JJ207 Thermodynamic Topic 2 First Law of Thermodynamics

JJ207-THERMODYNAMICS 1 Topic 2- First Law of Thermodynamics and its

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TOPIC 2 FIRST LAW OF THERMODYNAMICS AND ITS PROCESSES

At the end of the topic you will be able to:

Describe forms of energy. Describe energy transfer by heat and energy transfer by work. Classify the mechanical forms of work. Distinguish the mode of energy transfer between closed and open systems. . Defined of a closed system. Describe the non-flow energy equation. Explain the specific heat, internal energy, enthalpy of ideal gases, solids and

liquids. Calculate the energy balance for close systems. Definition of an open system. Describe control volumes; steadily flow processes and engineering devices,

unsteadily-flow processes. Calculate mass and volume flow rate and the continuity equation Determine and explain the steady flow energy equation (Negligible change in

kinetic or potential energy) leading to the concept of enthalpy - typical applications such as turbine, compressor, boiler, nozzle, diffuser and heat exchanger.

Calculate the energy balance for open systems.

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2.0 INTRODUCTION

he First Law of Thermodynamics is the most basic and fundamental law of nature. While its name makes it sound intimidating, it is actually the most intuitive law of nature as well. The First Law of Thermodynamics provides a method for accounting

for all energy inputs, outputs and stores within a system. One of the most fundamental laws of nature is the conservation of energy principle. It simply states that during an energy interaction, energy can change from one form to another but the total amount of energy remains constant. That is, energy cannot be created or destroyed.

T

2.1 The First Law of Thermodynamics

Figure 2.1 Pictures showing types of energy

The first thermodynamic law is the formulation of a more general law of physics (the law of conservation of energy) for thermodynamic processes. The first law of thermodynamics is simply a statement of conservation of energy principle and it asserts that total energy is a thermodynamic property. Energy can neither be created nor destroyed; it can only change forms. This principle is based on experimental observations and is known as the First Law of Thermodynamics. The First Law of Thermodynamics can therefore be stated as follows:

Energy cannot be created or destroyed; it can only be transformed from one form into another. ……The First Law of Thermodynamics

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Energy can exist in many forms such as thermal, kinetic, potential, electric, chemical,…


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During a transformation, the change in the internal energy of a system is equal with the sum of work and heat exchanged by the system with its surroundings.

dQ – dW = dU

Where: dQ are positive if they are transferred into the system and negative if they are released by the system.

To apply the first law of thermodynamics to a cyclic process we have to remember that internal energy is a system variable. In a cyclic process, the system returns at the same state, hence there is no change in its internal energy. In this case, work produced by the system equals the heat exchanged by the system. When a system undergoes a thermodynamic cycle then the net heat supplied to the system from its surroundings is equal to the net work done by the systems on its surroundings.

In symbols,

dQ = dW (2.1)

where represents the sum of a complete cycle.

2.2 Forms of Energy

Energy is the ability to do work. It is one of the basic human needs and is an essential component in any development programme. In this lesson, we are going to look at the forms that energy exists, namely: heat, light, sound, electrical, chemical, nuclear and mechanical. These forms of energy may be transformed from one form to the other, usually with losses.

a) Heat energy, also referred to as thermal energy, is really the effect of moving molecules. Matter is made up of molecules, which are in continual motion and in a solid, vibrate about a mean position. The motion of any molecule increases when the energy of the substance is increased. This may cause an increase in the temperature of the substance or lead to a change of state. The higher the temperature, the greater the internal energy of the substance.

b) Light energy, is a type of wave motion. That is, light is a form of energy caused by light waves. It enables us to see, as objects are only visible when they reflect light into our eyes.

c) Sound energy, is also a type of wave motion. Sound energy may be converted into electrical energy for transmission, and later the electrical energy can be converted

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back into sound energy at the receiving end. An example of such transformations could be seen in the microphone and the loudspeaker.

d) Electrical energy, is really the effect of moving electrical charges from one point to another in a conductor. Electrical charges moving through a conductor is called electricity. Electrical energy may be easily changed into other forms of energy to suit our particular needs. Lightning is an example of electrical energy. Electric current is the means by which electrical energy is most easily transported to places where it is needed and converted into other forms.

