thermodynamics solution

Dr. T.N. Shridhar, Professor, NIE, Mysore 1

1. a) An insulated rigid vessel contains some powdered coal and air at a pressure of 10 bar and atemperature of 200 C. The coal is ignited, there results a rise in the pressure and temperatureof the contents of the vessel, the final temperature is 5380 C. Taking the vessel and thecontents to be the system under consideration evaluate the increase in the energy of thesystem. b) The insulation is now removed. A heat transfer. of 50 kJ from the system causesthe temperature to fall to the initial value, 200 C. Evaluate the increase in the energy of thesystem during this process. c) Taking the initial energy of the system 32 kJ, write down theenergy values after process (a) and after process (b)

Solution: a) The vessel is rigid i.e., dv = 0 W = 0System boundary is insulated, Q = 0By 1st law Q = E + W E = 0

(Though there is a change in temperature and pressure of the system, there is no change inenergy. All that happens is chemical energy of the system is converted into internal energy).b) W = 0 ; Insulation removed: Q = - 50 kJ

-50 = 0 + E E = -50 kJ or decrease in energy = 50 kJc) During process (a) E = 0 Initial energy = energy after process = 32 kJFor process (b) initial energy = Final energy in process (a) = 32 kJE for process (b) = -50 kJi.e., Efinal Einitial = - 50 kJEfinal = -50 + 32

= 18 kJ

2. On a warm summer day, a housewife decides to beat the heat by closing the windows anddoors in the kitchen and opening the refrigerator door. At first she feels cool and refreshed,but after a while the effect begins to wear off. Evaluate the situation as it relates to first law,considering the room including the refrigerator as the system.

Solution:

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 non-conducting, Q = 0. For the operation of refrigerator, electricity is supplied from outside andhence electrical work We = is done on the system.From first law of TD Q = E + We

0 = E - We E = WePositive sign of energy indicates the increase in energy of the system with time. As the energy isincreasing, the temperature of air increases and hence effect of coolness gradually begins to wearoff.

.

.

+

-

We

Room

Ref

System


3. The average heat transfer from a person to the surroundings when he is not actively workingis about 950 kJ/hr. suppose that in the auditorium containing 1000 people the ventilationsystem fails. a) How much does the internal energy of air in the auditorium increase duringthe first 15 minutes after the ventilation fails? b) Considering the auditorium and all thepeople 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 airincreases?

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 kJAverage heat transfer for 15 min in the auditorium containing 1000 people

Q = 237.5 x 1000 = 237500 kJ/minFrom first law of TD, we have Q = E + W

237500 = E + 0E = 237.5 MJ

b) Considering the auditorium and all the people as system,Q = 0; W = 0Q = E + W0 = E + 0 E = 0

Increase in internal energy of the air due to increase in its temperature is compensated by thedecrease in internal energy of the people.

4. A household refrigerator is loaded with fresh food and closed. Consider the wholerefrigerator and the contents as a system. The machine uses 1 kWhr of electrical energy incooling the food and the internal energy of the food (system) decreases by 5250 kJ, as thetemperature drops. Find the magnitude and direction of heat transfer during the process.

Solution:

Given: Electrical work, = 1 kWhrWe = 860 Kcal= 860 (4.187) = - 3600.8 kJ

Given, E = - 5250 kJFrom 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

Ref+Conte-nts

PowerSupply


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

Solution:

Process 1-2: Constant volume process i.e., dv = 0 i.e., W1-2 = 0Q1-2 = (E2 E1) + W1-285 = (E2 E1) + 0(E2 E1) = 85 kJ

But E1 = 100 kJ E2 -100 = 85 E2 = 185 kJProcess 2-3: (Constant pressure process)

Q2-3 = (E3 E2) + W2-3-90 = (E3 E2) 15 (E3 E2) = -75 kJ

But E2 = 185 kJ E3 = -75 + 185 = 110 kJIn a cyclic process WQ Q1-2 + Q2-3 + Q3-1 = W1-2 + W2-3 + W3-185 90 + 0 = 0 -15 + W3-1W3-1 = 10 kJ6. 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 and50 kJ of work is done by the system. during a constant volume process, 125 kJ of heat isrejected from the system. find the work interaction if a rev. adiabatic process restores thesystem to the initial state.

