Video summary

ترمودینامیک یک جلسه 1 (مفاهیم اولیه) ، استاد امین روبراهان

Main summary

Key takeaways

Educational

Main ideas & concepts

Course/session purpose & format (by the presenter)

  • The video is a worked-solution style lecture for Thermodynamics Lesson 1.
  • At the start of each session, the presenter provides a brief outline of what will be covered.
  • Then the presenter solves examples with full steps.
  • A PDF of the slides matching the videos will be provided.
  • The presenter emphasizes that the teacher already covers the conceptual summary in class; the presenter’s “summary” is mainly a reminder.

Primary references used

  • Van Veylen, Thermodynamics + its solutions and exercises (available online).
    • Some editions may not show “Van Veylen” on the cover anymore, but the work is still known by that name.
  • Singleton, Thermodynamics (8th ed.)
    • Praised for strong problems/examples.
  • Moran & Shapiro, Fundamentals of Engineering Thermodynamics (5th ed.)
    • Also used; some problems/exercises differ from Singleton.

What thermodynamics is (core definition)

  • Word breakdown:
    • thermo = heat
    • dynamics = change
  • Thermodynamics is the study of changes caused by heat.
  • Examples of applications:
    • Gas turbines / power generation: heat → blade motion → mechanical work → electricity via generator
    • Refrigerators & heating/cooling systems
    • Aircraft engines
    • Building HVAC / air conditioning
    • Power plants (nuclear, gas/steam, etc.), which rely on heat-to-work/energy principles

Key prerequisite: choosing the “system”

  • To solve thermodynamics problems, you must:
    1. Select the system
    2. Correctly write the governing equations for that system

Two system types

  • Control Volume
    • A fixed region of space (volume) chosen as the system.
    • Cannot move (fixed in location).
    • Open system: matter and energy can enter and exit.
  • Control Mass
    • A specific mass chosen as the system.
    • Can move (moves with the material).
    • Closed system: exchanges energy but not matter (mass remains constant).

Fundamental variables & conversions introduced

  • Pressure

    • [ P = \frac{F}{A} ]

  • Specific volume

    • Inverse of density:

      • Density used earlier: [ \rho = \frac{M}{V} ]

      • Specific volume: [ v = \frac{V}{M} ] (“inverse of density”)

  • Temperature conversions mentioned

    • Kelvin: [ T(K) = T(°C) + 273 ]

    • Rankine: [ T(R) = T(°F) + 459 ]

    • Presenter notes the first conversion is the one to remember.

    • Unit consistency warning
    • When using (P \times A) to form forces, the pressure unit must match the force unit system:
      • Pa × m² = N
      • kPa × m² = kN (and then acceleration/mass units must be consistent)

Equilibrium vs. thermodynamics problems

  • The course uses static-equilibrium concepts as a stepping stone.
  • Even in thermodynamics (where equilibrium problems appear earlier), you repeatedly apply:
    • Force balance / no acceleration ideas under equilibrium or quasi-equilibrium.

  • Typical method for equilibrium-style tasks:

    1. Choose the system
    2. Draw a free-body diagram (FBD)

    3. Write equilibrium: [ \sum F = 0 ]

    4. Identify forces acting on the system, such as:

      • spring forces (e.g., (kx))
      • weight ((mg))
      • external applied forces (e.g., from hand/surface)

Worked problem methodology (structured steps)

A) Spring–piston–cylinder equilibrium approach (Van Veylen Q289)

Given (as described)

  • A 5 kg piston in a cylinder (radius given in the original problem).
  • A linear spring under the piston:
    • spring force stated as (F_s = kx), where (x) is deflection from equilibrium.
  • Outside air pressure: 100 kPa
  • Initial internal pressure: 4400 kPa
  • Initial air volume under piston: 14 liters
  • A valve is opened and the piston rises 2 cm

Procedure used

  1. Select the system: the piston (since it moves).
  2. Draw an FBD in the initial state, including forces such as:
    • internal pressure forces on the piston (as framed in the narration)
    • external atmospheric pressure on exposed faces
    • piston weight (mg)
    • spring force (kx)
  3. Determine unknown spring deflection using geometry:
    • use (V = A h) so displacement comes from (h = V/A)
    • convert liters to m³ and keep units consistent
  4. Write equilibrium equation
    • use (P = F/A \Rightarrow F = PA) for pressure forces
    • solve to find the spring-related constant/term (presenter shows the algebra leading to a numerical result)
  5. Apply changes after opening the valve
    • piston rise by 2 cm changes the deflection/geometry
    • write equilibrium again with unknown final pressure (P_2)

  6. Solve for final pressure

    • result stated:

      • [ P_2 \approx 517.5\ \text{kPa} ]

    • unit caution reiterated

B) Hydraulic lift equilibrium approach (Van Veylen Q288, hydraulic piston)

Given (as described)

  • Two cylinders connected by a T-shaped piston.
  • Cylinder A pumped to 5500 kPa.
  • Pressure inside A treated as 5000 kPa (as interpreted in the narration).
  • Mass of piston assembly: 25 kg
  • Atmospheric pressure acts on side surfaces; presenter argues it cancels because of equal pressure on both sides.
  • Asked to find the pressure on the “shoe/top” (unknown pressure (P_B)).

