Running ropes must be taken out of service if they have _____ broken wires in one strad in one lay
2 answers:
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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:
See explaination and attachment.
Explanation:
Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.
The starting point of the Navier-Stokes equations is the equilibrium equation.
The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.
The second step is to relate the deviatoric stress to viscosity in the fluid.
The final step is to impose any special cases of interest, usually incompressibility.
Please kindly check attachment for step by step solution.
Answer:
a) 28 stations
b) Rp = 21.43
E = 0.5
Explanation:
Given:
Average downtime per occurrence = 5.0 min
Probability that leads to downtime, d= 0.01
Total work time, Tc = 39.2 min
a) For the optimum number of stations on the line that will maximize production rate.
Maximizing Rp =minimizing Tp
Tp = Tc + Ftd
At minimum pt. = 0, we have:
dTp/dn = 0
Solving for n²:
The optimum number of stations on the line that will maximize production rate is 28 stations.
b)
Tp = 1.4 +1.4 = 2.8
The production rate, Rp =
The proportion uptime,