Engineering and Construction - thermodynamics and heat engines

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Assignment Brief (RQF) Assignment Brief (RQF) Higher National Certificate - General Engineering Student Name/ID Number: Unit Number and Title: Unit 13 Fundamentals of Thermodynamics and Heat Engines Academic Year: 2017-2018 Unit Assessor: Sumana Subramanyam Assignment Title: Solymar Thermal Solutions Issue Date: 9th July 2018 Submission Date: Internal Verifier Name: Date: Not Applicable Praneeta Phadke 23rd Aug 2018 Submission Format: The submission is in the form of individual written report/ booklets and calculations with linked written reponses.All work must be supported with research and referenced appropriately. The recommended word limit is around 3500 words, you will not be penalised for exceeding the total word limit. Unit Learning Outcomes: 1. Investigate fundamental thermodynamic systems and their properties. 2. Apply the Steady Flow Energy Equation to plant equipment. 3. Examine the principles of heat transfer to industrial applications. 4. Determine the performance of internal combustion engines. Assignment Brief and Guidance: You have recently been employed by an American company Solymar Thermal Solutions, which delivers thermal solutions, analyse and tests all thermodynamic system components to ensure their optimum performance. Your manager has asked you to complete the following tasks. Prepared by: Higher Education Qualifications (HEQ) Version 1.0 – November 2017 DCL 1 – Public (Unclassified) Approved by: HEQ, QDAM (HE) 1 Assignment Brief (RQF) As a new employee, you would be investigating the fundamental thermodynamic systems and their properties. Using your knowledge of steady flow energy equation, you would be producing specific steady flow energy equations based on stated assumptions for your company’s plant equipment. As a part of your work, you would be also working on heat exchangers and internal combustion engines. For all the above work, you would be submitting a written report, based on the tasks mentioned below by your boss. Task 1 Compile a training booklet or leaflet describing the thermodynamic systems and their properties. • • • Explain the application to the appropriate systems. Describe the relationships between system constants for perfect gas. Clarify the importance of expressions for the work done in Thermodynamic processes by applying first principles. Task 2 Calculate the polytropic index in a system using the given data • • Describe the parameters using non-flow energy equation and apply the steady flow energy equation to plant equipment. Also produce specific Steady Flow Energy Equations based on stated assumptions in plant equipment. PVn =C P1V1n=P2V2n find the value of n. TASK 3 Illustrate the Steady flow energy equation from first principles and apply it in an industrial plant. • Determine the heat transfer through composite walls • Apply heat transfer formulae to heat exchangers • Explore heat losses through lagged and unlagged pipes • 3 Distinguish the differences between parallel and counter flow recuperator heat exchangers Prepared by: Higher Education Qualifications (HEQ) Version 1.0 – November 2017 DCL 1 – Public (Unclassified) Approved by: HEQ, QDAM (HE) 2 Assignment Brief (RQF) Task 4 Compile a training leaflet or a booklet describing the operation the operation of both four stroke spark ignition and four stroke compression ignition engines. • • • • Describe with the aid of diagrams the operational sequence of four stroke spark ignition and four stroke compression ignition engines. Explain the mechanical efficiency of two and four stroke engines Review the relative efficiency of ideal heat engines operating on the Otto and Diesel cycles Produce a booklet evaluating the performance of both two-stroke and four stroke diesel engines, validate typical applications and discuss the options to improve their efficiency. Prepared by: Higher Education Qualifications (HEQ) Version 1.0 – November 2017 DCL 1 – Public (Unclassified) Approved by: HEQ, QDAM (HE) 3 Assignment Brief (RQF) Learning Outcomes and Assessment Criteria: Learning Outcome Pass Merit Distinction 1. Investigate fundamental thermodynamic systems and their properties. P1 Describe the operation of thermodynamic systems and their properties M1 Calculate the index of compression in polytrophic processes D1 Illustrate the importance of expressions for work done in thermodynamic processes by applying first principles M2 Derive the Steady Flow Energy Equation from first principles D2 Produce specific Steady Flow Energy Equations based on stated assumptions in plant equipment M3 Explore heat losses through lagged and unlagged pipes D3 Distinguish the differences between parallel and counter flow recuperator heat exchangers M4 Review the relative efficiency of ideal heat engines operating on the Otto and Diesel cycles D4 Evaluate the performance of two stroke and four stroke diesel engines P2 Explain the application of the first law of thermodynamics to appropriate systems 2. Apply the Steady Flow Energy Equation to plant equipment. P3 Explain the relationships between system constants for a perfect gas P4 Explain system parameters using the Non- Flow Energy Equation P5 Apply the Steady Flow Energy Equation to plant equipment 3. Examine the principles of heat transfer to industrial applications. P6 Determine the heat transfer through composite walls P7 Apply heat transfer formulae to heat exchangers 4. Determine the performance of internal combustion engines. P8 Describe with the aid of diesel engines diagrams the operational sequence of four stroke spark ignition and four stroke compression ignition engines P9 Explain the mechanical efficiency of two and four stroke engines Prepared by: Higher Education Qualifications (HEQ) Version 1.0 – November 2017 DCL 1 – Public (Unclassified) Approved by: HEQ, QDAM (HE) 4
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Surname 1
Name
Tutor
Course
Date
Thermodynamic Systems
Task One
A thermodynamic system has a wall or boundary as the cover while the region beyond is
the environment. Typically, a thermodynamic system is considered to be in a state of internal
thermal equilibrium rather than a state of disequilibrium. Hence, a thermodynamic system ought
to be enclosed in walls separating it from the external environment thus constraining the system.
Thermodynamic operations are external interventions subjected to the system. Consequently, the
system goes through thermodynamic processes as per the principles of thermodynamics. The
thermodynamic state of the system is dictated by its state parameters.
Additionally, a thermodynamic account needs a unique quantity termed as a state
function. A state function defines the state variables. For instance, if the state variables include
mole content, internal energy, and volume, then the state function serves as the entropy. The
quantities have associations with either one or more functional links termed as state equations.
Also, the system’s characteristic equation associates the quantities. Thermodynamics create
restrictions on the state equations which can arise and its characteristics. The limitations are
created by the laws applicable in thermodynamics (Ahmadi, Mohammad H., et al., 96-105).

