Advanced Thermodynamics: Lecture 4 - IIT Bombay · 2015. 8. 26. · Advanced Thermodynamics: Lecture 4 Shivasubramanian Gopalakrishnan [email protected] Shivasubramanian Gopalakrishnan

Advanced Thermodynamics: Lecture 4

Shivasubramanian [email protected]

Shivasubramanian Gopalakrishnan [email protected] ME 661

Specific heats for solids and liquids

The specific heats of incompressible substances depend ontemperature only.

Enthalpy change for a solid or an incompressible liquid.

h = u + Pv

dh = du + vdP +��*0Pdv

�h = �u + v�P ⇡ cavg�T + v�P

Constant–pressure processes, as in heaters(�P = 0) �h = �u = cavg�T

Constant–temperature processes, as in pump(�T = 0) �h = v�P


Specific heats for solids and liquids

The specific heats of incompressible substances depend ontemperature only.

Enthalpy change for a solid or an incompressible liquid.

h = u + Pv

dh = du + vdP +��*0Pdv

�h = �u + v�P ⇡ cavg�T + v�P

Constant–pressure processes, as in heaters(�P = 0) �h = �u = cavg�T

Constant–temperature processes, as in pump(�T = 0) �h = v�P


Cooling of an Iron Block by Water

A 50–kg iron block at 80C is dropped into an insulated tank thatcontains 0.5 m3 of liquid water at 25C. Determine the temperaturewhen thermal equilibrium is reached.


Temperature Rise due to SlappingIf you ever slapped someone or got slapped yourself, you probably remember the burning sensation. Recall theinfamous slapgate incident between Harbhajan Singh and Sreesanth. Assuming the temperature of the a↵ected

area of Sreesanth’s face to rose by 1.8OC (ouch!) and that the Bhajji’s slapping hand has a mass of 1.2 kg andabout 0.150 kg of the tissue on the face and the hand is a↵ected by the incident, estimate the velocity of Bhajji’s

hand just before impact. Take the specific heat of the tissue to be 3.8 kJ/kg OC.

what is the cooling e↵ect produced by Sreesanth’s tears? Assuming the properties of tears to be same as water andthe Sreesanth’s tear production rate was 10 mL per minute.


Analysis of open systems

The conservation of mass principle for a control volume can beexpressed as: The net mass transfer to or from a control volumeduring a time interval �t is equal to the net change (increase ordecrease) in the total mass within the control volume during �t.

min �mout = �mCV

where �mCV = mfinal �minitial is the change in the mass of thecontrol volume during the process. In the rate form this is given as

min � mout =dmCV

dt


Conservation of mass

Rate of change of mass with CVdmCV

dt=

d

dt

Z

CV⇢dV

Di↵erential mass flow rate �m = ⇢VndA = r(Vcos✓)dA = ⇢(V ·n)dA

Net mass flow rate �m =

Z

CSm =

Z

CS⇢(V · n)dA

General COMd

dt

Z

CV⇢dV +

Z

CS⇢(V · n)dA = 0

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th editionShivasubramanian Gopalakrishnan [email protected] ME 661

Flow work and Flow energy

Control volumes involve mass flow across their boundaries, andsome work is required to push the mass into or out of the controlvolume. This work is known as the flow work, or flow energy, andis necessary for maintaining a continuous flow through a controlvolume.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition

Wflow = FL = PAL = PV

On a per unit mass basis

wflow = Pv KJ/Kg


Total Energy of a flowing fluid

The total energy of a simple compressible system consists of threeparts: internal, kinetic, and potential energies

e = u + ke + pe = u +V 2

2+ gz

The fluid entering or leaving a control volume possesses anadditional form of energy–the flow energy Pv, as already discussed.Then the total energy of a flowing fluid on a unit-mass basis(denoted by ✓) becomes

✓ = Pv + e = Pv + (u + ke + pe)

✓ = h + ke + pe = h +V 2

2+ gz


Steady Flow Process

A process during which a fluid flows through a control volumesteadily. That is, the fluid properties can change from point topoint within the control volume, but at any point, they remainconstant during the entire process.Mass balance for Steady Flow system

X

in

m =X

out

m

Ein � Eout =��*0dEsystem

dt

Qin + Win +X

in

m✓ = Qout + Wout +X

out

m✓

Qin+Win+X

in

m

✓h +

V 2

2+ gz

◆= Qout+Wout+

X

out

m

✓h +

V 2

2+ gz

◆


Water Flow through a Garden Hose Nozzle

A garden hose attached with a nozzle is used to fill a 10-gal bucket.The inner diameter of the hose is 2 cm, and it reduces to 0.8 cmat the nozzle exit. If it takes 50 s to fill the bucket with water,determine (a) the volume and mass flow rates of water through thehose, and (b) the average velocity of water at the nozzle exit.


