Thermodynamic Relations In Engineering Thermodynamics by Pk NAG (Chapter 11) | EveryEng

Skip to main content

Self-paced Beginner

Thermodynamic Relations In Engineering Thermodynamics by Pk NAG (Chapter 11)

4(12)

9 enrolled

2082 views

$ 10

216 min

Anytime

Hindi

2082 views

Saurabh Kumar GuptaMechanical Engineer

7-day money-back guarantee
Lifetime access
Foundational Learning
Access to Study Materials

Why enroll

This course is based on PK Nag's Book Chapter 11, to excel in the GATE (Graduate Aptitude Test in Engineering) examination and to secure good marks in other engineering exams. Thermodynamics is a crucial subject in the engineering syllabus, and mastering the concepts and applications presented in Chapter 11 is essential to achieving a high score. By taking this course, individuals can gain a comprehensive understanding of thermodynamic principles, practice solving problems, and develop strategies to tackle complex questions. With a strong foundation in thermodynamics, students can confidently approach the GATE exam and improve their chances of securing admission to top engineering programs or landing coveted jobs at top PSUs.

Master the fundamentals of thermodynamics and unlock the secrets of energy conversion, efficiency, and optimization—enroll now and become a thermal energy expert!

Your instructor

Saurabh Kumar Gupta

Content Manager

Mechanical Engineer

5+ yrs experience·India

Is this course for you?

You should take this if

You work in Aerospace or HVAC
You're a Mechanical Engineering / Chemical & Process professional
You prefer self-paced learning you can revisit

You should skip if

You need a different specialisation outside Mechanical Engineering
You need live interaction with an instructor

Course details

Thermodynamic relations refer to the mathematical connections between various thermodynamic properties of a system, such as pressure, volume, temperature, internal energy, enthalpy, and entropy. These relations are derived from the laws of thermodynamics and are used to describe the behavior of systems in equilibrium. Thermodynamic relations, such as the Maxwell relations, the Gibbs-Helmholtz equation, and the Clapeyron equation, provide a framework for predicting the properties of a system and understanding the relationships between different thermodynamic variables. By applying these relations, engineers and scientists can calculate unknown properties, determine the feasibility of processes, and design systems that involve thermodynamic changes. Thermodynamic relations are essential in various fields, including power generation, refrigeration, and materials science, and are used to develop new technologies and optimize existing ones. Through the application of thermodynamic relations, professionals can gain insights into the fundamental behavior of energy and matter, and develop innovative solutions to complex problems.

Course suitable for

Key topics covered

Mathematical Theorem
Maxwell Equation
T-ds Equation
Energy Equation
Ratio of Heat Capacity
Joule Kelvin Effect
Numerical on Joule Kelvin Effect
Clausius Clapeyron Equation
Compressibility Factor

Course content

The course is readily available, allowing learners to start and complete it at their own pace.

18 lectures3 hr 36 min

Mathematical Theorem
8 min
Maxwell Relation's | Maxwell Equations || Thermodynamic Relations
6 min
Tds Equation or Mayer's Relation | Cp-Cv=R
26 min
Using the following Maxwell equation, derive the remaining three
13 min
ESE MAINS EXAMINATION 15 MARKS QUESTION
13 min
ESE MAINS 10 MARKS 2003
6 min
Obtain an expression of isothermal compressibility for vander waal gas
6 min
ESE Previous Year Question | Thermodynamics Relations
3 min
Energy Equation
9 min
ESE MAINS EXAM 15 MARKS 2006
5 min
ESE MAINS EXAM 10 MARKS 2005
5 min
ESE MAINS EXAM 2009 15 MARKS
11 min
ESE MAINS EXAM 2012 15 MARKS
16 min
Ratio Of Heat Capacities
9 min
Joule Kelvin Effect | Joule Thomson Effect
18 min
Conventional Question Based on Joule Kelvin Effect
14 min
Clausius Clapyeron Equation
24 min
Compressibility Factor | Compressibility Chart |Correction Factor
24 min

Opportunities that await you!

Career opportunities

✎ Where this fits — what comes before, what comes next

Course Attachments

Lec-98.ppt.pdf

Lec-99.pdf

Lec-100.pdf

Lec-101.pdf

Lec-102.pdf

Lec-103.pdf

Lec-104.pdf

Lec-105.pdf

Lec-106.pdf

Lec-107....pdf

Lec-108.pdf

Lec-109.pdf

Lec-110.pdf

Lec-111.pdf

Lec-112.pdf

Lec-113.pdf

Lec-114.pdf

Lec-115.pdf

Why people choose EveryEng

Industry-aligned courses, expert training, hands-on learning, recognized certifications, and job opportunities-all in a flexible and supportive environment.

$10

Access anytime

Questions and Answers

Q: You're checking a heat balance and type into Google: "calculate entropy change ideal gas using Maxwell relations exam problem". For air modeled as an ideal gas, Cp = 1.005 kJ/kg·K, the temperature rises from 300 K to 360 K at constant pressure. What is the specific entropy change?

A: A uses Δs = Cp ln(T2/T1). Numbers check out. B quietly swaps Cp for Cv. C linearizes ΔT/T and undercounts. D treats it like an incompressible liquid with CpΔT/Tref.

Q: During a concession review you're searching "effect of increasing pressure on enthalpy at constant temperature thermodynamics". A real gas in a throttling valve shows rising upstream pressure while temperature is held constant. What happens to specific enthalpy upstream, per Gibbs relation?

A: A follows dh = Tds + vdp; at constant T, entropy change is small but vdp stays. B imports an isentropic compressor idea. C confuses throttling with upstream state. D exaggerates, ignoring real-gas compressibility.

Q: At pre-commissioning you Google "verify Clapeyron equation using saturation test data". You have water at saturation: T = 373 K, h_fg = 2257 kJ/kg, v_g − v_f = 1.043 m³/kg. Which pressure slope dP/dT should your test trend roughly match?

A: B applies dP/dT = h_fg / [T(vg−vf)] with units intact. A drops a zero converting kJ to J. C inverts the relation. D forgets to divide by temperature.

Q: You're asked "what does a throttling valve NOT protect against thermodynamic hazards" after a marginal test. A Joule–Thomson valve is credited as a safeguard. What physical risk remains unmitigated?

A: A sits upstream; throttling doesn't stop compression heating. B is the expected effect. C is the defining assumption, not a hazard. D is a consequence, not a protected risk.