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As per the permeability of the walls of a thermodynamic system, energy transfer and
matter happens between it and the environment. The underlying assumption is that; the transfers
do not vary in time until the attainment of a state of equilibrium. Equilibrium states are taken to
be in equilibrium in thermodynamics. Classical thermodynamics contain thermodynamics at
equilibrium. At the state, the variable is not inclusive of fluxes as they have values at zero by
postulation. However, equilibrium thermodynamic processes might include fluxes, but they must
end upon completion.
On the other hand, non-equilibrium thermodynamics provide for the variables to have
fluxes which are not zero. The factor accounts for the transfer in either energy, mass, or entropy
between the environment and the system. A thermodynamic system is enclosed using walls
which bind and link it to its environment. Prevalently, the wall curtails passage from the system
in any energy (Izumida, et al.,180603). As a result, the association is not direct.
In some cases, a wall in an imaginary 2-dimensional enclosure which has direct contact
with the environment. A wall can be permanent for instance a constant volume reactor. Also, the
wall can be movable like in the case of a piston. For illustration, in a rotating engine, while the
piston is locked in at one position, then the wall is fixed, and an unchanging volume process may
happen. Similarly, while the piston is not locked, it can move in and out then the wall is not set.
Therefore, a wall can be semi-permeable, impermeable, permeable, adiabatic, or diathermal.
A thermodynamic system is limited by the boundaries across which allows the passage of
quantities in and out of the system. The environment is the space outside the system. The wall
dictates the nature of transfers that can happen. A permeable wall provides for quantities transfer.
The wall permeability is the criterion used is their classification. The transfer can be induced by
contact, for instance, heat conduction, or a broad range of forces like an electric field in the

