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3 years ago

Describe three parts of a fluid power system and the roles played by each to make the device work.

Engineering

1 answer:

artcher [175]3 years ago

5 0

Three parts of a fluid power system and the roles played by each to make the device work.

Explanation:

1.Fluid power is subdivided into hydraulics using a liquid such as mineral oil or water, and pneumatics using a gas such as air or other gases.

2.Fluid power systems perform work by a pressurized fluid bearing directly on a piston in a cylinder or in a fluid motor.

Hydraulics Pumps

Dynamic (non positive displacement) Pumps

This type is generally used for low-pressure, high volume flow applications. Since they are not capable of withstanding high pressures, there is little use in the fluid power field. Their maximum pressure is limited to 250-300 psi. This type of pump is primarily used for transporting fluids from one location to another. Centrifugal and axial flow propeller pumps are the two most common types of dynamic pumps.

Positive Displacement Pumps

This type is universally used for fluid power systems. With this pump, a fixed amount of fluid is ejected into the hydraulic system per revolution of pump shaft rotation. These pumps are capable of overcoming the pressure resulting from the mechanical loads on the system as well as the resistance to flow due to friction. These two features are highly desirable in fluid power pumps. These pumps also have the following advantages over non positive displacement pumps:
High-pressure capability (up to 12,000 psi)

Small compact size

high volumetric efficiency

small changes in efficiency throughout the design pressure range

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P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

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2 years ago

What is the reason for the development of new construction of materials for human use? (Select all that apply.)

salantis [7]

Answer:

All 4 could be justified.

Explanation:

They all represent ultimate improvement.

7 0

2 years ago

John has just graduated from State University. He owes $35,000 in college loans, but he does not have a job yet. The college loa

IRISSAK [1]

Answer:

The correct response is "821.88". A further explanation is given below.

Explanation:

According to the question,

The largest amount unresolved after five years would have been:

= $35000\times (\frac{F}{P}, 4 \ percent,5 )$

= $35000\times 1.216 7$

= $42584.50$

Now,

time (t) will be:

= $5\times 12$

= $60 \ monthly \ payments$

So, monthly payment will be:

= $42582.85\times (\frac{A}{P}, 0.5 \ percent,60 )$

= $42584.50\times 0.0193$

= $821.88$

6 0

3 years ago

ILL GIVE BRAINLIEST!!!

Sedbober [7]

Get the app socratic I saw the answer to your question on the app but I ran out of screen time to show you

6 0

2 years ago

Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)

Novosadov [1.4K]

Answer:

a) ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

The phase is; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0

3 years ago