e) Chemical energy, is the energy stored within chemical compounds. A chemical compound is formed by the rearrangement of atoms that is accompanied by energy loss or gain. This energy is the chemical energy gained or lost in the formation of the compound. Food, biomass, fuel and explosives have a store of chemical energy. The energy from food is released by chemical reactions in our bodies in the form of heat. Fuels like coal, oil and natural gas contain chemical energy that may be converted into other forms of energy like heat and light. The chemical energy present in a given fuel is determined by its calorific value – the heat liberated when 1 Kg of the fuel is burnt. Batteries and explosives also contain chemical energy that could be converted into other forms of energy, some beneficial, others harmful.

f) Nuclear energy, also known as atomic energy, is energy stored in the nucleus of an atom. It is this energy that holds the nucleus together and could be released when the nuclei are combined (fusion) or split (fission) apart. Nuclear energy can be used for peaceful purpose as well as destructive purposes (as in the atomic bomb). Considering peaceful purposes, nuclear energy is used to generate electricity in nuclear power plants, produce steam for driving machines, powering some submarines and spacecrafts. In these applications, the nuclei of uranium atoms are split in a process called fission. Nuclear energy is also the source of the sun’s energy. The sun combines the nuclei of hydrogen atoms into helium atoms in a process called fusion.

g) Mechanical energy, is the kind of energy that can do mechanical work directly. Naturally occurring sources of mechanical energy include winds, waterfalls and tides. There are two kinds of mechanical energy, namely kinetic energy and potential energy:

Kinetic energy, is the energy a body possesses by virtue of its motion. A moving body of mass, m kg and velocity C m/s possesses kinetic energy. Thus, the magnitude of the kinetic energy of an object depends both the mass and the velocity of the object. Flowing water and winds have kinetic energy.

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Potential energy, is the energy of a body due to its position or shape. This form of energy could be considered as energy stored in a body to be released when it begins to move or change its position or shape. Quite often, potential energy changes to kinetic energy. A ball at the top of a slope, water behind a dam, a compressed spring and a stretched elastic band possess potential energy.

Energy conversion. One important property of energy is its ability to change from one form to another form. For example, chemical energy from fossil fuels (coal, oil and natural gas) can be converted into heat energy when burned. The heat energy may be converted into kinetic energy in a gas turbine and finally into electrical energy by a generator. The electric energy may subsequently be converted into light, sound or kinetic energy in our homes through various household appliances. During any energy conversion, the amount of energy input is the same as the energy output. This concept is known as the law of conservation of energy and sometimes referred to as the First Law of Thermodynamics. This law states: energy cannot be created nor destroyed but can be transformed from one form to another. Thus, the total energy of an isolated system is always constant and when energy of one form is expended an equal amount of energy in another form is produced.

2.3 Energy Transfer by Heat and Work

Historically, heat was considered to be a fluid that can spontaneously flow from a hot body to a cold body. Heat is a form of energy which crosses the boundary of a system during a change of state produced by the difference in temperature between the system and its surroundings. The unit of heat is taken as the amount of heat energy equivalent to one joule or Nm. The joule is defined as the work done when the point of application of a force of one newton is displaced through a distance of one meter in the direction of the force.

Work transfer is defined as a product of the force and the distance moved in the direction of the force. When a boundary of a close system moves in the direction of the force acting on it, then the system does work on its surroundings. When the boundary is moved inwards the work is done on the system by its surroundings. The units of work are, for example, Nm or J. If work is done on unit mass of a fluid, then the work done per kg of fluid has the units of Nm/kg or J/kg. Consider the fluid expanding behind the piston of an engine. The force F (in the absence of friction) will be given by:

F = pA (2.2)where

p is the pressure exerted on the piston and A is the area of the piston

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If dx is the displacement of the piston and p can be assumed constant over this displacement, then the work done W will be given by,

W = F x dx= pA x dx= p x Adx= p x dV= p(V2 – V1) (2.3)

where dV = Adx = change in volume.

Figure 2.3 Work transfer

When two systems at different temperatures are in contact with each other, energy will transfer between them until they reach the same temperature (that is, when they are in equilibrium with each other). This energy is called heat, or thermal energy, and the term "heat flow" refers to an energy transfer as a consequence of a temperature difference.

2.4 Sign Convention for Work Transfer

It is convenient to consider a convention of sign in connection with work transfer and the usual convention adopted is:

if work energy is transferred from the system to the surroundings, it is donated as positive.

if work energy is transferred from the surroundings to the system, it is donated as negative.