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environment. An isolated wall prevents all forms of transfers. The conception is ideal as some
transfer is bound to happen. An axiom of thermodynamics holds that even an isolated system at
the end attains a thermodynamic state of equilibrium and variations stop. An open system wall
provides for the transfer of energy and matter. Any form of quantities that is lost across the
boundary and has an impact on the system ought to be accounted for in a suitable equation
balance. The volume can be described as the area surrounding an atom producing energy. Max
Planck in 1900 defined the region to be a body of steam or the air in steam ran the engine. Sadi
Carnot on the other hand in 1824 termed it to be just one nuclide as supported by the quantum
thermodynamics hypotheses (Ahmadi, Mohammad H., et al., 96-105). The system is defined as
the universe under study while the environment is the remaining part of the universe outside the
walls. As per the nature of the system, there might be an interaction allowing the exchange of
mass, forms of energy like work and heat, electric charge, motion or other conserved properties.
Closed System
In the case of a closed thermodynamic system, there is no mass transfer in and out of the
walls. As a result, the system does not record changes in matter. However, work and heat can be
transferred across the system’s boundaries. The capacity to exchange work, heat or both rely on
the nature of the boundary. There are two forms of boundaries namely; adiabatic and rigid
boundary. An aquatic boundary can be thermally isolated not providing for the exchange of heat.
A rigid boundary restricts work exchange making it be mechanically isolated. For illustration,
when a fluid is under compression by a piston inside a cylinder. Also, a bomb calorimeter is an
excellent example of a closed system. The calorimeter is utilized in the measurement of
combustion of heat in a particular reaction. Electrical energy moves across the wall to create a
spark between the electrodes which induces combustion (Izumida, et al.,180603). Heat is lost

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across the wall after the combustion process while no mass is transferred. The first law of
thermodynamics is indicated below.

In this case, U is the internal energy while Q is the heat added to the system as no mass is
lost from the system. The two expressions which involve a flow of mass are at point zero which
derives the first law of thermodynamics for a closed thermodynamic system. The first law of
thermodynamics for a closed system holds that an increase in the internal energy of the system is
equal to the quantity of heat put in the system less the work carried out in the system. Regarding
the little variations, the first law for closed thermodynamic systems states:

If the work is as a result of an increase in volume, by V at pressure P then:

In the case of a homogenous system, the execution of a reversible process, the second law states
that:

Thus, T is the absolute temperature.
S is the system’s entropy.
As per the relations above the fundamental thermodynamic relation used in the
calculation of variations in the internal energy is denoted as:

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For a simple thermodynamic system containing one nature of the particle, that is, an atom
or molecule, a closed system produces a constant amount of particles. But, for a system in the
process of chemical reactions, there is a broad range of molecules generated and destroyed
within the system. Thus, the fact that the system is closed is denoted by indicating the aggregate
number for every elemental atom is conserved. Consequently, it is regardless of the nature of the
molecule it is part of. It can be illustrated mathematically as:

Therefore,
N is the amount of j-type molecules,
aij is the number of atoms contained in element i in molecule j
Whereas, bi0 is the total number of atoms of element i in the system.
The thermodynamic system stays at a constant state while the, which remains constant while it is
closed. Every element in the system has one such equation (Izumida, et al.,180603).
Isolated System
An isolated system has a higher number of restrictions compared to a closed system as
there is no interaction with the surrounding environment. The system has a constant mass and
energy concentration. Furthermore, no transfers are happening across the wall. With the variation
in time, the difference between the system and the environment tends to attain equilibrium by

