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Fudgin [204]
3 years ago
7

Calculate the diameter of pneumatic cylinder to drive a mechanism. The force exerted by the mechanism is 80 N. The return force

is negligible. The working pressure is 5 bars.
Engineering
1 answer:
sashaice [31]3 years ago
5 0

Answer:

14.27 mm

Explanation:

Force = 80 Newton

Pressure exerted = 5 bar = 5×10⁵ Pa

Pressure exerted = Force/Area

Area = Force/Pressure

Area=\frac{80}{500000}\\\Rightarrow Area=0.00016\ m^2

Area of the cylinder=πd²/4

\Rightarrow \frac{\pi}{4}d^2=0.00016\\\Rightarrow d^2=\frac{4\times 0.00016}{\pi}\\\Rightarrow d^2=0.0002037\\\Rightarrow d=0.01427\ m

Hence diameter of cylinder 14.27 mm

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Answer:

See attachment for chart

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The IPO chart implements he following algorithm

The expressions in bracket are typical examples

<u>Input</u>

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

<u>Processing</u>

Assign variable to the input number (x)

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2 years ago
1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
QveST [7]

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3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
3 years ago
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