Summary of "2026년 1회 소방설비기사 기계 필기 기출문제 2. 소방유체역학"
Main ideas / concepts covered
The video is a solutions/teaching walkthrough of multiple-choice problems for:
- Firefighting equipment engineer (소방설비기사)
- Mechanical theory (기계)
- Fluid mechanics (소방유체역학)
The lecturer repeatedly:
- states a concept (e.g., hydrostatics, flow measurement, ideal gas/thermodynamics),
- checks options option-by-option to determine which statements are correct/incorrect,
- derives formulas and substitutes given numerical values.
Key topic areas appearing across the subtitles
- Fluid properties & hydrostatic pressure directions
- Gauge vs. absolute pressure relations
- Buoyancy / specific gravity
- Flow rate & velocity conversions (e.g., gravimetric flow rate, cross-sectional area)
- Pitot-static / wind tunnel airspeed using ideal gas + dynamic pressure
- Specific gravity measurement using buoyancy (displacement / Archimedes)
- Bernoulli / Darcy–Weisbach type reasoning for losses vs. velocity
- Pump shaft power and cavitation prevention
- Polytropic processes ((PV^n=\text{constant})), including work sign interpretation
- Reynolds number and minimum pipe diameter for onset conditions
- Orifice discharge using coefficients + Torricelli
- Water compressibility to estimate applied pressure
- Thermodynamic definitions/classification via polytropic exponent (n)
- Heat required for phase change (sensible heat + latent heat)
- Surface tension formulas for soap bubbles
- Momentum/impact force on a plate at an angle
- Isothermal compression using Boyle’s law
- Thermal efficiency of the Carnot cycle
- Specific heat ratio correctness checks ((\gamma))
Methodologies & instruction-style steps (as implied by the lecturer)
A) Hydrostatics: choosing correct pressure-direction/property statements
Pressure direction on a surface
- Pressure on a surface acts in the normal (perpendicular) direction—not “horizontally across the surface of action” (as framed in the lecture).
Pressure in open containers
- Check how pressure relates to:
- density (argued to be proportional, not inversely),
- depth (argued to be proportional, not inversely).
Pressure in closed containers
- Pressure at a single point has the same magnitude in all directions.
B) Gauge pressure vs. absolute pressure
Remember the standard relations:
-
Gauge pressure [ P_g = P_{abs} - P_{atm} ]
-
Absolute pressure [ P_{abs} = P_{atm} + P_g ]
Vacuum pressure framing (as stated in the explanation):
- “Subtract vacuum pressure from atmosphere” → yields absolute pressure.
C) Buoyancy and predicting weights / specific gravity
Meaning of buoyancy
- Buoyancy is the upward supporting force.
Weight/force relationships
- The lecturer rejects a wrong claim like:
- “subtracting weight in water from buoyancy gives weight in air”
- Instead, the lecture uses appropriate accounting for:
- buoyancy vs. apparent/immersed weight.
D) Gravimetric flow rate → average velocity in a pipe
Problem type
- Given a “weight flow rate” (gravimetric flow rate) in N/s, find average pipe velocity (v).
Steps used
- Treat the given value as gravimetric flow rate.
-
Use the relation involving specific weight:
- [ \gamma = SG \times \gamma_{water} ]
-
Compute pipe cross-sectional area: [ A = \frac{\pi}{4}d^2 ]
-
Solve for (v) using the relation between:
- gravimetric flow rate,
- (\gamma),
- (A),
- (v).
- Perform a units check (e.g., ensure N/s is consistent with (\gamma) in N/m³).
E) Pitot-static / airspeed from pressure difference using ideal gas
Steps shown
-
Use an airspeed form: [ v = \sqrt{\frac{2\,\Delta P}{\rho}} ]
-
Compute air density using ideal gas: [ \rho = \frac{P}{RT} ] Convert temperature from °C to Kelvin.
-
Compute the pressure difference: [ \Delta P = \gamma_{fluid}\,h ]
-
Use specific gravity: [ \gamma_{fluid} = SG \times \gamma_{water} ]
-
Convert (h) (e.g., cm → m). 4. Substitute into the velocity formula and calculate.
-
F) Specific gravity from an object weighed in air vs. in liquid
Steps shown
-
Specific gravity relative to water: [ S = \frac{\gamma_{liquid}}{\gamma_{water}} ]
-
Determine buoyancy from weight difference:
- Buoyancy (=) (weight in air) − (apparent weight in liquid)
-
Compute liquid specific weight from buoyancy: [ B = \gamma_{liquid}V_{submerged} \Rightarrow \gamma_{liquid}=\frac{B}{V} ]
-
Divide by (\gamma_{water}) to obtain (S).
G) Carnot thermal efficiency
Use the lecturer’s formula:
[ \eta = 1 - \frac{T_{low}}{T_{high}} ]
Steps:
- Convert temperatures to Kelvin.
- Substitute into (\eta).
- Convert to percent by multiplying by 100.
H) Specific heat ratio ((\gamma)) correctness checking
Key points used:
- Static (constant volume) specific heat definition relates to internal energy change.