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F

WORK W2

+ veSYSTEM

SURROUNDINGS BOUNDARY

Figure 2.4 Sign Convention for work transfer

WORK W1

- ve

PRESSURE

dx


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2.5 Sign Convention for Heat Transfer

The sign convention usually adopted for heat energy transfer is such that : if heat energy flows into the system from the surroundings it is said to be

positive. if heat energy flows from the system to the surroundings it is said to be

negative. It is incorrect to speak of heat in a system since heat energy exists only when it flows across the boundary. Once in the system, it is converted to other types of energy.

Figure 2.5 Sign convention for heat transfer

2.6 Specific heat capacityThe specific heat capacity (symbol C) is the amount of heat, expressed in kJ, which

must be transferred to or from 1 kg of a given material to cause its temperature to change by 1°C. Hence, the unit of measurement is kJ/kg.°C or in thermodynamics scale the unit is kJ/kg.K. The word specific, used in this and other terms we will use later, refers to 1kg.

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HEAT ENERGY Q2

-ve

SURROUNDINGS

HEATENERGY

Q1

+ ve

SYSTEM

BOUNDARY


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2.7 Internal Energy

Internal energy is the sum of all the energies a fluid possesses and stores within itself. The molecules of a fluid may be imagined to be in motion thereby possessing kinetic energy of translation and rotation as well as the energy of vibration of the atoms within the molecules. In addition, the fluid also possesses internal potential energy due to inter-molecular forces.

Suppose we have 1 kg of gas in a closed container as shown in Figure 2.7. For simplicity, we shall assume that the vessel is at rest with respect to the earth and is located on a base horizon. The gas in the vessel has neither macro kinetic energy nor potential energy. However, the molecules of the gas are in motion and possess a molecular or 'internal' kinetic energy. The term is usually shortened to internal energy. If we are to study thermal effects then we can no longer ignore this form of energy. The symbol for internal energy is U and in the International System (SI) it is measured in joules (J) or kilojoules (kJ). 1kJ = 1000J. Also, we shall denote the specific (per kg) internal energy as u J/kg.

Now suppose that by rotation of an impeller within the vessel, we add work dW to the closed system and we also introduce an amount of heat dQ. The gas in the vessel still has zero macro kinetic energy and zero potential energy. The energy that has been added has simply caused an increase in the internal energy.

The change in internal energy is determined only by the net energy that has been transferred across the boundary and is independent of the form of that energy (work or heat) or the process path of the energy transfer. In molecular simulations, molecules can of course be seen, so the changes occurring as a system gains or loses internal energy are apparent in the changes in the motion of the molecules. It can be observed that the molecules move faster when the internal energy is increased. Internal energy is, therefore, a thermodynamic property of state. Equation 2.4 is sometimes known as the non-flow energy equation and is a statement of the First Law of Thermodynamics.

or,

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(2.4)


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Figure 2.7 Added work and heat raise the internal energy of a close system

2.8 Enthalpy (H) and Specific Enthalpy (h)

A more useful system variable for this course is enthalpy. Enthalpy (H), unit kJ is the sum of internal energy and the product of pressure and volume of the system.

H = U + p⋅ VIn a thermodynamic system with fixed volume and pressure, enthalpy has the same meaning as internal energy and it is also measured in kilojoules (kJ). The reference condition regarding enthalpy differs from one substance to another. For water (H2O), the reference condition is defined at a temperature of approximately 0.01°C and a pressure of approximately 0.61 kPa(a). This is the only combination of pressure and temperature at which ice, water and vapour can stably exist together. The condition is referred to as the triple point. Every substance has its own triple point. At the triple point enthalpy of the substance is considered to be 0.

The enthalpy contained in one kg of a substance is known as specific enthalpy (h). Specific enthalpy depends on the material, its pressure, temperature and state. Specific enthalpy data for light and heavy water can be found in steam tables. Specific enthalpy (h) is measured in kJ/kg.

h = u + p⋅ vTherefore, taking the value of (h) and multiplying it by the mass, one can obtain the value of enthalpy (H).

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dQ

dW

The figure above shows a certain process, which undergoes a complete cycle of operations. Determine the value of the work output for a complete cycle, Wout.

A system is allowed to do work amounting to 500 kNm whilst heat energy amounting to 800 kJ is transferred into it. Find the change of internal energy and state whether it is an increase or decrease.


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Example 2.1

Solution to Example 2.1

Q = Qin + Qout = (10) + (-3)

= 7 kJ

W = Win + Wout = (-2) + (Wout)

Hence Q - W = 0 W = Q

(-2) + (Wout) = 7 Wout = 9 kJ

Example 2.2


U2 – U1 = Q12 – W12

now,W12 = +500 kNm = 500 kJQ12 = +800 kJ

U2 – U1 = 800 – 500

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Qin = +10 kJ Wout = (+) ?