Discharge of Water from a Tank

A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is opento the atmosphere is initially filled with water. Now the dischargeplug near the bottom of the tank is pulled out, and a water jetwhose diameter is 0.5 in streams out. The average velocity of thejet is given by V =

p2gh, where h is the height of water in the

tank measured from the center of the hole (a variable) and g is thegravitational acceleration. Determine how long it will take for thewater level in the tank to drop to 2 ft from the bottom.


Energy Transport by Mass

Steam is leaving a 4-L pressure cooker whose operating pressure is150 kPa. It is observed that the amount of liquid in the cooker hasdecreased by 0.6 L in 40 min after the steady operating conditionsare established, and the cross-sectional area of the exit opening is8 mm2. Determine (a) the mass flow rate of the steam and theexit velocity, (b) the total and flow energies of the steam per unitmass, and (c) the rate at which energy leaves the cooker by steam.


Deceleration of Air in a Di↵userAir at 10C and 80 kPa enters the di↵user of a jet engine steadilywith a velocity of 200 m/s. The inlet area of the di↵user is 0.4 m2.The air leaves the di↵user with a velocity that is very smallcompared with the inlet velocity. Determine (a) the mass flow rateof the air and (b) the temperature of the air leaving the di↵user.


Acceleration of Steam in a NozzleSteam at 250 psia and 700F steadily enters a nozzle whose inletarea is 0.2 ft2. The mass flow rate of steam through the nozzle is10 lbm/s. Steam leaves the nozzle at 200 psia with a velocity of900 ft/s. Heat losses from the nozzle per unit mass of the steamare estimated to be 1.2 Btu/lbm. Determine (a) the inlet velocityand (b) the exit temperature of the steam.


Compressing Air by a Compressor

Air at 100 kPa and 280 K is compressed steadily to 600 kPa and400 K. The mass flow rate of the air is 0.02 kg/s, and a heat lossof 16 kJ/kg occurs during the process. Assuming the changes inkinetic and potential energies are negligible, determine thenecessary power input to the compressor.


Power Generation by a Steam Turbine

The power output of an adiabatic steam turbine is 5 MW, and theinlet and the exit conditions of the steam are as indicated in figure.

1 Compare the magnitudes of �h, �ke, and �pe.

2 Determine the work done per unit mass of the steam flowingthrough the turbine.

3 Calculate the mass flow rate of the steam.


Expansion of Refrigerant-134a in a Refrigerator

Refrigerant-134a enters the capillary tube of a refrigerator assaturated liquid at 0.8 MPa and is throttled to a pressure of 0.12MPa. Determine the quality of the refrigerant at the final stateand the temperature drop during this process.


Mixing of Hot and Cold Waters in a Shower

Consider an ordinary shower where hot water at 140F is mixedwith cold water at 50F. If it is desired that a steady stream ofwarm water at 110F be supplied, determine the ratio of the massflow rates of the hot to cold water. Assume the heat losses fromthe mixing chamber to be negligible and the mixing to take placeat a pressure of 20 psia.


Cooling of Refrigerant-134a by Water

Refrigerant-134a is to be cooled by water in a condenser. Therefrigerant enters the condenser with a mass flow rate of 6 kg/minat 1 MPa and 70C and leaves at 35C. The cooling water enters at300 kPa and 15C and leaves at 25C. Neglecting any pressure drops,determine (a) the mass flow rate of the cooling water required and(b) the heat transfer rate from the refrigerant to water.


Electric Heating of Air in a House

The electric heating systems used in many houses consist of asimple duct with resistance heaters. Air is heated as it flows overresistance wires. Consider a 15-kW electric heating system. Airenters the heating section at 100 kPa and 17C with a volume flowrate of 150 m3/min. If heat is lost from the air in the duct to thesurroundings at a rate of 200 W, determine the exit temperature ofair.


Heat engines

They receive heat from a high-temperature source (solarenergy, oil furnace, nuclear reactor, etc.).

