PHYS 101 Lecture Notes - Lecture 31: Hydroelectricity, Kelvin, Thermodynamics

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28 Mar 2017

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Ethermal (input) = work + ethermal(lost, exhaust) (this is just conservation of energy) A heat engine takes advantage of the natural tendency of energy flow: from hotter to colder environments. The second law of thermodynamics implies that the exhaust term is always greater than zero. This means that we are unable to use all of the input thermal energy! Energy efficiency = work output (j) thermal energy input. Heat engines are always less than 100% efficient! This is a new way to state the second law of thermodynamics: Any process that uses thermal energy as the input to do work must also have a thermal energy output (exhaust). Heat engines are always less that 100% efficient. The best possible efficiency is given by (tinput - texhaust)/tinput with the temperature in kelvins. The size of a kelvin and of a degree celsius is the same. 273 k = 0 c (note: 0 k, zero kelvin, is called absolute zero.

Related Questions

A firebox is at 750 K, and the ambient temperature is
300 K. The efficiency of a Carnot engine doing 150 J of work
as it transports energy between these constant- temperature
baths is 60.0%. The Carnot engine must take in energy
150 J/0.600 5 250 J from the hot reservoir and must put
out 100 J of energy by heat into the environment. To follow
Carnotâs reasoning, suppose some other heat engine
S could have an efficiency of 70.0%. (a) Find the energy
input and exhaust energy output of engine S as it does 150 J
of work. (b) Let engine S operate as in part (a) and run the
Carnot engine in reverse between the same reservoirs. The
output work of engine S is the input work for the Carnot
refrigerator. Find the total energy transferred to or from
the firebox and the total energy transferred to or from the
environment as both engines operate together. (c) Explain
how the results of parts (a) and (b) show that the Clausius
statement of the second law of thermodynamics is violated.
(d) Find the energy input and work output of engine S as it
puts out exhaust energy of 100 J. Let engine S operate as in
part (c) and contribute 150 J of its work output to running theCarnot engine in reverse. Find (e) the total energy the
firebox puts out as both engines operate together, (f) the
total work output, and (g) the total energy transferred to
the environment. (h) Explain how the results show that
the KelvinâPlanck statement of the second law is violated.
Therefore, our assumption about the efficiency of engine S
must be false. (i) Let the engines operate together through
one cycle as in part (d). Find the change in entropy of the
Universe. (j) Explain how the result of part (i) shows that
the entropy statement of the second law is violated.

The compression ratio of an Otto cycle, as shown in Figure 22.13, is V_A/V_B = 8.00. At the beginning A of the compression process, 500 cm3 of gas is at 100 kPa and 20.0°C. At the beginning of the adiabatic expansion, the temperature is T_C = 750°C. Model the working fluid as an ideal gas with E_int = nC_VT = 2.50nRT and y = 1.40.

(a) Fill in the table below to follow the states of the gas:

(c) Identify the energy input Q_h, the energy exhaust Q_c, and the net output work W_eng. (d) Calculate the thermal efficiency. (e) Find the number of crankshaft revolutions per minute required for a one-cylinder engine to have an output power of 1.00 kW =1.34 hp. Note that the thermodynamic cycle involves four piston strokes.

amaranthlocust933

Kinetic energy is a microscopic concept. The equivalent cacroscopicconcept is:
a. mass b. entropy c. temp d. heat

the second law of thermodynamics has a number of equivalentdefinitions, all of which give an indication of which way natureflows . which of these statements is not generally derivable fromthe concept that that a particular engine is themost efficientengine possible and that it is reversible?
a. heat will not flow spontaneously from a cold reservoir to hotone
b. an engine or combination of engines alone cannot make theat flowfrom a cold reservoir to a hot one.
c. heat will flow spontaneoulsy from a cold reseroir to hotone
d. it is not possible to convert 100 % of heat into work.

PHYS 101 Lecture Notes - Lecture 31: Hydroelectricity, Kelvin, Thermodynamics

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