Solution: Process 1-2: Constant pressure processWe have Q12 = 125 W12 = + 50

Q12 = E2 E1 + W12125 = (E2 E1) + 50 (E2 E1) = 75 kJ

Process 2-3: Constant volume process, i.e., dv = 0 W23 = 0We have Q23 = -125

Q2-3 = (E3 E2) + W2-3-125 = (E3 E2) + 0 (E3 E2) = - 125 kJ

Q3-1 = 0For a cyclic process WQ

125 125 + 0 = 50 + 0 + W3-1

1

23

p

V


W3-1 = - 50 kJor Process 3-1: (Rev. Adiabatic process) dQ = 0Q3-1 = E1 E3 + W3-10 = (E1 E2) + (E2 E3) + W3-1

= -75 + 125 + W3-1 W3-1 = -50 kJ7. 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

Q (kJ)Work transfer

W(N-m)Change in internal

energy U (kJ)1-2 10 0 102-3 -25 15 x 103 -40 kJ3-4 60 23.5 kJ 36.5 kJ4-1 -15 -8.5 x 103 -6.5 kJ

Find the magnitude of the unknown quantities in kJ (Hint: For the process 3-4, use 0dU8. 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) Thestone is above to enter the water. b) The stone has just come to rest in the bucket and c) Heathas been transferred to the surroundings in such an amount that the stone and water are at thesame temperature they were initially.

Solution: The 1st law of THERMODYNAMICS. is

Q1-2 = U2 U1 + 2m (v22 v12) + mg (h2 h1) + W1-2 --- (1)

a) The stone is about to enter the waterAssuming no heat transfer. to or from the stone a sit falls,Q1-2 = 0 W1-2 = 0 U = 0Equation: 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 = 0Equation 1) reduce to pE = -U = mg (h2 h1) = - 1kJ

U = 1kJ & pE = - 1kJc) After heat has been transferred so that stone & water are at the same temperature they

were initially, U = 0 In this caseU = 0 kE= 0 W1-2 = 0Equation 1) reduces to Q1-2 = PE = mg (h2 h1) = -1kJ


The Pure SubstanceThe system encountered in thermodynamics is often quite less complex and consists of fluidsthat don not change chemically, or exhibit significant electrical, magnetic or capillary effects.These relatively simple systems are given the generic name the Pure Substance.

DefinitionA 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 issame as the composition of any other part. Homogeneous in chemical aggregation implies thatthe 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 withrespect to time.

(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 andwater. 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 fromsteam or water. The same is through for system (ii) consisting of water and uncombined mixtureof hydrogen and oxygen. System (iii) however is not homogeneous in chemical compositionbecause in the upper part of the system hydrogen and oxygen are present in the ratio 1:1 where asin the bottom portion they are present in the ratio 2:1.

System (i) also satisfies condition (ii), because both hydrogen and oxygen have combinedchemically 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 havechemically combined where as in the upper part of the system the (ii) elements appear as amixture of two individual gases.

Invariable in chemical aggregation means that the state of chemical combination of the systemshould not change with time. Thus the mixture of hydrogen and oxygen, if it is changing intosteam during the time the system was under consideration, then the systems chemical

Steam

Water

H2 + O2(Gas)

Water

H2 + O2(Gas)

Water


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 twoindependent 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 thevalues for all other properties are fixed. This means that there is a definite relation between thetwo independent properties and each of the other properties. Each of these relations is calledEquation of state for a pure substance. The equation of state for a pure substance can be in anyone 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, CWhen interaction of heat takes place between a closed system and its surroundings, the

internal energy of the system changes. If Q is the amount of heat transferred to raise thetemperature of 1 kg of substance by dT, then, specific heat C = Q/dT

As we know, the specific heat of gas depends not only on the temperature but also upon the typeof the heating process. i.e., specific heat of a gas depends on whether the gas is heated underconstant 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 pressureFor a perfect gas, Cp & CV 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 evaluatethis function, let 1 kg of gas be heated at constant volumeFrom non-flow energy equation, Q = dU + W

W = 0 since volume remains constantQ = dU = CV. dTInt. U = CVT + k where k is a constant

For mass m, Int. energy = m CVTAny 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

Cp Cv = R ; ;V

p

CC

1 rRCv & 1 r

rRCP


Enthalpy: Consider a system undergoing a quasi equilibrium constant pressure process. Wehave from 1st law of thermodynamics for a non-flow process,Q1-2 = U2 U1 + W1-2W1-2 = 21 pdvSince 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 definethis sum as a property called Enthalpy (H)

i.e., H = U + pVIn a constant pressure quasi equilibrium process, the heat transfer is equal to the change inenthalpy which includes both the change in internal energy and the work for this particularprocess.

