An important fluid property is the kinematic viscosity which determines the viscous, or frictional, forces acting in a flow. The
kinematic viscosity of air multiplied by 106 is given at 350, 450, 500, 550, and 650 K as 20.92, 32.39, 38.79, 45.57, and 60.21 m2/s, respectively. Using any suitable interpolation method, compute the intermediate values at 400 and 600 K. Compare the results obtained with the values given in the literature as 26.41 × 10−6 m2/s and 52.69 × 10−6 m2/s, respectively. Also, solve the problem using the interp1 command in MATLAB, and compare the results with those obtained earlier
1 answer:
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Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re =
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re =
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s
Answer:
radius = 9.1 × m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = .................1
here F is applied load and is length
put here value and we get
radius =
solve it we get
radius = 9.1 × m