They convert part of this heat to work (usually in the form ofa rotating shaft).

They reject the remaining waste heat to a low-temperaturesink (the atmosphere, rivers, etc.).

They operate on a cycle.


Heat engines

6–3 ■ HEAT ENGINESAs pointed out earlier, work can easily be converted to other forms of energy,but converting other forms of energy to work is not that easy. The mechani-cal work done by the shaft shown in Fig. 6–8, for example, is first convertedto the internal energy of the water. This energy may then leave the water asheat. We know from experience that any attempt to reverse this process willfail. That is, transferring heat to the water does not cause the shaft to rotate.From this and other observations, we conclude that work can be converted toheat directly and completely, but converting heat to work requires the use ofsome special devices. These devices are called heat engines.

Heat engines differ considerably from one another, but all can be charac-terized by the following (Fig. 6–9):

1. They receive heat from a high-temperature source (solar energy, oil fur-nace, nuclear reactor, etc.).

2. They convert part of this heat to work (usually in the form of a rotatingshaft).

3. They reject the remaining waste heat to a low-temperature sink (theatmosphere, rivers, etc.).

4. They operate on a cycle.

Heat engines and other cyclic devices usually involve a fluid to and fromwhich heat is transferred while undergoing a cycle. This fluid is called theworking fluid.

The term heat engine is often used in a broader sense to include work-producing devices that do not operate in a thermodynamic cycle. Enginesthat involve internal combustion such as gas turbines and car engines fall intothis category. These devices operate in a mechanical cycle but not in athermodynamic cycle since the working fluid (the combustion gases) doesnot undergo a complete cycle. Instead of being cooled to the initial tempera-ture, the exhaust gases are purged and replaced by fresh air-and-fuel mixtureat the end of the cycle.

The work-producing device that best fits into the definition of a heatengine is the steam power plant, which is an external-combustion engine.That is, combustion takes place outside the engine, and the thermal energyreleased during this process is transferred to the steam as heat. Theschematic of a basic steam power plant is shown in Fig. 6–10. This is arather simplified diagram, and the discussion of actual steam power plantsis given in later chapters. The various quantities shown on this figure areas follows:

Qin ! amount of heat supplied to steam in boiler from a high-temperaturesource (furnace)

Qout ! amount of heat rejected from steam in condenser to a low-temperature sink (the atmosphere, a river, etc.)

Wout ! amount of work delivered by steam as it expands in turbineWin ! amount of work required to compress water to boiler pressure

Notice that the directions of the heat and work interactions are indicatedby the subscripts in and out. Therefore, all four of the described quantitiesare always positive.

282 | Thermodynamics

WATER

Heat

Work

WATER

Heat

No work

FIGURE 6–8Work can always be converted to heatdirectly and completely, but thereverse is not true.

Wnet,out

Low-temperatureSINK

Qout

Qin

HEATENGINE

High-temperatureSOURCE

FIGURE 6–9Part of the heat received by a heatengine is converted to work, while therest is rejected to a sink.

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Schematic of a Power Plant

The net work output of this power plant is simply the difference betweenthe total work output of the plant and the total work input (Fig. 6–11):

(6–1)

The net work can also be determined from the heat transfer data alone. Thefour components of the steam power plant involve mass flow in and out, andtherefore they should be treated as open systems. These components, togetherwith the connecting pipes, however, always contain the same fluid (not count-ing the steam that may leak out, of course). No mass enters or leaves this com-bination system, which is indicated by the shaded area on Fig. 6–10; thus, itcan be analyzed as a closed system. Recall that for a closed system undergoinga cycle, the change in internal energy !U is zero, and therefore the net workoutput of the system is also equal to the net heat transfer to the system:

(6–2)

Thermal EfficiencyIn Eq. 6–2, Qout represents the magnitude of the energy wasted in order tocomplete the cycle. But Qout is never zero; thus, the net work output of a heatengine is always less than the amount of heat input. That is, only part of theheat transferred to the heat engine is converted to work. The fraction of theheat input that is converted to net work output is a measure of the perfor-mance of a heat engine and is called the thermal efficiency hth (Fig. 6–12).

