1. A certain fluid at 10 bar is contained in a cylinder behind a piston, the initial volume being 0.05 m3. Calculate the work done by the fluid when it expands reversibly: a. at constant pressure to a final volume of 0.2 m3;

b. according to a linear law to a final volume of 0.2 m3 and a final pressure of 2 bar; c. according to a law Pv = constant to a final volume of 0.1 m3; d. according to a law Pv3 = constant to a final volume of 0.06 m3; e. according to a law, P = (A/v2) – (B/v), to a final volume of 0.1 m3 and a final pressure of 1 bar, where A and B are constants. f. Sketch all processes on a Pv diagram.

2. 1 kg of a fluid expands reversibly according to a linear law from 4.2 bar to 1.4 bar; the initial and final volumes are 0.004 m3 and 0.02 m3. The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law Pv = constant back to the initial conditions of 4.2 bar and 0.004 m3. Calculate the work done in each process and the net work of the cycle. Sketch the cycle on a Pv diagram.

3. A fluid at 0.7 bar occupying 0.09 m3 is compressed reversibly to a pressure of 3.5 bar according to a law Pvn = constant. The fluid is then heated reversibly at constant volume until the pressure is 4 bar; the specific volume is then 0.5 m3/kg. A reversible expansion according to a law Pv2 = constant restores the fluid to its initial state. Sketch the cycle on a Pv diagram and calculate: g. the mass of the fluid present;

h. the value of n in the first process;
i. the net work of the cycle.

4. A fluid is heated reversibly at a constant pressure of 1.05 bar until it has a specific volume of 0.1 m3/kg. It is then compressed reversibly according to a law Pv = constant to a pressure of 4.2 bar, then allowed to...

...Biomolecule BCH 3101
Thermodynamics of Biological Systems • Movement, growth, synthesis of biomolecules, and the transport of ions and molecules across membranes all requires energy. • All organisms acquire energy from their surroundings and utilize that energy efficiently to carry out life processes. • In order to study these bio-energetic phenomena we will require knowledge of thermodynamics.
BCH3101 1
• Thermodynamics: defined as a collection of laws and principles describing the flows and interchanges of heat, energy and matter in systems of interest . • Thermodynamics also allows us to determine whether or not chemical processes and reactions can occur spontaneously.
BCH3101
2
•
The ease for a reaction to occur and the direction of reaction is determined by the Laws of Thermodynamics (refer to Mathews and van Holde, 1996 ed. p.62).
BCH3101
3
• Several basic thermodynamic principles considered including the analysis of heat flow, entropy production, and free energy functions and the relationship between entropy and information.
BCH3101
4
• In any consideration of thermodynamics, a distinction must be made between the system and the surroundings. The system is that portion of the universe with which we are concerned, whereas the surroundings include everything else in the universe. The nature of the system must also be specified....