Win= -2 kJQout = -3 kJ

SYSTEM

Qin is +10 kJ

Qout is –3 kJ

Win is –2 kJ

Wout is +ve

During a complete cycle operation, a system is subjected to the following:Heat transfer is 300 kJ supplied and 150 kJ rejected.Work done by the system is 200 kJ.

Calculate the work transferred from the surrounding to the system.


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= 300 kJ (Since U2 U1, the internal energy has increased)Example 2.3

Solution to Example 2.3Q = Qin + Qout = (300) + (-150) = 150 kJ

W = Win + Wout = (Win) + (200)

Hence Q - W = 0W = Q

(Win) + (200) = 150Win = - 50 kJ

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2.9 STEADY FLOW PROCESSES

In heat engine it is the steady flow processes which are generally of most interest. The conditions which must be satisfied by all of these processes are:

i. The mass of fluid flowing past any section in the system must be constant with respect to time.

ii. The properties of the fluid at any particular section in the system must be constant with respect to time.

iii. All transfer of work energy and heat which takes place must be done at a uniform rate.

A typical example of a steady flow process is a steam boiler, operating under a constant load as shown diagrammatically in Fig. 2.9. In order to maintain the water level in the boiler, the feed pump supplies water at exactly the same rate as that at which steam is drawn off from the boiler. To maintain the production of steam at this rate at a steady pressure, the furnace will need to supply heat energy at a steady rate. Under these conditions, the properties of the working fluid at any section within the system must be constant with respect to time.

Figure 2.9 Steam Boiler

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STEAMOUT

WATER LEVEL

BOUNDRY

WATERIN

FURNACE


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2.10 Equation of ContinuityThis is an equation which is often used in conjunction with the steady flow energy

equation. It is based on the fact that if a system is in a steady state, then the mass of fluid passing any section during a specified time must be constant. Consider a mass flow rate (in symbol, ) kg/s flowing through a system in which all conditions are steady as illustrated in Fig.2.10.

Figure 2.10 Mass flowing through a system

Let A1 and A2 represent the flow areas in m2 at the inlet and outlet respectively.Let v1 and v2 represent the specific volumes in m3/kg at the inlet and outlet respectively.Let C1 and C2 represent the velocities in m/s, at the inlet and outlet respectively.

Then mass flowing per sec = volume flowing per sec m 3 /s volume per kg m3/kg

= at inlet

= at outlet

i.e. = (2.5)

2.11 STEADY FLOW ENERGY EQUATION

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1

1

2

2

C1 C2

AREA A2

AREA A1


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This equation is a mathematical statement on the principle of Conservation of Energy as applied to the flow of a fluid through a thermodynamic system.

The various forms of energy which the fluid can have are as follows:

a) Potential energyIf the fluid is at some height Z above a given datum level, then as a result of its mass it possesses potential energy with respect to that datum. Thus, for unit mass of fluid, in the close vicinity of the earth,

Potential energy = g Z 9.81 Z

b) Kinetic energyA fluid in motion possesses kinetic energy. If the fluid flows with velocity C, then, for unit mass of fluid,

Kinetic energy =

c) Internal energyAll fluids store energy. The store of energy within any fluid can be increased or decreased as a result of various processes carried out on or by the fluid. The energy stored within a fluid which results from the internal motion of its atoms and molecules is called its internal energy and it is usually designated by the letter U. If the internal energy of the unit mass of fluid is discussed this is then called the specific internal energy and is designated by u.

d) Flow or displacement energyIn order to enter or leave the system, any entering or leaving volume of fluid must be displaced with an equal volume ahead of itself. The displacing mass must do work on the mass being displaced, since the movement of any mass can only be achieved at the expense of work.

Thus, if the volume of unit mass of liquid (its specific volume) at entry is v1

and its pressure is P1, then in order to enter a system it must displace an equal specific volume v1 inside the system. Thus work to the value P1v1 is done on the specific volume inside the system by the specific volume outside the

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system. This work is called flow or displacement work and at entry it is energy received by the system.