For heat engines, the desired output is the net work output, and therequired input is the amount of heat supplied to the working fluid. Then thethermal efficiency of a heat engine can be expressed as

(6–3)Thermal efficiency "Net work output

Total heat input

Wnet,out " Qin # Qout¬¬1kJ 2

Wnet,out " Wout # Win¬¬1kJ 2

Chapter 6 | 283

System boundary

Boiler

Pump Turbine

Qout

WinWout

Qin

Energy source(such as a furnace)

Energy sink(such as the atmosphere)

Condenser

FIGURE 6–10Schematic of a steam power plant.

HEATENGINE

Wout Wnet,out

Win

FIGURE 6–11A portion of the work output of a heatengine is consumed internally tomaintain continuous operation.

Heat input100 kJ 100 kJ

1Net

workoutput20 kJ

2Net

workoutput30 kJ

SINKWaste heat80 kJ

Waste heat70 kJ

ηth,1 = 20% ηth,2 = 30%

SOURCE

FIGURE 6–12Some heat engines perform better thanothers (convert more of the heat theyreceive to work).

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Total work output

Wnet,out = Wout �Win



Thermal E�ciency

Defined as

Thermal E�ciency =Net work output

Total heat input

⌘th =Wnet,out

Qh

SinceWnet,out = QH � QL

⌘th = 1� QL

QH


Second Law: Kelvin Planck StatementIt is impossible for any device that operates on a cycle to receiveheat from a single reservoir and produce a net amount of work.

The Second Law of Thermodynamics:Kelvin–Planck StatementWe have demonstrated earlier with reference to the heat engine shown inFig. 6–15 that, even under ideal conditions, a heat engine must reject someheat to a low-temperature reservoir in order to complete the cycle. That is,no heat engine can convert all the heat it receives to useful work. This limi-tation on the thermal efficiency of heat engines forms the basis for theKelvin–Planck statement of the second law of thermodynamics, which isexpressed as follows:

It is impossible for any device that operates on a cycle to receive heat from asingle reservoir and produce a net amount of work.

That is, a heat engine must exchange heat with a low-temperature sink as wellas a high-temperature source to keep operating. The Kelvin–Planck statementcan also be expressed as no heat engine can have a thermal efficiency of100 percent (Fig. 6–18), or as for a power plant to operate, the working fluidmust exchange heat with the environment as well as the furnace.

Note that the impossibility of having a 100 percent efficient heat engine isnot due to friction or other dissipative effects. It is a limitation that appliesto both the idealized and the actual heat engines. Later in this chapter, wedevelop a relation for the maximum thermal efficiency of a heat engine. Wealso demonstrate that this maximum value depends on the reservoir temper-atures only.

6–4 ■ REFRIGERATORS AND HEAT PUMPSWe all know from experience that heat is transferred in the direction ofdecreasing temperature, that is, from high-temperature mediums to low-temperature ones. This heat transfer process occurs in nature without requir-ing any devices. The reverse process, however, cannot occur by itself. Thetransfer of heat from a low-temperature medium to a high-temperature onerequires special devices called refrigerators.

Refrigerators, like heat engines, are cyclic devices. The working fluidused in the refrigeration cycle is called a refrigerant. The most frequentlyused refrigeration cycle is the vapor-compression refrigeration cycle, whichinvolves four main components: a compressor, a condenser, an expansionvalve, and an evaporator, as shown in Fig. 6–19.

Chapter 6 | 287

To supply energy at this rate, the engine must burn fuel at a rate of

since 19,000 Btu of thermal energy is released for each lbm of fuel burned.Discussion Note that if the thermal efficiency of the car could be doubled,the rate of fuel consumption would be reduced by half.

m#

!689,270 Btu>h

19,000 Btu>lbm! 36.3 lbm/h

HEATENGINE

Wnet,out = 100 kW

QH = 100 kW

QL = 0

Thermal energy reservoir

·

·

·

FIGURE 6–18A heat engine that violates theKelvin–Planck statement of thesecond law.

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In other wordsNo heat engine can have a thermal e�ciency of 100 percent, or asfor a power plant to operate, the working fluid must exchange heatwith the environment as well as the furnace.Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition


Refrigeration and heat pumps

The refrigerant enters the compressor as a vapor and is compressed to thecondenser pressure. It leaves the compressor at a relatively high temperatureand cools down and condenses as it flows through the coils of the condenserby rejecting heat to the surrounding medium. It then enters a capillary tubewhere its pressure and temperature drop drastically due to the throttling effect.The low-temperature refrigerant then enters the evaporator, where it evapo-rates by absorbing heat from the refrigerated space. The cycle is completed asthe refrigerant leaves the evaporator and reenters the compressor.