-
[ \gamma = \frac{c_p}{c_v} ]
-
Since generally (c_p > c_v), typically:
- (\gamma > 1)
Therefore, options stating “(\gamma) is always < 1” are judged incorrect.
I) Pump shaft power
Key emphasis
- Shaft power differs from electrical power due to efficiency factors (not the focus of the lecture).
Formula style (as presented)
- The lecturer uses a simplified proportional form such as: [ p \approx \frac{9.8\,q\,h}{\eta} ] (with unit conversions implied)
Steps used
- Compute total head:
- includes actual lift + pipeline losses
- Convert flow units if needed (e.g., m³/min → m³/s by dividing by 60).
- Substitute and compute the kW result.
J) Darcy–Weisbach loss relationship to velocity
Referenced form: [ h_f = f\frac{L}{D}\frac{v^2}{2g} ]
Key relationship asserted:
- Since (v^2) is in the numerator, head loss is proportional to the square of velocity.
K) Polytropic process work using (PV^n=\text{constant})
Steps used
-
Start with: [ P_1V_1^n = P_2V_2^n ]
-
Solve for (V_2).
-
Use a polytropic work expression like: [ W=\frac{P_2V_2 - P_1V_1}{1-n} ] (equivalent forms may be used)
-
Interpret the sign:
- Negative work → compression case → energy absorbed from outside.
L) Reynolds number / minimum pipe diameter
Steps used
-
Reynolds number: [ Re=\frac{\rho vD}{\mu} ]
-
Use a form with kinematic viscosity (\nu): [ Re \sim \frac{vD}{\nu} ]
-
Express velocity using (Q) and area (A).
- Substitute and solve algebraically for minimum diameter (D).
- Substitute the given numerical values.
M) Orifice discharge with coefficients + Torricelli
Steps used
-
Compute the discharge coefficient product (as described): [ C = C_v \times C_c ]
-
Use Torricelli’s ideal velocity: [ v=\sqrt{2gh} ]
-
Compute flow rate: [ Q = C\,A\,\sqrt{2gh} ]
-
Orifice area: [ A = \frac{\pi}{4}d^2 ]
-
Substitute (h) (water depth to orifice center) and calculate.
N) Applied pressure from water compressibility
Steps shown
-
Use compressibility/bulk modulus relation: [ K=\frac{1}{\beta} ]
-
Relate pressure change to volumetric strain: [ \Delta P = K\left|\frac{\Delta V}{V}\right| ]
-
Convert percent volume decrease to decimal:
- e.g., (0.5\% \to 0.005)
- Compute (\Delta P) (Pa).
O) Polytropic classification by exponent (n)
For the general relation (PV^n=\text{constant}), classification depends on (n):
- (n=0): lecture labels a specific “isothermal-like/equilibrium-equivalent” case (subtitle wording is garbled but associated to (n=0))
- (n=1): isothermal
- (n=k) (ratio of specific heats): adiabatic
- (n=\infty): constant volume (isochoric)
The lecture concludes which multiple-choice option matches each case.
P) Heat required for phase change (ice to steam; multi-stage)
(As described in 4 stages.)
-
Warm ice from (-15^\circ C) to (0^\circ C): [ Q_1 = mc_{ice}\Delta T ]
-
Melt ice at (0^\circ C): [ Q_2 = mL_f ]
-
Heat water from (0^\circ C) to (100^\circ C): [ Q_3 = mc_{water}\Delta T ]
-
Vaporize water at (100^\circ C): [ Q_4 = mL_v ]
Total: [ Q = Q_1+Q_2+Q_3+Q_4 ]
Q) Pump cavitation prevention (multiple-choice logic)
- Correct idea emphasized: shorten suction lift → reduces suction losses → helps prevent cavitation.
- Incorrect idea explicitly rejected: increase pump rotational speed (makes cavitation more likely).
- Another option treated as preventive in the lecture: using two or more pumps (as stated).
R) Thermal equilibrium definition
Thermal equilibrium is defined as:
- after contact of two bodies, no further heat flow occurs,
- temperatures become equal and remain unchanged.
S) Surface tension of a soap bubble from internal excess pressure
Soap bubble vs droplet distinction (as stated):
- soap bubble uses a denominator form like 8 instead of 4 for a droplet.
Formula structure used: [ \sigma = \frac{\Delta P \, d}{8} ]
Steps:
- Convert diameter from mm to m.
- Substitute (\Delta P) and (d) to get (\sigma).
T) Force on a plate from a jet at an angle
Momentum/force logic used:
- perpendicular component depends on (\sin\theta), e.g.: [ F_\perp \propto \rho A v^2 \sin\theta ]
Steps implied:
- Use (\rho = 1000\,\text{kg/m}^3),
- compute area from diameter,
- substitute (v) and (\theta = 60^\circ),
- compute perpendicular force.
U) Isothermal compression: Boyle’s law
Steps:
-
Isothermal: [ P_1V_1 = P_2V_2 ]
-
Solve: [ \frac{V_1}{V_2}=\frac{P_2}{P_1} ]
-
Substitute pressures to get the ratio.
Speakers / sources featured (as visible in subtitles)
- Single lecturer/instructor (unnamed; uses “I/we” narration style)
- Music cue appears (no identifiable artist is named)
Category
Educational
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