Similarly, at exit, in order to leave, the flow work must be done by the fluid inside the system on the fluid outside the system. Thus, if the pressure at exit, is P2 and the specific volume is v2 the equation is then,

Flow or displacement work rejected = P2v2

e) Heat received or rejectedDuring its passage through the system the fluid can have direct reception or rejection of heat energy through the system boundary. This is designated by Q. This must be taken in its algebraic sense.Thus,

Q is positive when heat is received.Q is negative when heat is rejected.Q = 0 if heat is neither received nor rejected.

f) External work doneDuring its passage through the system the fluid can do external work or have external work done on it. This is usually designated by W. This also must be taken in its algebraic sense.Thus if,

External work is done by the fluid then W is positive.External work is done on the fluid then W is negative.If no external work is done on or by the fluid then W = 0.

Figure 2.11 illustrates some thermodynamic system into which is flowing a fluid with pressure P1, specific volume v1, specific internal energy u1 and velocity C1. The entry is at height Z1 above some datum level. In its passage through the system, external work W may be done on or by the fluid and also heat energy Q may be received or rejected by the fluid from or to the surroundings.

The fluid then leaves the system with pressure P2, specific volume v2, specific internal energy u2 and velocity C2. The exit is at height Z2 above some datum level.

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Figure 2.11 A schematic of a steady flow system

The application of the principle of energy conservation to the system is,

Total energy entering the system = Total energy leaving the system

or, for unit mass of substance,

(2.6)

This is called the steady flow energy equation.

This equation is not applicable to all energy forms. In such cases, the energy forms concerned are omitted from the energy equation.

In equation 2.6, it was stated that the particular combination of properties of the form, h = u + Pv is called specific enthalpy and is designated as h. Thus, the steady flow energy equation is written as

or, in the easy way the equations becomes

(2.7)

Note: q = specific heat (kJ/kg)w = specific work (kJ/kg)

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P2 ,v2

u2 ,C2

P1, v1

U1,C1


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and with the flow rate, (kg/s) the equation may be written as

(2.8)

Note: Q = heat flow (kJ/s or kW)W = work (kJ/s or kW)

2.12 APPLICATION OF STEADY FLOW EQUATION

The steady flow energy equation may be applied to any apparatus through which a fluid is flowing, provided that the conditions stated previously are applicable. Some of the most common cases found in engineering practise are dealt with in detail as below.

a) BoilersIn a boiler operating under steady conditions, water is pumped into the boiler along the feed line at the same rate as which steam leaves the boiler along the steam main, and heat energy is supplied from the furnace at a steady rate.

Figure 2.12.1 Steam Boiler

The steady flow energy equation gives

(2.9)

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STEAMOUT

BOUNDRY

WATERIN

FURNACE

SYSTEM

Q

1

1

2

2


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In applying this equation to the boiler, the following points should be noted :i. Q is the amount of heat energy passing into the fluid per secondii. W is zero since a boiler has no moving parts capable of affecting a

work transferiii. The kinetic energy is small as compared to the other terms and may

usually be neglectediv. The potential energy is generally small enough to be neglected.v. (kg/s) is the rate of the flow of fluid.

Hence the equation is reduced to

(2.10)

b) Condensers

In principle, a condenser is a boiler reverse. In a boiler, heat energy is supplied to convert the liquid into vapour whereas in a condenser heat energy is removed in order to condense the vapour into a liquid. If the condenser is in a steady state then the amount of liquid, usually called condensate, leaving the condenser must be equal to the amount of vapour entering the condenser.

Figure 2.12.2 Condenser


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VAPOUR

CONDENSATE

BOUNDARY

WATERIN

WATEROUT

SYSTEM

A boiler operates at a constant pressure of 15 bar, and evaporates fluid at the rate of 1000 kg/h. At entry to the boiler, the fluid has an enthalpy of 165 kJ/kg and on leaving the boiler the enthalpy of the fluid is 2200 kJ/kg. Determine the heat energy supplied to the boiler.


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Points to note,i. Q is the amount of heat energy per second transferred from the systemii. W is zero in the boileriii. The kinetic energy term may be neglected as in the boileriv. The potential energy is generally small enough to be neglectedv. is the rate of the flow of fluid.

Thus the equation is reduced to

(2.11)

Example 2.4



Q = heat energy per hour entering systemW = work energy per hour leaving system = 0

= fluid flow rate = 1000 kg/h = 0.278 kg/sh2 = 2200 kJ/kgh1 = 165 kJ/kgC1& C2 = neglectedZ1& Z2 = neglected

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BOUNDARYWATER

IN

SYSTEM

Q

1

1

STEAM OUT2

2

If 65 % of the heat energy supplied to the boiler in example 2.4 is used in evaporating the fluid, determine the rate of fuel consumption required to maintain this rate of evaporation, if 1 kg of fuel produces 32000 kJ of heat energy.

