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UkoKoshka [18]
3 years ago
11

WHAT IS A TOROID IN HYDRAULUCS?

Engineering
2 answers:
slamgirl [31]3 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]3 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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A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0
3 years ago
A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
Illusion [34]
Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/



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6 0
3 years ago
How to study thermodynamics?​
expeople1 [14]

It is study of the relationships between heat, temprature, work and energy

7 0
3 years ago
Read 2 more answers
HOW TO CALCULATE MARGINAL RATE
Basile [38]

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

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Explanation:

8 0
3 years ago
Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following
torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches 5\times 10^5

(a).  \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16  \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\

(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}

7 0
3 years ago
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