The enthalpy of a fluid is the property of the fluid, since it consists of the sum of a property andthe 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 aflow 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 = mCpTFor any process, Q = dH

= mCpdTFor a process between states 1 & 2

Change in enthalpy = (H2 H1) = mCp (T2 T1)

Specific heat at Constant Volume:When heat interaction takes place at constant volume, W = 0 and from 1st law ofthermodynamics, for unit mass, (q)V = dU

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

Or CV =VdT

Q

VdT

dU

Or dU = CV dT

Specific heat at Constant pressureWhen heat interaction is at constant pressure, (q)p = dh


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

Or Cp =pdT

Q

pdT

dh

Or dh = Cp. dT

Application of 1st law of thermodynamics to non-flow or closed system:a) Constant volume process (V = constant)

Applying 1st law of thermodynamics to the process,Q1-2 = U2 U1 + W1-2

= U2 U1 + 0i.e., Q1-2 = CV (T2 T1)

For mass m of a substance, Q = mCV (T2 T1)

b) Constant pressure (p = Constant)Applying 1st law of thermodynamics to the process,Q1-2 = u2 u1 + W1-2The work done, W1-2 = 21 p dV = p (V2 V1)i.e., Q1-2 = u2 u1 + p (V2 V1) = (u2 + pV2) (u1 + pV1)

= h2 h1i.e., Q = Cp (T2 T1)For mass m of a substance, Q = mCp (T2 T1)

c) Constant temperature process (Isothermal process, T = constant)Applying 1st law of thermodynamics to the process,

Q1-2 = U2 U1 + W1-2= CV (T2 T2) + W1-2

i.e., Q1-2 = W1-2 21 TT Q1-2 = p1V1 lnV2/V1

= p1 V1 ln p1/p2

d) Reversible adiabatic process (pV) = constantApplying 1st law of thermodynamics to the process,

Q1-2 = U2 U1 + W1-2O = u2 U1 + W1-2 --- (1)

Or (U1 U2) = 12211

VpVp

(U1 U2) = 1 21

TTR

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


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

To derive pV = C: For a reversible adiabatic process

We have q = du + uFor a reversible process, w = p dVq = du+ p dV

= O For an adiabatic process q = 0Also for a perfect gas, pV = RT or p =

VRT

dU + RTVdV

Also, u = CV T or du = CV dT

CV dT + RT VdV

or CV 0 VdVR

TdT

Int., CV ln T + R ln V = constantSub. T = pV/R

Cv ln vRRPv ln = constant

Or ln VCR

RpV

V

ln = constant

Also, CV = 11 VC

Ror

R

ln VR

pV ln)1( = constant

ln 1ln VR

pV= constant

or lnR

pVxV 1= constant

i.e., lnR

pV r= constant

orR

pV r= econstant = constant

i.e., pV = constant

we have pV = RT

or p =VRT

sub. This value of p in pV = C


VRT V = C or TV-1 = constant --- (a)

Also, V =P

RTsub. This in equation pressure = C

p

pRT

= constant

12

pT

= constant or

1

p

T= constant --- (b)

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

p1V1 = p2V2 orr

VV

pp

2

1

1

2--- (c)

T1V1-1 = T2V2-1 or1

2

1

1

2

VV

TT

--- (d)

rr

rr

p

T

p

T1

2

21

1

1 or

r

r

pp

TT

1

1

2

1

2

--- (e)

The work done in an adiabatic process is W = u1 u2The gain in I.E. of a perfect gas, is u2 u1 = CV (T2 T1)

W = CV (T1 T2)But CV = 1

R

W =1

)( 21

TTR

Using pV = RT, W =1

2211

VpVp

e) Poly tropic process (pVn = constant)Applying 1st law of thermodynamics, Q1-2 = u2 u1 + W1-2

= (u2 u1) + 1 21

n

TTR

i.e., Q = 1

21

n

TTR- CV (T1 T2)

Also CV = 1R

sub. & simplifying Q = Wn

n

1

In a poly tropic process, the index n depends on the heat and work quantities during theprocess.


9. A cylinder contains 0.45 m3 of a gas at 1 bar & 800C. The gas is compressed to a volume of0.13 m3, the final pressure being 5 bar. Determine i) the mass of the gas, ii) the value ofindex n for composition, iii) the increase in internal energy of the gas and iv) the heatreceived or rejected by the gas during compression. (Take = 1.4, R = 294.2 J/kg-K).