Fluid enters a condenser at the rate of 35 kg/min with a specific enthalpy of 2200 kJ/kg, and leaves with a specific enthalpy of 255 kJ/kg. Determine the rate of heat energy loss from the system.


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Thus, the steady flow energy equation becomes

Example 2.5


Heat energy from fuel required per seconds =

= 870.4 kJ/s @ kW

Heat energy obtained from the fuel = 32 000 kJ/kg

Fuel required =

= 0.0272 kg/s

Example 2.6

Solution to Example 2.6 = 35 kg/min = 0.58 kg/s


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For a condenser, W = 0, and the term representing the change in kinetic and potential energy may be neglected. Therefore the equation is reduced to

From the above equation

= - 1 128.1 kJ/s

ACTIVITY 1)

2.1) In an air conditioning system, air is cooled by passing it over a chilled water coil condenser. Water enters the coil with an enthalpy of 42 kJ/kg and leaves the coil with an enthalpy of 80 kJ/kg. The water flow rate is 200 kg/h. Find the rate of heat absorption by the water in kilowatts.

2.2) In a steady flow system, a substance flows at the rate of 4 kg/s. It enters at a pressure of 620 kN/m2, a velocity of 300 m/s, internal energy 2100 kJ/kg and specific volume 0.37 m3/kg. It leaves the system at a pressure of 130 kN/m2, a velocity of 150 m/s, internal energy 1500 kJ/kg and specific volume 1.2 m3/kg. During its passage through the system the substance has a loss by heat transfer of 30 kJ/kg to the surroundings. Determine the power of the system in kilowatts, stating whether it is from or to the system. Neglect any change in potential energy.

Answer to ACTIVITY 1)

2.1) Data: = 200 kg/h =

h1 = 42 kJ/kg; h2 = 80 kJ/kg

Q = ?

The diameter of the water tube in a cooler is normally constant. Therefore, there is no change in water velocity and kinetic energy. In general the change in potential energy is also negligible.

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The equation of steady flow is therefore reduced to

= 0.056(42 – 80) = - 2.13 kJ/s or kW

The rate of heat absorption by the water is 2.13 kW

2.2) By neglecting the change in potential energy, for a unit mass of substance, the steady flow energy equation becomes:

(1)

q is written negative since 30 kJ/kg are lost to the surroundings.From equation (1)

Specific work, w =

Worki in kilojoules (kJ)

Specificwork, w = (2100-1500) + (620x0.37 -130x1.2) + ( ) +(-30)

= 676.75 kJ/kg.

The substance flows at the rate of, = 4 kg/s

Output (since W is positive), W = = 4 x 676.75 = 2707 kJ/s or kW

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c) Turbine

A turbine is a device which uses a pressure drop to produce work energy which is used to drive an external load.

Figure 2.12.3 Turbine


Points to note :i. The average velocity of flow of fluid through a turbine is normally high, and

the fluid passes quickly through the turbine. It may be assumed that, because of this, heat energy does not have time to flow into or out of the fluid during its passage through the turbine, and hence Q = 0 .

ii. Although velocities are high the difference between them is not large, and the term representing the change in kinetic energy may be neglected.

iii. Potential energy is generally small enough to be neglected.iv. W is the amount of external work energy produced per second.The steady flow energy equation becomes

-

or (2.12)

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Qout

Wout

FLUID OUT

FLUID IN

SYSTEM

1

2BOUNDARY

A fluid flows through a turbine at the rate of 45 kg/min. Across the turbine the specific enthalpy drop of the fluid is 580 kJ/kg and the turbine loss 2100 kJ/min in the form of heat energy. Determine the power produced by the turbine, assuming that changes in kinetic and potential energy may be neglected.


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Example 2.7

Solution to Example 2.7The steady flow energy equation gives

Q = heat energy flow into system = -2100 kJ/min = -35 kJ/sW = work energy flow from system kJ/min

= fluid flow rate = 45 kg/min = 0.75 kg/sh2 – h1 = -580 kJ/kgC1& C2 = neglectedZ1& Z2 = neglected

Therefore the steady flow energy equation becomes

-35 kJ/s – W = 0.75 kg/s (-580 kJ/kg) W = (-35) + (435) kJ/s

= 400 kJ/s = 400 kW

d) Nozzle

A nozzle utilises a pressure drop to produce an increase in the kinetic energy of the fluid.