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equalizing pressures and temperatures similar to density changes. Thermodynamic equilibrium is
attained when the equalizing processes are complete. Isolated systems do not exist in real life.
The aspect is, due to the rationale that gravity exists between the system’s mass and masses in
other places. But, actual systems can act similar to their isolated counterparts for a very long time
frame. Thus, the paradigm of an isolated system can be used as an excellent model for estimating
a broad range of actual situations. The idea is used in the construction of many mathematical
models regarding particular natural situations.
In a bid to make a justification on the hypothesis of entropy increase in the second law of
thermodynamics, Boltzmann’s H-theorem made use of equations which assumed that a system,
for instance, gas was under isolation. In other words, the entire mechanical degrees of freedom
can be specified thus treating the walls as conditions of mirror boundary. The factor led to the
Loschmidt paradox. But, under the consideration of the stochastic behavior in real walls,
alongside the randomizing impact of the ambient, thermal radiation in the thermal radiation, the
assumption of molecular disarrangement is justifiable. The second law of thermodynamics for
isolated thermodynamic systems indicates that the entropy of the system is not in equilibrium
tends to increase as time changes. The positive variation moves towards attaining the maximum
value at a state of equilibrium (Bejan, Adrian. Advanced engineering thermodynamics).
Commonly, in an isolated system, the internal energy remains constant, and entropy does not
decrease. For a closed system, entropy can reduce for example upon the extraction of heat from
the system. It is vital to note that isolated systems are not equal to their closed counterparts.
Closed systems cannot exchange matter with the environment, but energy transfers can happen.
It is critical to note that isolated systems are not equal to their thermodynamic counterparts.
Closed dynamic systems do not allow the exchange of quantities with the environment but allow

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energy transfer. Isolated systems cannot exchange matter or energy making them ideal in the
entire planet earth. It is notable that, a closed system is commonly applied in thermodynamics.
Open System
In most cases, the borders of a thermodynamic system provide for transfer in not only
heat but also substance. The general case renders the system open. A system is termed as closed
if the walls are not impermeable. The open thermodynamic system cannot be in equilibrium. The
description of the deviation of a thermodynamic system makes use of a set of internal variables.
are the internal variables added to the ones used above. The equilibrium state is taken
to be stable and the key characteristic of the internal variables. Below is the illustration:

In that case,

is the relaxation period of the respective variables. It is

appropriate to consider the initial value to be zero. Prigogine brought a substantial contribution
to the thermodynamics of systems not in an equilibrium state. The proponent and his colleagues
studied the systems while substances are undergoing a chemical reaction. The stable conditions
of similar systems occur due to the transfer of quantities with the surrounding. Prigogine has
made specifications on three contributions to the changes in entropy of the system taken to be
open at a certain volume and unchanging temperature T. The positive change in entropy S can be
calculated using the formula below.

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The initial term on the right side of the equation introduces a streak of thermal energy into the
thermodynamic system (O'hayre, Ryan, et al. Fuel cell fundamentals). The second term
represents the energy stream which gets into the system with the substances’ particles.
Adiabatic System
An adiabatic system does not provide for the transfer of heat in and out of the system.
PV^(gamma) = C equation is valid only for an adiabatic system which also undergoes a process
which is reversible as long as the system is closed and has an ideal gas. If the system fails to
meet the conditions then, dQ=0 is justified. As a result, in that event, it cannot be denoted using
an equation like PV^(gamma)=C.
Task Two
A polytropic process involves a relationship where the system’s volume and pressure has
an association with the equation, Pvn=C. Polytropic processes are determined by the constant
parameter in the process. In the equation, p stands for pressure, v represents volume, n stands for
the polytropic index while C is the constant parameter. The equation can also be expressed as
PVn=K and P=K/Vn while Vn is the volume. The specific volume of the thermodynamic system
can be found by dividing the term by the mass. Consequently, for specific volume: P= K/m /
Vn/m. Thus, a polytropic process can be linked to work by the following equation: W= P2V2P1V1/ 1-n. The processes have shapes while some have names like linear, or hyperbolic to
mention a few. Open and closed systems can follow a polytropic path. In the context of the
Solymar Thermal Solutions, the following steady flow energy equation applies to the plant’s
equipment (Ahmadi, Mohammad H., et al., 96-105).
∆𝑞 = ∆𝑞 + 𝑉𝑑𝑝 − 𝑑(𝑐2⁄2)

