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UkoKoshka [18]
4 years ago
11

WHAT IS A TOROID IN HYDRAULUCS?

Engineering
2 answers:
slamgirl [31]4 years ago
8 0

Answer:

Expl

A hydraulic actuator comprised of a toroidal piston within a toroidal enclosure, with differential fluid pressure alternatively applied to upper and lower surfaces of the piston to cause a reciprocating motion, and with plural double ended piston rods extending in parallel above and below the piston, and slidably extending in fluid sealed relation through end caps of the toroidal enclosure to distribute the vibrational force produced by the reciprocating piston over plural points of a surface area of the mass to be vibrated, thereby reducing the likelihood of stress, strain, or harmonics in the mass.anation:

xenn [34]4 years ago
3 0

Answer:

A hydraulic actuator comprised of a toroidal piston within a toroidal enclosure, with differential fluid pressure alternatively applied to upper and lower surfaces of the piston to cause a reciprocating motion, and with plural double ended piston rods extending in parallel above and below the piston, and slidably extending in fluid sealed relation through end caps of the toroidal enclosure to distribute the vibrational force produced by the reciprocating piston over plural points of a surface area of the mass to be vibrated, thereby reducing the likelihood of stress, strain, or harmonics in the mass.

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Is it a problem that the stress intensity factor, K, from Irwin's near-tip approximation approaches infinity as you get close to
valina [46]

Answer:

No it is not a problem

Explanation:

It is not a problem because the stress intensity factor K would approach infinity as you get close to a crack tip and the intensity factor would approach Zero as you get too far away from the crack tip and this is simply because a crack is a notch with zero tip radius .

and The application of stress intensity factor k in respect to present fatigue crack tip is termed " linear elastic fracture mechanics "

3 0
3 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
3 years ago
Which of the following is true regarding screw gauges and shank?
agasfer [191]
correct me if i’m wrong i’m pretty sure it’s B i’ve had the same question
6 0
3 years ago
18. What is being shown in the above Figure?
slavikrds [6]

D. Camshaft gear backlash is being checked​

hope this helps :)

8 0
3 years ago
Read 2 more answers
A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps,
masha68 [24]

Answer:

(a). max possible efficiency = 55.62%

(b). max power output = = 133.5 MW

Explanation:

From the question we were given the Maximum temperature in the system as

Tmax = 500°C

Minimum temperature in the system Tmin = 70°C

the Heat supplied to the boiler Qb = 240000 KJ/s

we use the temperature conversion factor from °C to K

given T(K) = T (°C) + 273

⇒ Tmax = 500 + 273 = 773 K

⇒ Tmin = 70 + 273 = 343 K

(a). we are to determine the maximum possible thermal efficiency;

(Πth)max = 1 - Tmin/Tmax

(Πth)max = 1 - 343/773  = 0.5562

(Πth)max = 55.62%

(b). to determine the maximum possible power output for the plant we have;

(Πth)max = Wmax/Qb

where Wmax rep the maximum power output

(Πth)max = 0.5562

Qb = 240000

∴ Wmax = 0.5562 × 240000 = 133505.6 Kw

Wmax = 133.5 MW

cheers i hope this helps

3 0
3 years ago
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