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tatyana61 [14]
3 years ago
11

Implement the function F(A, B, C, D) = Σm(1, 2, 3, 4, 6, 9, 12, 13, 15) as a minimal hazard-free SOP circuit. In your solution,

clearly indicate which product terms are required for minimal implementation and which ones are present to prevent hazards.

Engineering
1 answer:
Marina86 [1]3 years ago
3 0

Answer:

See attachment below

Explanation:

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yep

Explanation:

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A sleeve made of SAE 4150 annealed steel has a nominal inside diameter of 3.0 inches and an outside diameter of 4.0 inches. It i
irga5000 [103]

Answer:

Class of fit:

Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).

Here minimum shaft diameter will be greater than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

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3 years ago
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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0
3 years ago
Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Figure, with an initia
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The heat transferred to and the work produced by the steam during this process  is 13781.618 kJ/kg

<h3>​How to calcultae the heat?</h3>

The Net Change in Enthalpy will be:

= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg

Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)

= 1/2 x ( 75 + 225) x (5 - 2)

W = 450 KJ

From the First Law of Thermodynamics, Q = U + W

So, Heat Transfer = Change in Internal Energy + Work Done

= 13331.618 + 450

Q = 13781.618 kJ/kg

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Which is not required when working in a manufacturing facility?
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Flip flops are not required
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