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2 years ago

Summarize the difference in hydraulic and pneumatic systems.

Engineering

1 answer:

Naily [24]2 years ago

7 0

Answer:

The main difference between a pneumatic system and a hydraulic system lies in how the energy necessary to make these systems work is generated: thus, while in the pneumatic system this energy is generated through the compression of gases, in the hydraulic system the energy is obtained Through the insertion of pressurized fluids, such as water. Furthermore, hydraulic systems are closed circuits, where the water constantly circulates within the system; while in pneumatic systems the air that is compressed and used is constantly renewed.

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago

Water at 200C flows through a pipe of 10 mm diameter pipe at 1 m/s. Is the flow Turbulent ? a. Yes b. No

Degger [83]

Answer:

Yes, the flow is turbulent.

Explanation:

Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.

Given:

Diameter of pipe is 10mm.

Velocity of the pipe is 1m/s.

Temperature of water is 200°C.

The kinematic viscosity at temperature 200°C is $1.557\times10^{-7}$ m2/s.

Calculation:

Step1

Expression for Reynolds number is given as follows:

$Re=\frac{vd}{\nu}$

Here, v is velocity, $\nu$ is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

$Re=\frac{vd}{\nu}$

$Re=\frac{1\times(10mm)(\frac{1m}{1000mm})}{1.557\times10^{-7}}$

Re=64226.07579

Thus, the Reynolds number is 64226.07579. This is greater than 2000.

Hence, the given flow is turbulent flow.

5 0

3 years ago

For welding the most important reason to use jigs and fixtures in a welding shop is to

diamong [38]

Answer:

Reduce manufacturing costs.

Explanation:

Hope This Helps

Have A Great Day

7 0

2 years ago

Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and

Alborosie

Answer:

$V_{pp}=2V$

Explanation:

Source Voltage $V= 120V$

Frequency $f=60Hz$

Peak output voltage $Vp=15V$

Peak Output Voltage with filter $V_p'=14V$

Generally the equation for Peak to peak voltage is mathematically given by

$V_p'=V_p-\frac{V_{pp}}{2}$

Therefore

$V_{pp}=2(V_p-v_p')$

$V_{pp}=2(15-14)$

$V_{pp}=2V$

5 0

2 years ago