ENTHALPY and INTERNAL ENERGY in 12 Minutes!

Enthalpy ~ Heat added or removed at constant pressure
Internal Energy ~ Heat added or removed at constant volume
Quality and Saturated Liquid & Saturated Vapor values for u and h.
0:00 Constant Volume vs. Constant Pressure
0:33 Heating at a Constant Volume
1:12 T-v Diagram at Constant Pressure
1:26 T-v Diagram at Constant Volume
2:10 First Law for Constant Volume
2:37 Q and Internal Energy Discussion
3:31 Change in Internal Energy
4:15 First Law for Constant Pressure
5:26 Enthalpy
5:57 Property Tables for u and h
6:21 Saturated Liquid & Saturated Vapor
6:50 Quality for u and h
7:16 Example for a Constant Volume Process
7:54 Solution
Example 1: • Saturated to Superheat...
Example 2: • Enthalpy from SATURATE...
Example 3: • ENTHALPY of a Superhea...
Previous Lecture:
7. Interpolating and Quality Derivation: • INTERPOLATION for Ther...
Next Lecture:
9. Compressed Liquid Properties: • Thermodynamics COMPRES...
_______________________________________
Other Thermodynamics Lectures:
1. Thermodynamics Intro and Units Review: • THERMODYNAMICS Basic U...
2. Zeroth Law of Thermodynamics: • ZEROTH LAW of Thermody...
3. Closed vs. Open Systems & Intensive vs. Extensive Properties: • CONTROL VOLUMES - Clos...
4. First Law and Energy Types: • Energy Conservation Eq...
5. Efficiency in Thermodynamic Systems: • EFFICIENCY of Thermody...
6. Five Regions of T-v Diagrams: • T-v Diagrams and PROPE...
7. Interpolating and Quality Derivation: • INTERPOLATION for Ther...
8. Enthalpy and Internal Energy: • ENTHALPY and INTERNAL ...
9. Compressed Liquid Properties: • Thermodynamics COMPRES...
10. Ideal Gas and Compressibility Factor: • Ideal Gas Equation and...
11. Specific Heats: • Thermodynamics SPECIFI...
12. Specific Heat Relationships of Ideal Gases and Incompressible Substances: • Specific Heat and Ener...
13. Polytropic Process Work: • Work Expressions for P...
14. Mass and Volumetric Flow Rate: • Volumetric and MASS FL...
15. Flow Work and Energy Conservation in Open Systems: • FLOW WORK & Energy Con...
16. Pipe Flow, Nozzles, Diffusers, and Throttling Devices: • Pipe Flow, Nozzles, Di...
17. Turbines, Compressors, and Pumps: • Thermodynamics - Turbi...
18. Mixing Chambers and Heat Exchangers: • Heat Exchangers and Mi...
19. Transient Systems:
20. Second Law Intro and Power Cycles: • 2nd Law Intro and POWE...
21. Refrigerators and Heat Pumps: • REFRIGERATION and Heat...
22. Carnot Cycles and Reversible Processes: • Reversible Processes a...
23. Entropy as a Thermodynamic Property: • Thermodynamics - ENTRO...
24. Reference Entropy and Specific Heats: • REFERENCE ENTROPY and ...
25. Isentropic Efficiency: • Turbines, Compressors,...
26. Brayton Cycle: • Ideal BRAYTON CYCLE Ex...
27. Intercooling, Reheating, and Regenerators: • Regeneration, Intercoo...
28. Otto Cycle:
29. Standard Diesel Cycle:
30. Rankine Cycle:
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________________________________________
Other Engineering Courses (Playlists):
/ lessboringlectures
Statics: • Statics: Engineering M...
Mechanics of Materials: • Mechanics of Materials
Machine Design: • Mechanical Engineering...
SolidWorks (CAD): • SolidWorks (CAD)
Fluid Mechanics: • Fluid Mechanics

Пікірлер: 18

  • @robertbengel2689
    @robertbengel2689Ай бұрын

    You are a wizard with this note-taking software

  • @ayushchoubey635
    @ayushchoubey6354 ай бұрын

    Sir, I have never on the Internet found lectures so precise and concise,you are truly amazing,* hats off *

  • @lucasfc4587
    @lucasfc45878 ай бұрын

    What???? It was so easy to follow im totally impressed, now I think I may understand enthalpy

  • @diyaazaghloul756
    @diyaazaghloul7562 жыл бұрын

    Amazing video thank you

  • @sohansingh2022
    @sohansingh202210 ай бұрын

    thank you

  • @seunmabogunje4586
    @seunmabogunje4586 Жыл бұрын

    Great

  • @diwaahkar
    @diwaahkar Жыл бұрын

    Thanks

  • @foxhound1008
    @foxhound10089 ай бұрын

    I love your videos, Question: If I was to take a fluid element a sea level, and raise it up to a higher altitude (which is at a lower pressure) the fluid element would expand, pushing back the atmosphere. The work done pushing back the atmosphere would cause the internal energy of the fluid element to decrease. The transferred energy, out of the fluid element, would be in the rest of the atmosphere. Of course, the atmosphere is an infinite reservoir, so you would not be able to see any temperature increase. However, what about flow thru a closed duct system, with a compressor? The "work" done by the compressor on a fluid element would appear as an increase in internal energy, an increase in kinetic energy (which we will assume is negligible) and some of the work will go into the fluid element "pushing" into the system. Where does the energy from the "pushing" go? The duct system is not an infinite reservoir like the atmosphere, does that energy appear spread out in the system as an increase in internal or kinetic energy?

  • @gooddeedsleadto7499
    @gooddeedsleadto74992 ай бұрын

    Why is in a steam drum not a constant volume process, it is heated with burners the pressure should also go up. How it is heated at constant pressure in a steam drum under constant volume? Thank u

  • @elisiopossamaifilho4998
    @elisiopossamaifilho4998 Жыл бұрын

    niiiiiice

  • @gooddeedsleadto7499
    @gooddeedsleadto74992 ай бұрын

    Line ( dome line) represents saturated liquid or saturated vapor? and everything within the dome is also saturated liquid or saturated vapor? And Left & right of the dome is sub cooled liquid and super heated vapor respectively? Please verify? Thank u

  • @ichbinangie
    @ichbinangie18 күн бұрын

    Which note taking software is this?

  • @alexanderlee7276
    @alexanderlee72762 ай бұрын

    So you aren't changing the potential energy by compressing water vapor inside a rigid tank?

  • @JMac___
    @JMac___Ай бұрын

    i love you

  • @iRul3TheSky1995
    @iRul3TheSky199511 ай бұрын

    1 bar is 100 kPa not 101.325 kPa

  • @uhafi

    @uhafi

    10 ай бұрын

    it is 101.325 kpa. 1 bar = 101325 Pa for converting it into kilo we divide by 100. 1 bar = 101325 ÷ 10³ = 101.325. you can round of that value to 100 Kpa. have a great day!🤍

  • @sundaydiego5939

    @sundaydiego5939

    9 ай бұрын

    @@uhafi le

  • @stoscherzando

    @stoscherzando

    7 ай бұрын

    ​@@uhafi No 1 bar = 10^5 Pa you're talking about 1 atm= 101 325 Pa