Solution: V1 = 0.45 m3 p1 = 1 x 105 Pa V2 = 0.13 m3 T1 = 353 Kp2 = 5 x 105 Pa

i) We have p1V1 = mRT1 m = 3532.29445.0101 5

x

xx= 0.433 kg

ii) p1V1n = p2V2n i.e.,1

2

2

1

pp

VV

n

Or

15

13.045.0 n n = 1.296

iii)296.01

2

1

1

2

13.045.0

n

VV

TT T2 = 509.7 K

Increase in int. energy, U = mCv (T2 T1)= 0.433 x

1rR (T2 T1)

= 0.433 x 3537.50914.12.294

= 49.9 kJiv) We have Q = U + W

W =1

2211

n

VpVp=

1

21

n

TTmR=

296.0

7.50935327.294433.0

= - 67.44 kJQ = 49.9 67.44 = - 17.54 kJHeat rejected = - 17.54 kJ

10. The properties of a certain fluid are related as followsU = 196 + 0.718 tpv = 0.287 (t + 273) where u is the sp. Internal energy (kJ/kg), t is in 0C,

p is pressure (kN/m2) and v is sp. Volume (m3/kg).For this fluid, find Cv & Cp

Solution: By definition sp. Heat at constant volume Cv = dtdu

dtdu

v

CV = dtd (196 + 0.718 t)

= 0.718 kJ/kg0C

Also, Cp = pvudtd

dtdh

p


= pvdtd

dtdu

= tdtd 718.0196 273287.0287.0 xt

dtd

ttdtd 287.0718.0

= 1.005 kJ/kg0C

11. A fluid system consisting of 4.17 kg of a pure substance has an energy E of 85 kJ. Thekinetic energy of the system is 17 kJ and its gravitational potential energy is 5 kJ. The systemundergoes an adiabatic process in which the final sp. i.e., is 150 kJ/kg, the final kineticenergy is 1.9 kJ and the final gravitational potential energy is 1.1 kJ. The effects due toelectricity, capillary and magnetism are assumed to be absent. a) Evaluate the initial value ofthe sp. i.e., of the fluid. b) Determine the magnitude and sign of the work done during theprocess.

Solution: Total initial energy E1 =KE1 + PE1 +U185 = 17 + 5 + U1 U1 = 63 kJ

Initial sp. i.e., = kgkJ /108.1517.4

63 Final state: E2 = k2 + P2 + U2 = 1.9 + 1.1 + 4.17 (150) = 628.5 kJFrom 1st law, Q = E2 E1 + W

0 = 628.5 63 + W = - 565.5 kJ12. A mass of 0.2 kg of a pure substance at a pressure of 1 bar and a temperature of 313 k

occupies a volume of 0.15 m3. Given that the int. energy of the substance is 31.5 kJ, evaluatethe sp. Enthalpy of the substance.

Solution: m = 0.2 kg P = 1 x 105 N/m2 T = 313 k v = 0.15 v = 31.5 kJWe have, enthalpy = U + Pv

= 31.5 x 103 + 1 x 105 x 0.15= 46.5 kJ

sp. Enthalpy = 46.5/0.2 = 232.5 kJ/kg13. A gas enters a system at an initial pressure of 0.45 MPa and flow rate of 0.25 m3/s and leaves

at a pressure of 0.9 MPa and 0.09 m3/s. During its passage through the system the increase ini.e., is 20 kJ/s. Find the change of enthalpy of the medium.

Solution: p1 = 0.45 x 106 Pa V1 = 0.25 m3/sp2 = 0.9 x 106 Pa V2 = 0.09 m3/s(u2 u1) = 20 x 103 J/s

We have from 1st law for a constant pressure quasi static process

Q1-2 = (u2 + p2V2) (u1 + p1V1)= (H2 H1)= Change in enthalpy

= (u2 u1) + p2V2 p1V1= 20 x 103 + 0.9 x 106 x 0.09 0.45 x 106 x 0.25


(H2 H1) = - 11.5 kJ/sThere is a decrease in enthalpy during the process14. A closed system of constant volume experiences a temperature rise of 200C when a certain

process occurs. The heat transferred in the process is 18 kJ. The specific heat at constantvolume for the pure substance comprising the system is 1.2 kJ/kg0C, and the system contains2 kg of this substance. Determine the change in the internal energy and the work done.

