Class 11 - Unit # 07 : Fluid Dynamics - Solved Numericals
Physics Numerical Sheet: Fluid Dynamics
Physics Numerical Sheet Fluid Dynamics
1. Two spherical raindrops of equal size are falling through air at a velocity of 0.08 m/s. If the drops join together forming a large spherical drop, what will be the new terminal velocity?
Data:
Terminal velocity of small drop (v) = 0.08 m/s
Number of small drops (n) = 2
Terminal velocity of large drop (v') = ?
Solution:
When two small drops of radius r combine, the total volume remains conserved, forming a large drop of radius R:
Volume of Large Drop = 2 × (Volume of Small Drop)
(4/3)πR³ = 2 × (4/3)πr³
R³ = 2r³ ➔ R = 2^(1/3) × r
Terminal velocity is directly proportional to the square of the radius ($v \propto r^2$):
v' / v = (R / r)²
v' / v = [2^(1/3) × r / r]² = 2^(2/3)
v' = v × 2^(2/3)
v' = 0.08 × 1.5874 ≈ 0.127 m/s
RESULT: The terminal speed of the big combined drop is 0.13 m/s.
2. Calculate the viscous drag on a drop of oil of 0.1 mm radius falling through air and its terminal velocity. (Viscosity of air = 1.8 × 10⁻⁵ Pa·s; density of oil = 850 kg/m³)
RESULT: The terminal velocity of the oil drop is 1.03 m/s and the viscous drag force is 3.5 × 10⁻⁸ N.
3. What area must a heating duct have if air moving 3.0 m/s along it can replenish the air every 15 minutes in a room of volume 300 m³? Assume air density remains constant.
The volumetric flow rate can be computed via both volume delivery over time and cross-sectional dynamics:
Volume Flow Rate = V / t
Volume Flow Rate = A × v
Equating the two relationships:
A × v = V / t
A = V / (v × t)
A = 300 / (3.0 × 900) = 300 / 2700
A ≈ 0.111 m²
RESULT: The cross-sectional area of the heating duct must be 0.11 m².
4. Water circulates throughout a house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm diameter pipe on the second floor 5.0 m above? Assume the pipes do not divide into branches.
RESULT: The volume flow rate of the water escaping the faucet is 4.12 × 10⁻³ m³/s.
6. The stream of water emerging from a faucet 'necks down' as it falls. The cross-sectional areas are 1.2 cm² and 0.35 cm². The two levels are separated by a vertical distance of 45 mm. At what rate does water flow from the tap?
Data:
Upper Area (A₁) = 1.2 cm² = 1.2 × 10⁻⁴ m²
Lower Area (A₂) = 0.35 cm² = 0.35 × 10⁻⁴ m²
Vertical distance (h) = 45 mm = 0.045 m
Acceleration due to gravity (g) = 9.8 m/s²
Volume flow rate (Q) = ?
Solution:
From the Equation of Continuity, $v_2 = (A_1 / A_2)v_1$:
v₂ = (1.2 / 0.35)v₁ ≈ 3.4285v₁
Applying Bernoulli’s Equation across the falling stream vertical drop:
P₁ + ½ρv₁² + ρgh = P₂ + ½ρv₂² + 0
Since the falling stream is fully exposed to open atmospheric air, $P_1 = P_2$:
RESULT: The precise volumetric flow rate at the tap is 34.3 cm³/s (or 3.43 × 10⁻⁵ m³/s).
7. Water leaves the jet of a horizontal hose at 10 m/s. If the velocity of water within the hose is 0.40 m/s, calculate the pressure within the hose. Density of water is 1000 kg/m³ and atmospheric pressure is 100000 Pa.
Since the hose setup is perfectly horizontal, potential energy changes vanish ($h_1 = h_2$). Apply Bernoulli's Theorem:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ = P₂ + ½ρ(v₂² - v₁²)
P₁ = 100000 + ½(1000)(10² - 0.40²)
P₁ = 100000 + 500(100 - 0.16)
P₁ = 100000 + 500(99.84) = 100000 + 49920
P₁ = 149920 Pa
RESULT: The static pressure inside the upstream hose channel is 1.5 × 10⁵ Pa.
8. What is the maximum weight of an aircraft with a wing area of 50 m² flying horizontally, if the velocity of the air over the upper surface of the wing is 150 m/s and that over the lower surface is 140 m/s? Density of air is 1.29 kg/m³.
Data:
Wing Planform Area (A) = 50 m²
Air speed on upper surface (v_top) = 150 m/s
Air speed on lower surface (v_bottom) = 140 m/s
Density of air (ρ) = 1.29 kg/m³
Maximum aircraft weight sustained (W) = ?
Solution:
According to Bernoulli's equation, the aerodynamic lift pressure difference ($\Delta P$) equals:
For steady horizontal cruise flight, total upward Lift Force must balance aircraft Weight:
Weight (W) = ΔP × Area
W = 1870.5 × 50 = 93525 N
RESULT: The maximum weight structural capacity of the aircraft is 93525 N.
9. A liquid flows through a pipe with a diameter of 0.50 m at a speed of 4.20 m/s. What is the rate of flow in L/min?
Data:
Diameter of pipe (d) = 0.50 m
Radius (r) = d / 2 = 0.25 m
Flow speed (v) = 4.20 m/s
Volumetric Flow Rate (Q) = ? (in L/min)
Solution:
First, evaluate the localized cross-sectional profile area:
Area (A) = πr² = 3.1416 × (0.25)² = 0.19635 m²
Calculate the fundamental standard flow rate metric in cubic meters per second:
Q = A × v = 0.19635 × 4.20 = 0.82467 m³/s
Convert the metric unit expression from m³/s to Liters/minute ($1\text{ m}^3 = 1000\text{ L}$ and $1\text{ min} = 60\text{ s}$):
Q = 0.82467 × 1000 × 60 = 49480 L/min
RESULT: The volumetric rate of flow is approximately 49500 L/min.
10. Calculate the average speed of blood flow in the major arteries of the body, which have a total cross-sectional area of about 2.1 cm². Use the data of standard physiological reference volume flow rate (from example flow context: Q = 26.46 cm³/s).
Data:
Total cross-sectional profile area (A) = 2.1 cm²
Total reference circulatory flow rate (Q) = 26.46 cm³/s
Average flow velocity (v) = ?
Solution:
According to the basic physical definition inside the Equation of Continuity:
Q = A × v
Rearranging the formula parameters to directly resolve velocity:
v = Q / A
v = 26.46 / 2.1
v = 12.6 cm/s
RESULT: The average speed of blood flow inside the major structural arterial paths is 12.6 cm/s.
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