...Example 1
Nozzle Flow - Steam Steam at 1.5 bar and 150 deg C enters a nozzle at 10 m/s and exits at 1 bar. Assuming the flow is reversible and adiabatic, determine the exit temperature and velocity. If the exit nozzle area is 0.001 m2, evaluate the mass flow rate of the steam through the nozzle. P1 = 1.5 bar T1 = 90 deg C V1 = 10 m/s A2 = 0.001 m2
P2 =1 bar T2 = ? V2 = ? mdot = ?
V2 = sqrt(V1*V1+2*(h1-h2)) mdot = rho*A2*V2 = A2*V2/v2 T2
382.60 m/s 0.22 kg/s 111.81 degC
Superheated steam at nozzle entry T1 150 deg C P1 1.5 bar s1 7.420 kJ/kg K h1 2773 kJ/kg K V1 10 m/s P2 A2 1 bar 0.001 m2
At Nozzle exit P2 = 1 bar Rev & Adiabatic s1 = s2 = 7.420 kJ/kg K h2 (Superheated because s2 > sg2 (7.359)
s 7.360 7.614 7.420 s 7.360 7.614 7.420 s 7.360 7.614 7.420
h 2676 2777 2699.86 v 1.696 1.937 1.753 T 100 150 111.81
v2
T2
Example 2
Nozzle Flow - Air Air enters a nozzle at 1.5 bar and 90 deg C at a velocity of 10 m/s and exits the nozzle at 1 bar. Assuming the flow is reversible and adiabatic, determine the exit temperature and velocity. If the exit nozzle area is 0.001 m2, evaluate the mass flow rate of the air through the nozzle. P1 = 1.5 bar T1 = 90 deg C V1 = 10 m/s A2 = 0.001 m2
P2 =1 bar T2 = ? V2 = ? mdot = ? R Cp Cv gama = Cp/Cv
0.287 kJ/kg K 1.005 kJ/kg K 0.718 kJ/kg K 1.40
(gama-1)/gama = exp T2 = T1*(P2/P1)^exp process for reversible and adiabatic
0.29
323.31 K 50.31 deg C 282.62 m/s 1.50 0.30 kg/s 0.93 m3/kg
T1 P1 V1...

...Thermodynamics Lab
Introduction:
Thermodynamics is the study of energy which can exist in many forms, such as heat, light, chemical energy, and electrical energy. The variables that thermodynamics can be used to define include temperature, internal energy, entropy, and pressure. Temperature, relating to thermodynamics, is the measure of kinetic energy in the particles of a substance. Light is usually linked to absorbance and emission in thermodynamics while pressure, linked with volume, can do work on an entire system. The entropy is the measure of the flow of heat through a system whose equation is for a thermodynamically reversible process as
Regarding thermodynamics, there are three laws the first of which is appropriately named the First Law of Thermodynamics. The First Law of Thermodynamics states that energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe must remains constant at all times, however it can change from one form to another. In other words energy is always conserved with the amount remaining constant. The Second Law of Thermodynamics...

...ENGINEERING
Lecture Outlines
2000
Ian A. Waitz
THERMODYNAMICS:
COURSE INTRODUCTION
Course Learning Objectives:
To be able to use the First Law of Thermodynamics to estimate the potential for thermomechanical energy conversion in aerospace power and propulsion systems.
Measurable outcomes (assessment method):
1) To be able to state the First Law and to define heat, work, thermal efficiency and
the difference between various forms of energy. (quiz, self-assessment, PRS)
2) To be able to identify and describe energy exchange processes (in terms of
various forms of energy, heat and work) in aerospace systems. (quiz, homework,
self-assessment, PRS)
3) To be able to explain at a level understandable by a high school senior or nontechnical person how various heat engines work (e.g. a refrigerator, an IC engine,
a jet engine). (quiz, homework, self-assessment, PRS)
4) To be able to apply the steady-flow energy equation or the First Law of
Thermodynamics to a system of thermodynamic components (heaters, coolers,
pumps, turbines, pistons, etc.) to estimate required balances of heat, work and
energy flow. (homework, quiz, self-assessment, PRS)
5) To be able to explain at a level understandable by a high school senior or nontechnical person the concepts of path dependence/independence and
reversibility/irreversibility of various thermodynamic processes, to represent these
in terms of...

...first established principle of thermodynamics (which eventually became the Second Law) was formulated by Sadi Carnot in 1824. By 1860, as found in the works of those such as Rudolf Clausius and William Thomson, there were two established "principles" of thermodynamics, the first principle and the second principle. As the years passed, these principles turned into "laws." By 1873, for example, thermodynamicist Josiah Willard Gibbs, in his “Graphical Methods in the Thermodynamics of Fluids”, clearly stated that there were two absolute laws of thermodynamics, a first law and a second law.
Over the last 80 years or so, occasionally, various writers have suggested adding Laws, but none of them have been widely accepted.
[edit] Overview
* Zeroth law of thermodynamics
A \sim B \wedge B \sim C \Rightarrow A \sim C
* First law of thermodynamics
\mathrm{d}U=\delta Q-\delta W\,
* Second law of thermodynamics
\oint \frac{\delta Q}{T} \ge 0
* Third law of thermodynamics
T \rightarrow 0, S \rightarrow C
* Onsager reciprocal relations - sometimes called the Fourth Law of Thermodynamics
\mathbf{J}_{u} = L_{uu}\, \nabla(1/T) - L_{ur}\, \nabla(m/T) \!;
\mathbf{J}_{r} = L_{ru}\, \nabla(1/T) - L_{rr}\,...