Solution: T = 200C Q = + 18 kJ Cv = 1.2 kJ/kg0C m = 2kg U= ? W = ?Change in int. energy, v = mCv T

= 2 (1.2) (20) = 48 kJFrom 1st law of thermodynamics Q = U + W

+ 18 = 48 + WW = - 30 kJ

15. The stationary mass of gas is compressed without friction from an initial state of 2 m3 and 2 x105 N/m2 to a final state of 1 m3 and 2 x 105 N/m2, the pressure remaining the same. There isa transfer of 360 kJ, of heat from the gas during the process. How much does the internalenergy of the gas change?

Solution: p1 = p2 = 2 bar V1 = 2m3 V2 = 1m3 Q = -360 kJ U = ?W = pdV = 2 x 105 (1-2) = - 2 x 105 J

From 1st law of thermodynamics, Q = U + W- 360000 = U 2 x 105

U = - 160 kJ16. The internal energy of a certain substance is given by the following equation

u = 3.56 pv + 84 where u is given in kJ/kg, p is in KPa and v in m3/kg. A system composedof 3 kg of this substance expands from an initial pressure of 500 KPa and a volume of 0.22m3 to a final pressure of 100 KPa in a process pv1.2 = constant. i) If the expansion is quasi-static, find Q, U, and W for the process. ii) In another process the same system expandsaccording to the same pressure volume relationship as in part (i) and from the same initialstate to the same final state as in part (i) but the heat transfer in this case is 30 kJ. Find thework transfer for this process. iii) Explain the difference in work transfer in parts (i) and (ii).

Solution: internal energy equation is, u = pv + 84, V1 = 0.22 m3 p1 = 500 kPa p2 = 100 kPa,Process is pv1.2 = C

i) u = 3.56 pv + 84u = u2 u1 = 3.56 (p2 v2 p1 v1) per kgU = U2 U1 = 3.56 (p2 V2 p1 V1) for 3 kg

We have p1 V11.2 = p2 V21.2 V2 = (p1/p2)1/1.2 V1 = 0.8412 m3U = 3.56 (100 x 103 x 0.8412 500 x 103 x 0.22)= - 92.134 kJ

For a quasi static process, W = pdv =1

2211

n

VpVp

2.022.0105008412.010100 33 xxxx

= + 129.4 kJ


From 1st law of thermodynamics, Q = U + W= - 92.134 + 129.4= 37.27 kJ

ii) Here Q = 30 kJQ = U + W30 = -92.134 + W W = 122.134 kJ

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 isgiven by the following equation, U = 34 + 3.15 pV where U is in kJ, p is in kPa and V in m3.If the fluid changes from an initial state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06m

3, 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 + bVOr 170 = a + b (0.03)

400 = a + b (0.06)Solve above two equations230 = b (0.03) b = 7666.67 kN/m2a = - 60 kN/m2Work transfer involved during the process W1-2 = 21 pdv = 21 (a + bV) dV= a (V2 V1) + 2.

21

22 VVb

= - 60 (0.06-0.03) + 7666.67 2

03.006.0 22

= 8.55 kJFrom 1st 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.

18. A piston cylinder arrangement has a gas in the cylinder space. During a constant pressureexpansion to a larger volume the work effect for the gas are 1.6 kJ, the heat added to the gasand cylinder arrangement is 3.2 kJ and the friction between the piston and cylinder wallamounts 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 - Qf3.2 = U + 1.6 0.24U = 1.84 kJ

19. A system receives 42 kJ of heat white expanding with volume change of 0.123 m3 against anatmosphere of 12 N/m2. A mass of 80 kg in the surroundings is also lifted a distance of 6 mts.


i) Find the change in energy of the system. ii) The system is returned to its initial volume byan 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 system.

Solution: Q = 42 kJ, V = 0.123 m3, p = 12 x 104 N/m2 m = 80 kgd = 6 mt W during adiabatic process = - 100 kJ

i) Q = E + WNow, W = pV + W

= 1.2 x 104 x 0.123 + 80 (9.81) (6)= 19.469 kJ

E = Q W = 42 19.469 = 22.531 kJii) Q = 0, W = - 100 kJ

Q = E + W0 = E 100

E = 100 kJiii) 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 constantvolume. During a 1 hr test it is found that a current of 50A and 2v flows while thetemperature increases from 200C to 32.50C. Find the change in internal energy of the cellduring the period of operation.

Solution: Q = 0 We = 50 x 2 x 3600= 36 x 104 J

Q = U + We0 = U + 36 x 104

U = - 36 x 104 Joules

thermodynamics solution

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