A dash pot 10 cm diameter and 12.5 cm long slides vertically down in ...
Since the space between the dash pot and the cylinder is very small, i.e., the oil film is thin, we can presume that
du / dy = u / t
Where u is the piston speed and t is the oil film thickness.
Shear stress τ = μ du / dy = μ u / t
Shear force = Shear stress × area
= μ u / t (2πrl)
Given,
r = 10 / 2 = 5 cm = 0.05 m
u = 0.8 poise = 0.08 Nsec/m2
t = 10.05 − 10 / 2
= 0.025 cm = 0.00025 m
Viscous force equals the load of 10 N
∴ 10 = 0.08 × u / 0.00025 × (2π × 0.05 × 0.125)
Hence, piston speed u = 0.796 m/s
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A dash pot 10 cm diameter and 12.5 cm long slides vertically down in ...
Problem Statement: Find the speed with which the piston slides down if load on the piston is 10 N.
Given:
- Diameter of dash pot, D = 10 cm
- Length of dash pot, L = 12.5 cm
- Diameter of cylinder, d = 10.05 cm
- Viscosity of oil, μ = 0.80 poise
- Load on piston, W = 10 N
Solution:
The force acting on the piston is given by:
$$F = W - A\cdot P$$
where A is the area of annular space and P is the pressure of oil.
The velocity of piston is given by:
$$v = \frac{dx}{dt} = \frac{W}{c}$$
where c is the coefficient of viscous resistance.
The coefficient of viscous resistance is given by:
$$c = \frac{1}{\mu}\cdot \frac{S}{L}$$
where S is the area of the piston.
The area of annular space is given by:
$$A = \pi\cdot (\frac{d}{2})^2 - \pi\cdot (\frac{D}{2})^2$$
The area of the piston is given by:
$$S = \pi\cdot (\frac{D}{2})^2$$
Putting the values in the above equations, we get:
$$A = \pi\cdot (\frac{10.05}{2})^2 - \pi\cdot (\frac{10}{2})^2 = 0.181 cm^2$$
$$S = \pi\cdot (\frac{10}{2})^2 = 78.54 cm^2$$
$$c = \frac{1}{0.80}\cdot \frac{78.54}{12.5} = 7.789 N\cdot s/cm$$
$$F = 10 - 0.181\cdot P$$
Equating the forces, we get:
$$10 - 0.181\cdot P = 7.789\cdot \frac{dx}{dt}$$
Integrating both sides, we get:
$$t = \frac{8.92}{v} - 0.1133\cdot ln(\frac{8.92}{v})$$
where t is the time taken to slide down.
Differentiating both sides with respect to time, we get:
$$\frac{dt}{dx} = -\frac{8.92}{v^2} + \frac{0.1133}{v}$$
Putting the values, we get:
$$\frac{dt}{dx} = -\frac{8.92}{v^2} + \frac{0.1133}{v} = 0.0787$$
Solving for v, we get:
$$v = 0.78\ to\ 0.80\ cm/s$$
Answer: The speed with which the piston slides down is in the range of 0.78 to 0.80 cm/s.
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