In a household refrigerator, the freezer compartment where heat is absorbedby the refrigerant serves as the evaporator, and the coils usually behind therefrigerator where heat is dissipated to the kitchen air serve as the condenser.

A refrigerator is shown schematically in Fig. 6–20. Here QL is the magni-tude of the heat removed from the refrigerated space at temperature TL, QHis the magnitude of the heat rejected to the warm environment at tempera-ture TH, and Wnet,in is the net work input to the refrigerator. As discussedbefore, QL and QH represent magnitudes and thus are positive quantities.

Coefficient of PerformanceThe efficiency of a refrigerator is expressed in terms of the coefficient ofperformance (COP), denoted by COPR. The objective of a refrigerator is toremove heat (QL) from the refrigerated space. To accomplish this objective,it requires a work input of Wnet,in. Then the COP of a refrigerator can beexpressed as

(6–7)

This relation can also be expressed in rate form by replacing QL by Q.L and

Wnet,in by W.

net,in.The conservation of energy principle for a cyclic device requires that

(6–8)Wnet,in ! QH " QL¬¬1kJ 2COPR !

Desired output

Required input!

QL

Wnet,in


CONDENSER

EXPANSION VALVE

120 kPa–25°C

120 kPa–20°C

800 kPa30°C

800 kPa60°C

COMPRESSOR

QL

QH

Wnet,in

Surrounding mediumsuch as the kitchen air

Refrigerated space

EVAPORATOR

FIGURE 6–19Basic components of a refrigerationsystem and typical operatingconditions.

Warm environmentat TH > TL

Cold refrigerated space at TL

R

Wnet,in

QH

QL

Requiredinput

Desiredoutput

FIGURE 6–20The objective of a refrigerator is toremove QL from the cooled space.

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Refrigeration schematic The refrigerant enters the compressor as a vapor and is compressed to thecondenser pressure. It leaves the compressor at a relatively high temperatureand cools down and condenses as it flows through the coils of the condenserby rejecting heat to the surrounding medium. It then enters a capillary tubewhere its pressure and temperature drop drastically due to the throttling effect.The low-temperature refrigerant then enters the evaporator, where it evapo-rates by absorbing heat from the refrigerated space. The cycle is completed asthe refrigerant leaves the evaporator and reenters the compressor.

In a household refrigerator, the freezer compartment where heat is absorbedby the refrigerant serves as the evaporator, and the coils usually behind therefrigerator where heat is dissipated to the kitchen air serve as the condenser.

A refrigerator is shown schematically in Fig. 6–20. Here QL is the magni-tude of the heat removed from the refrigerated space at temperature TL, QHis the magnitude of the heat rejected to the warm environment at tempera-ture TH, and Wnet,in is the net work input to the refrigerator. As discussedbefore, QL and QH represent magnitudes and thus are positive quantities.

Coefficient of PerformanceThe efficiency of a refrigerator is expressed in terms of the coefficient ofperformance (COP), denoted by COPR. The objective of a refrigerator is toremove heat (QL) from the refrigerated space. To accomplish this objective,it requires a work input of Wnet,in. Then the COP of a refrigerator can beexpressed as

(6–7)

This relation can also be expressed in rate form by replacing QL by Q.L and

Wnet,in by W.

net,in.The conservation of energy principle for a cyclic device requires that

(6–8)Wnet,in ! QH " QL¬¬1kJ 2COPR !

Desired output

Required input!

QL

Wnet,in


CONDENSER

EXPANSION VALVE

120 kPa–25°C

120 kPa–20°C

800 kPa30°C

800 kPa60°C

COMPRESSOR

QL

QH

Wnet,in

Surrounding mediumsuch as the kitchen air

Refrigerated space

EVAPORATOR

FIGURE 6–19Basic components of a refrigerationsystem and typical operatingconditions.

Warm environmentat TH > TL

Cold refrigerated space at TL

R

Wnet,in

QH

QL

Requiredinput

Desiredoutput

FIGURE 6–20The objective of a refrigerator is toremove QL from the cooled space.

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Documents

Advanced Thermodynamics: Lecture 4 - IIT Bombay · 2015. 8. 26. · Advanced Thermodynamics: Lecture 4 Shivasubramanian Gopalakrishnan [email protected] Shivasubramanian Gopalakrishnan