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SYSTEMBOUNDARY

FLUIDOUT

FLUIDIN

1

1

2

2

Fluid with a specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with negligible velocity at the rate of 14 kg/s. At the outlet from the nozzle the specific enthalpy and specific volume of the fluid are 2250 kJ/kg and 1.25 m3/kg respectively. Assuming an adiabatic flow, determine the required outlet area of the nozzle.


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Figure 2.12.4 Nozzle


Points to note :i. The average velocity of flow through a nozzle is high, hence the fluid

spends only a short time in the nozzle. For this reason, it may be assumed that there is insufficient time for heat energy to flow into or out of the fluid during its passage through the nozzle, i.e. Q = 0.

ii. Since a nozzle has no moving parts, no work energy will be transferred to or from the fluid as it passes through the nozzle, i.e. W = 0.

iii. Potential energy is generally small enough to be neglected.

Hence the equation becomes

or (2.13)

Often C1 is negligible compared with C2. In this case the equation becomes

(2.14)

Example 2.8

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When applied to the nozzle, this becomes

Since the inlet C1 is negligible, this may be written as

=

= 1049 m/s

Applying the equation of continuity at outlet gives

From ,

= 0.01668 m2

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ACTIVITY 2

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

2.3 Steam enters a turbine with a velocity of 16 m/s and specific enthalpy 2990 kJ/kg. The steam leaves the turbine with a velocity of 37 m/s and specific enthalpy 2530 kJ/kg. The heat lost to the surroundings as the steam passes through the turbine is 25 kJ/kg. The steam flow rate is 324000 kg/h. Determine the work output from the turbine in kilowatts.

2.4 In a turbo jet engine the momentum of the gases leaving the nozzle produces

the propulsive force. The enthalpy and velocity of the gases at the nozzle entrance are 1200 kJ/kg and 200 m/s respectively. The enthalpy of the gas at exit is 900 kJ/kg. If the heat loss from the nozzle is negligible, determine the velocity of the gas jet at exit from the nozzle.

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ANSWER TO ACTIVITY 2)

2.3 Neglecting the changes in potential energy, the steady flow energy equation is

Q is negative since heat is lost from the steam to the surroundings

specific W = - Q

= (2999-2530) +

= 434.443 kJ/kg

The steam flow rate = 324000/3600 = 90 kg/s

W = 434.443 x 90 = 39099.97 kJ/s or kW 39100 kW 39.1 MW

2.4 The steady energy flow equation for nozzle gives

On simplification,

=

= 800 m/s

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e) Throttling process

A throttling process is one in which the fluid is made to flow through a restriction , e.g. a partially opened valve or orifice, causing a considerable drop in the pressure of the fluid.

Figure 2.12.5 Throttling process


Points to note:i. Since throttling takes place over a very small distance, the available area

through which heat energy can flow is very small, and it is normally assumed that no energy is lost by heat transfer, i.e. Q = 0.

ii. Since there are no moving parts, no energy can be transferred in the form of work energy, i.e. W = 0.

iii. The difference between C1 and C2 will not be great and consequently the term representing the change in kinetic energy is normally neglected.

iv. The potential energy is generally small enough to be neglected.

The steady flow energy equation becomes

0 = (h2 – h1) or h2 = h1 (2.15)

i.e. during a throttling process the enthalpy remains constant.

Example 2.9

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1

2

A fluid flowing along a pipeline undergoes a throttling process from 10 bar to 1 bar in passing through a partially open valve. Before throttling, the specific volume of the fluid is 0.3 m3/kg and after throttling is 1.8 m3/kg. Determine the change in specific internal energy during the throttling process.


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For a throttling process, the steady flow energy equation becomes

0 = (h2 - h1) or h2 = h1

From h = u + Pv,u2 = h2 - P2v2

u1= h1 - P1v1

Therefore the change in specific internal energy

u2 – u1 = ( h2 - P2v2 ) - ( h1 – P1v1) = ( h2 - h1 ) – ( P2v2 - P1v1 ) = 0 – ( 1 x 102 x 1.8 – 10 x 102 x 0.3 ) kN/m2 x m3/kg = 120 kNm/kg = 120 kJ/kg

f) Pump

The action of a pump is the reverse of that of a turbine, i.e. it uses external work energy to produce a pressure rise. In applying the steady flow energy equation to a pump, exactly the same arguments are used as for turbine and the equation becomes

- (2.16)

Since h2 > h1, W will be found to be negative.