Procedure used

  1. Select system: the moving T-shaped piston (control mass).
  2. Draw an FBD
    • upward force from pressure on one face: (P_A A_A)
    • downward forces from pressure on other face(s): (P_B A_B)
    • include (mg)
  3. Handle atmospheric pressure
    • if atmospheric pressure acts with equal effective areas on opposing faces, net contribution is zero → omit it
  4. Use area difference to reduce unknowns
    • employ:
      • (A’ = A_A - A_B)
    • (presenter indicates this helps eliminate an extra variable)
  5. Write equilibrium and solve for (P_B)
    • result stated:
      • [ P_B \approx 5996\ \text{kPa} \approx 6\ \text{MPa} ]

C) Manometer (multi-fluid) method & sign convention

Core definitions

  • Absolute pressure
    • referenced to vacuum

  • Gauge pressure (relative pressure)

    • [ P_{gauge} = P_{absolute} - P_{atmospheric} ]

  • Manometer purpose

    • infer pressure differences using hydrostatic relations

General solving method

  • Choose a reference level on the left (or consistently choose one side).
  • Move through the fluid columns using sign conventions:
    • Moving down increases pressure:
      • treat as (+\rho g\Delta h)
    • Moving up decreases pressure:
      • treat as (-\rho g\Delta h)

  • Use fluid properties:

    • either absolute density ( \rho )
    • or specific weight: [ \gamma = \rho g ]

  • For SG (specific gravity / relative density):

    • “oil density is 79% SG” means: [ \gamma_{oil} = 0.79\,\gamma_{water} ]

Example 1 (Sengel / multi-fluid manometer)

  • Given multiple fluids (water, oil, mercury), asked for air gauge pressure.

  • Steps (as described):

    • start from left reference level
    • apply sign changes while stepping through each fluid segment
    • combine terms
    • convert from absolute to gauge using: [ P_{gauge} = P_{absolute} - P_{atmospheric} ]

  • Final numeric values were not cleanly preserved in the transcription.

Example 2 (Sengel / U-tube manometer with pipeline gas)

  • Given manometer gauge reading 370 kPa, asked for gas line relative pressure.
  • Steps (as described):
    • choose left reference level and interpret that pressure level as the gas’s relative pressure
    • traverse fluids using the sign convention for hydrostatic changes
    • convert SG values into relative specific weights (water reference; mercury high; gas contributions small)
    • neglect air density effects (very small)
  • Result stated:
    • [ P \approx 345.6\ \text{kPa} ]

Example 3 (Maram & Shapiro / two-tank absolute pressure using barometer)

  • Setup: tank A inside tank B; both contain air.
  • A barometer/manometer reading is given as a fraction of a bar (transcription approximations like “1 and 14th of the bar”).
  • Asked:
    • absolute pressure in tank A and tank B (in bar)
    • relative to atmospheric
Procedure & unit conversion
  • Use reference level method again with hydrostatic balance.

  • Emphasize bar → Pascal conversion:

    • [ 1\ \text{bar} = 10^5\ \text{Pa} = 100\ \text{kPa} ]

  • Convert mercury properties via (\gamma) / specific weight.

  • Compute absolute pressures and express them in bar.

  • Use relationship:

    • [ P_{abs,A} = P_{abs,B} + P_{relative,A} ]

  • Final answers stated (in transcription):

    • tank B absolute pressure converted to about (1.27\times10^{-2}) bar
    • tank A obtained using the relative-pressure relation

Closing points from the presenter

  • The session is described as mostly foundational, covering:
    • system/control volume vs control mass
    • pressure and specific volume
    • equilibrium/FBD basics
    • manometer techniques and unit conversions
  • More “serious” thermodynamics topics will be covered in the next session.

Speakers / sources featured

  • Speaker/Presenter: Amin Ruban (also referenced as “استاد امین روبراهان” in the title)
  • Textbooks used (sources):
    • Van Veylen, Thermodynamics (and solutions)
    • Singleton, Thermodynamics (8th edition)
    • Moran & Shapiro, Fundamentals of Engineering Thermodynamics (5th edition)
    • Sengel (referenced for manometer-related problems; book title not fully captured)
    • Maram & Shapiro (referenced for pressure units/barometer-related problems)

Original video