...like on a
pressure volume diagram?
So we're starting to put some of our
tools together.
The PV, pressure volume diagram are state
diagrams, we're going to discuss those
quite a bit in the coming lectures.
But we also talked about work as being,
in it's most general form, an integral
expression.
And then, we said, okay, there's a
specific form of that expression for
expansion and compression work.
So, let's answer that question now.
Okay, so we talked about the PV diagram.
And again we said that work in its most
general form could be expressed as an
intergral moving from one state condition
to a second state condition so that's S1
to S2 and the most general form of that
expression is one where we use just some
force.
Times the distance, that it, that force
has executed in that process.
So from state one to state two forced
integrated over that distance, over that
process distance.
And we said for expansion and compression
work in particular, we would move from
state one to state two and that the form
of the expression was pressure times the
differential D volume so remember the V
with the cross hash is a volume
expression okay so that's our expression
if we have a constant pressure process.
We know that that pressure is constant,
and we know from our math that we can
take the pressure outside of that
integral.
The state conditions here are...

...COLLEGE
CHENNAI
DEPARTMENT OF MECHANICAL ENGINEERING
113301 – ENGINEERING THERMODYNAMICS
UNIT – 1 : ASSIGNMENT QUESTIONS
1. A piston and cylinder machine contains a fluid system which passes through a complete cycle of four processes. During a cycle, the sum of all heat transfers is -170kJ. The system completes 100 cycles per min. Complete the following table showing the method for each item, and compute the net rate of work output in kW.
|Process |Q (kJ/min) |W (kJ/min) |∆E (kJ/min) |
|a-b |0 |2,170 |- |
|b-c |21,000 |0 |- |
|c-d |-2,100 |- |-36,600 |
|d-a |- |- |- |
(P.K.Nag Pg.No.66)
2. Air at 1.02 bar, 22ᵒC, initially occupying a cylinder volume of 0.015 m3 is compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate : (i) the final temperature, (ii) the final...

...UEME1112 Goh Sing Yau (May 2014), FES, UTAR
5
Temperature
5.1 Definition of Temperature Equality
If 2 bodies are brought into contact, after a period of time, there is no observable change in
their physical properties (example: length, electrical resistance, density etc). The bodies are
said to be in thermal equilibrium and to be equal in temperature.
The 2 systems are equal in temperature when no change in any property occurs when they are
brought into communication.
5.2 The Zeroth Law of Thermodynamics
(This law was formulated after the First Law of Thermodynamics. Since it is more
fundamental, it is called the Zeroth Law)
It was observed experimentally that 2 systems that are equal in temperature to a third
system are also equal in temperature to each other.
This may appear obvious or trivial but in general 2 systems that behave in the same way to a
third system DO NOT necessarily behave in the same way with respect to each other.
For example:
or
5.2
20
UEME1112 Goh Sing Yau (May 2014), FES, UTAR
5.3 The use of the Zeroth Law for Temperature measurement.
(1) S3 is brought into contact with S2. After a
period of time, the temperature S2 and S3
become equal.
(2) Measure a physical property of S3. (say the
length of the mercury After a period of time, the
temperatures of S2 and S3 become equal.
(3) S3 is brought into contact with S1. If there is
no change in properties of S3, then S1 and S2
are...

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