 x 2 x 3600 4 = 36 x 10 J Q = U + We 4 0 = U + 36 x 10 U = - 36 x 104 Joules

Dr. T.N. Shridhar, Professor, NIE, Mysore

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