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𝑑(𝑐2⁄2) = −𝑉𝑑𝑝
2𝐶
2

∗ 𝑑𝑐 = −𝑉𝑑𝑝

The rate of flow in mass is constant.
Therefore,
𝑚 = 𝑦𝐴𝐶 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐴.𝐶
𝑉
𝑑𝐴
𝐴
𝑑𝐴
𝐴

= log 𝐴 + log 𝐶 − log 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

=

𝑑𝑐
𝑐



𝑉𝑑𝑃
𝑐2

𝑑𝑣
𝑣

+

=0

𝑑𝑣
𝑣

In that case,
A= Area
C = Velocity
V- volume
Q – W = ΔU
Where Q is the heat transferred to the system.
W is the work done by the system
ΔU is the variation in internal energy.
Task Three
There are three forms of work namely:

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a) Boundary work (Wb) as illustrated by the operations of a cylinder and piston.
b) Shaft work (Ws) like in the case of a paddle wheel.
c) Electrical work (We) which is denoted by V.I.t
There are three assumptions which come into play. The assumptions are held in the precise list
below.
a) Amount of mass flowing in and out of the system.
b) There are zero losses due to friction.
c) As the piston moves, every incremental motion by the piston maintains a state of
equilibrium.
Therefore,
2

2

2

W1-2 = ∫1 𝐹𝑑𝑥 = ∫1 𝑃𝐴𝑑𝑥 = ∫1 𝑃𝑑𝑣
Also, the area under the curve of P against V
2

𝑊1 − 2 = ∫ 𝑃𝑑𝑣 = 𝑚
1
2

∫ 𝑃𝑑𝑣 = 𝑚 ∗ 𝑊1 − 2
1

For the case of a closed thermodynamic system, both shaft and electrical work have a negative
value.
The steady flow energy equation that can be used by the plant is shown below/
𝑚𝑣12
𝑀𝑣22⁄
𝐻1 +
+ 𝑀𝑔21 + 𝑄 − 𝑊 = 𝐻2 +
2 + 𝑀𝑔𝑧2
2

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𝑀𝑣22 𝑀𝑣12
𝑄 − 𝑊 = 𝐻2 − 𝐻1 +

+ 𝑀𝑔𝑍2 − 𝑀𝑔𝑍1
2
2
𝜕𝑇

Every layer must meet ∇2 𝑇1 = 𝜕𝑥2 = 0
Layer 1
∇2 𝑇2 =

𝜕2 𝑇1
𝜕𝑥 2

= 0 thus, 𝑇1 (𝑥) = 𝑎1 + 𝑏1 𝑥

Layer 2
∇2 𝑇2 =

𝜕2 𝑇2
𝜕𝑥 2

= 0 thus, 𝑇2 (𝑥) = 𝑎2 + 𝑏2 𝑥

Constants a1, b1, and a2, b2, can only be evaluated when the following conditions a, b, and c
have been met.
A) Connective boundary conditions x=0
−𝐾1 𝑏1 = ℎ1 (𝑎1 − 𝑇∞, 1)
B) The interface between walls one and two 𝑥 = 𝑥1

𝑇1 (𝑥1 ) = 𝑇2 (𝑥1 ) = 𝑎1 + 𝑏1 𝑥1 = 𝑎2 + 𝑏2 𝑥1
𝑏1 𝐾1 = 𝑏2 𝑥2
C) The connective boundary conditions 𝑥 = 𝑥2
−𝐾2 𝑏2 = ℎ2 (𝑎1 + 𝑏2 𝑥2 − 𝑇∞, 2)

