9. A concrete highway is built of slabs 14 m long (20Cº). How wide should the expansion cracks be (at 20Cº) between the slabs to prevent buckling if the temperature range is -30Cº to +50Cº? I know I use the formula:

L = LoT
L = ? (Change in length of the slabs. We are solving for this.) = 12e-6 (Coefficient of expansion. I looked it up on page 388.) Lo = 14 m (Initial length of slabs.)
T = 30 Cº (50Cº - 20Cº) (Change in temperaure. You only care about the hottest number since you are dealing with expansion.) I then had the formula:
L = e14 m 30Cº
L = .00504 m
L = .50 cm
The space between the slabs must be the same as the expansion of the the slabs. Therefore .50 cm is the correct answer. 17. A quartz sphere is 14.5 cm in diameter. What will its change in volume if it is heated from 30Cº to 200Cº? I had the formula:

V = VoT
V = ? (Volumetric change in the sphere. We are solving for this.) = 1e -6 (Cº)-1 (Coefficient of volumetric expansion. It is on page 388 listed as .) Vo = (4/3) * (.145 m/2)3 (Initial volume. You must take the diameter and divide by to to get the radius. Then you must stick it in the formula for volume of a sphere: (4/3)r3 Vo = .001569 m3

T = 170Cº (Change in temperature. 200Cº - 30Cº.) I then had the formula:
V = 1e-6*.001569m3*170Cº
V = 2.7 e-7 m3 *1e6 (Conversion from m3 to cm3.)
V = .27 cm3
The answer is therefore .27 cm3.

19. If the fluid is contained in a long, narrow vessel so it can expand in essentially one direction only, show that the effective coefficient of linear expansion is approximately equal to the coefficient of volume expansion (in the IB packet.)? Since it can only expand in one direction, is is really linear expansion, not volumetric expansion. Therefore the two coefficients would be the same.

23. A 23.4 kg solid aluminum wheel of radius 0.52 m is rotating about its axel in frictionless bearings with an angular velocity 32.8...

...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 states that in...

...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}\,...

...Objective: to show the relationship between pressure and temperature of saturated steam
Apparatus: a Marcet boiler (Figure 1) is used. It is provided with a pressure gauge, a digital thermometer and a safety valve. An aneroid barometer is used to determine atmospheric pressure.
1 2 3 Drain valve Heater Overflow
Variation of saturation temperature with pressure
Thermodynamics Laboratory Manual
Marcet boiler
4 5
6 7 8 9
Temperature sensor Pressure relief valve Filler opening with plug Pressure gauge Master switch Boiler with insulating jacket
11 Temperature display
10 Heater switch
The drain valve (1) can be used to drain the vessel. An electric heater (2) is bolted into the floor of the boiler in such a way that the heating element protrudes from below into the boiler. A pressure gauge (8) is fitted to provide a direct indication of the boiler pressure. There is also a Pt-100 temperature sensor (4) to measure the boiler temperature, and a safety valve (5) to prevent excess pressure build-up in the boiler. If the Safety valve is activated, the excess pressure is discharged to the rear of the unit via a drain pipe. The boiler temperature can be read from the digital display (11) fitted into the switch box. The unit is switched on at the master switch (9). The additional
8 Thermo_Lab_Manual_BE_Year_2_r008.docx
The main element of the apparatus is the stainless steel steam boiler (7). It has a mineral wool insulating jacket. The...

...Thermodynamics - Assignment 4
D3
SS Sultana – a Mississippi River steamboat paddle wheeler that exploded on April 27, 1865 in the greatest maritime disaster in United States history.
Around 1,600 of Sultanas 2,400 passengers were killed when THREE of the ships four boilers exploded and Sultana sank near Memphis, Tennessee.
The cause of the explosion was a leaky and poorly repaired steam boiler. There was reason to believe allowable working steam pressure was exceeded in an attempt to overcome the spring river current. The boiler (or boilers) gave way when the steamer was 7 to 9 miles north of Memphis at 2:00 am. The enormous explosion flung some of the passengers on deck into the water, and destroyed a large section of the ship. Hot coals scattered by the explosion soon turned the remaining superstructure into an inferno, the glare of which was visible as far away as Memphis.
The official cause of Sultana disaster was determined to be mismanagement of water levels in the boiler, worsened by the fact that Sultana was severely overcrowded and top heavy. As the steamship made its way north, Sultana listed severely to one side then the other. Sultana's four boilers were interconnected and mounted side-by-side, so that if the ship tipped sideways, water would tend to run out of the highest boiler. With the fires still going against the empty boiler, this created hot spots. When the ship tipped the other way, water rushing back into the empty boiler would hit...