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Q

W

SYSTEM

1

2

BOUNDARY

OUTLET

INLET


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Figure 2.12.6 PumpExample 2.10


The flow rate of fluid, = 45 kg/min = 0.75 kg/s

The steady flow energy is

Q = - 105 kJ/min = - 1.75 kJ/sW = work energy flow (kJ/s)h1 = 46 kJ/kgh2 = 1.27 kJ/kgm = 0.75 kg/sThe kinetic and potential energy may be neglected

Substituting the data above with the steady flow energy equation gives Q - W = (h2 – h1)

- 1.75 – W = 0.75 (175 – 46) W = -1.75 – (0.75 x 129) = - 98.5 kJ/s = - 98.5 kW

(N.B. The negative sign indicates work energy required by the pump)

Since the efficiency of the drive is 85 %

Power required by the compressor = - 98.5 x

= - 114.8 kW

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A pump delivers fluid at the rate of 45 kg/min. At the inlet to the pump the specific enthalpy of the fluid is 46 kJ/kg, and at the outlet from the pump the specific enthalpy of the fluid is 175 kJ/kg. If 105 kJ/min of heat energy are lost to the surroundings by the pump, determine the power required to drive the pump if the efficiency of the drive is 85 %.


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TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

2.3 A rotary air pump is required to deliver 900 kg of air per hour. The enthalpy at the inlet and exit of the pump are 300 kJ/kg and 500 kJ/kg respectively. The air velocity at the entrance and exit are 10 m/s and 15 m/s respectively. The rate of heat loss from the pump is 2500 W. Determine the power required to drive the pump.

2.4 In Activity 2, for question No. 2.4, if the diameter of the nozzle at exit is 500 mm, find the mass flow rate of gas. The gas density at the nozzle inlet and exit are 0.81 kg/m3 and 0.39 kg/m3 respectively. Also determine the diameter of the nozzle at the inlet.

2.3 Data : = 0.25 kg/s

h1 = 300 kJ/kgh2 = 500 kJ/kgC1= 10 m/sC2= 15 m/sQ = 2500 W = 2.5 kWW = ?


Neglecting the change in Potential energy since it is negligible

-W = 0.25 [( 500- 300) + ( )] + 2.5

W = - 52.5 kW

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2.4 Data : A2 = = 0.196 m2

1 = 0.81 kg/m3

2 = 0.39 kg/m3

= ?d = ?

Mass flow rate at exit,

= A2 C2 2

= 61.2 kg/s

From the mass balance,Mass entering the nozzle = mass leaving the nozzle = m

= A1 C1 1 = A2 C2 2

On substitution A1 x 200 x 0.81 = 61.2

On simplificationA1 = 0.378 m2

ord1 = 0.694 m = 694 mm

SELF ASSESMENT)

You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self-Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck.

1 Steam flows through a turbine stage at the rate of 4500 kJ/h. The steam velocities at inlet and outlet are 15 m/s and 180 m/s respectively. The rate of heat energy flow from the turbine casing to the surroundings is 23 kJ/kg of steam flowing. If the specific enthalpy of the steam decreases by 420 kJ/kg in passing through the turbine stage, calculate the power developed.

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2 A rotary pump draws 600 kg/hour of atmospheric air and delivers it at a higher pressure. The specific enthalpy of air at the pump inlet is 300 kJ/kg and that at the exit is 509 kJ/kg. The heat lost from the pump casing is 5000 W. Neglecting the changes in kinetic and potential energy, determine the power required to drive the pump.

3 A nozzle is supplied with steam having a specific enthalpy of 2780 kJ/kg at the rate of 9.1 kg/min. At outlet from the nozzle the velocity of the steam is 1070 m/s. Assuming that the inlet velocity of the steam is negligible and that the process is adiabatic, determine:a) the specific enthalpy of the steam at the nozzle exitb) the outlet area required if the final specific volume of the steam is

18.75 m3/kg.

4 Fluid at 10.35 bar having a specific volume of 0.18 m3/kg is throttled to a pressure of 1 bar. If the specific volume of the fluid after throttling is 0.107 m3/kg, calculate the change in specific internal energy during the process.

Have you tried the questions????? If “YES”, check your answers now.

1. 476 kW

2. 353 kW

3. 2208 kJ/kg; 2660 mm2

4. 175.7 kJ/kg

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JJ207 Thermodynamic Topic 2 First Law of Thermodynamics

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