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1

𝑞𝑥 = − ∪ ∆𝑇or 𝑞𝑥 = −(𝑅)∆𝑇
∆𝑇 = 𝑇∞, 2 − 𝑇∞, 1
1

Where 𝑅 = ∪
U is the effective heat transfer coefficient
Qx is the heat flux
R is thermal resistance
ΔT is the gradient in temperature
Heat flux in every layer is, therefore;
𝑞𝑥 = −ℎ1 (𝑇𝑎 − 𝑇∞, 1)

𝑞𝑥 = −

𝐾1
(𝑇 − 𝑇𝐴 )
𝐿2 𝐵

𝑞𝑥 = −

𝐾2
(𝑇 − 𝑇𝐵 )
𝐿2 𝐶

𝑞𝑥 = −

𝐾3
(𝑇 − 𝑇𝐶 )
𝐿3 𝐷

𝑞𝑥 = −ℎ2 (𝑇∞, 2 − 𝑇𝐷 )
The following can be applied in the heat exchanger
The heat transfer rate per unit of length is given by the following.

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𝑄=

2𝜋𝐾(𝑇𝐴 − 𝑇𝐵 )
𝐾
𝑟1 ℎ1

+𝑟

𝐾

2 ℎ2

𝑟2

+ 1𝑛(𝑟 1 )

TA is the fluids bulk temperature
H0 is the coefficient for heat transfer
1
𝑟2
𝑟2
𝑟2
1
=
+ 𝐼𝑛 ( 1 ) +
ℎ0 𝑟1 ℎ1 𝐾
𝑟
ℎ2
𝑄 = 2𝜋𝑟2 ℎ0 (𝑇𝐴 − 𝑇𝐵 )
The logging of pipes has the effect of retaining the heat within the pipes. A special type of
insulation should be fitted around the pipes of water.
Task Four
A four stroke engine makes use of four different strokes that is: intake, exhaust, power,
and compression to execute one complete cycle. Thus, the piston can take two full passes to
finish one cycle. The operating cycle takes double revolutions of the crankshaft. The following
image holds a description of the ignition and compression process in a four-stroke engine (Bejan,
Adrian. Advanced engineering thermodynamics).

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Intake Stroke
The intake pertains to the introduction of the fuel and air combination into the cylinder.
The operation happens while the piston is in motion from the Top Dead Corner to the Bottom
Dead Corner while the intake valve is open. The motion of the Bottom Dead Corner leads to low
pressure in the cylinder. The pressure in the environment makes the air and fuel mixture get into
the open intake valve to make up for the created pressure. The cylinder proceeds to get filled past
the Bottom Dead Corner as the mixture flows by inertia while the piston is changing in direction.
The intake valve then closes thus sealing the mixture into the cylinder.

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Compression Stroke
The stroke occurs when the air and fuel mixture is locked and compressed in the cylinder.
The sealing of the combustion chamber is done to make the charge. The charge is translated as
the amount of air and fuel compressed in the chamber and ready for ignition. The compression
provides for the release of a substantial amount of energy upon ignition. The outlet and inlet
valves should be closed to warrant that the cylinder is fully sealed. Therefore, compression is the
operation undertaken in reducing or increasing the charge in the four-stroke engine cylinder by
changing the volume. The flywheel plays a key role in the maintaining an effective motion to
keep a working charge. The heat generated by the work done by the piston in compressing the
mixture leads to heat production. The compression and temperature increase leads to heating
making the fuel evaporate (Izumida, et al.,180603). The positive variation in temperature
happens evenly in the combustion chamber to provide for hastened oxidation of the fuel after
ignition.
Ignition Stroke
The ignition event happens when the charge is sparked, and oxygen is supplied rapidly
using a chemical reaction to result in heat. Consequently, combustion can be defined as the fast,
fuel oxidation reaction where it mixes with oxygen from the systems’ surrounding and produces
heat ener...


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