- Numericals on Steady Flow Energy Equation (SFEE)
- Lecture 2: Second Law Of Thermodynamics
- Lecture 3: Carnot Engine
- Lecture 4: Heat Engines and COP for Heat Pumps and Refrigerators
- Lect 5: Numericals on COP and Efficiency- II
- Numericals on Efficiency and COP -III
What you'll learn
- To study the aspects of Second law of Thermodynamics
- To understand the statements of Second law of thermodynamics
- To get the concept of Equivalence of Kelvin-Planck and Clausius Statement
- To get the idea of COP of Heat pumps and Refrigeration
- To know the methods of solving numericals on Heat Engine, Heat Pump and Refrigerator
- Get idea about Carnot’s theorem and Carnot Cycle.
- Know the working principle of Heat Engines, Heat Pumps and Refrigerators .
Description
Course Contents as per syllabus :
Equivalence of Clausius and Kelvin Planck Statement, PMM I and II, Concept of Reversibility and Irreversibility.
Course Objectives:
•To study the aspects of Second law of Thermodynamics
•To understand the statements of Second law of thermodynamics
•To get the concept of Equivalence of Kelvin-Planck and Clausius Statement
•To define Perpetual Machine of IInd Kind (PMM-II)
•To Study Carnot theorem and Carnot Cycle
•To get the knowledge of Heat engine and its efficiency
•To compare Heat pumps and Refrigeration system
•To get the idea of COP of Heat pumps and Refrigeration
•To know the methods of solving numericals on Heat Engine, Heat Pump and Refrigerator
Brief description about Heat Engine, Refrigerator and Heat pump :
HEAT ENGINE
Based on Kelvin- Planck’s Statement.
It is a device that converts heat to work.
In case of heat engine, efficiency is defined as ratio of work done to heat energy supplied.
Here important criteria is net work done.
Work done by the engine =W = (Q1 – Q2)
Efficiency = Work done / Heat supplied
Efficiency = W / Q1 = (Q1 – Q2) / Q1
= (T1-T2 ) / T1
= 1 – ( T2 / T1)
REFRIGERATOR
Based on Clausius Statement.
In case of a refrigerator ,COP is defined which is the ratio of heat absorbed (refrigeration effect) to work done on the system ,Here the important criteria is heat absorbed as refrigeration effect.
Work Done on system = (Q1 – Q2)
COPR = Desired Effect / Work done
= Q2 / ( Q1 – Q2)
= T2 / ( T1-T2)
COP is reverse of Efficiency
HEAT PUMP
Based on Clausius Statement.
In case of a heat pump ,COP is defined which is the ratio of heat rejected (refrigeration effect) to work done on the system. Here the important criteria is heat rejected as refrigeration effect.
Work Done on system = (Q1 – Q2)
COPHP = Desired Effect / Work done
= Q1 / ( Q1 – Q2)
= T1 / ( T1-T2)
COP HP = COPR + 1
Course Outcomes :
Learner will be able to
•Get idea about limitations of First law of Thermodynamics.
•Can compare between Kelvin and Clausius Statements.
•Understand various statements of Second law of Thermodynamics.
•Have clear idea about Kelvin and Clausius Statement.
•Get idea about Carnot’s theorem and Carnot Cycle.
•Know the working principle of Heat Engines, Heat Pumps and Refrigerators .
•Find out Efficiency of Heat Engines and COP of heat pumps and refrigerators.
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About the instructors
- 4.48 Calificación
- 9215 Estudiantes
- 18 Cursos
Basawraj Racheti
Mechanical Engineer Ex- Executive Engineer MAHAGENCO Nashik
Myself B.S. Racheti ME (Mech), MBA (HR), Energy Auditor (BEE). I am having industrial experience of 31 years and 4 years of teaching experience in reputed Engineering colleges of SPPU, Pune. I have also trained number of Junior Engineers, Technicians in Thermal Power plant. I am having hobby / passion of teaching mechanical subjects particularly Basic mechanical engineering, Thermodynamics, Power plant Engineering, Energy Engineering etc.