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3 years ago

Are the wooden pillars shown in the image below, a dead load?

Engineering

1 answer:

Brrunno [24]3 years ago

6 0

Answer:

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

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Traffic at a roundabout moves

sweet-ann [11.9K]

Answer:

b-counter-clockwise

Explanation:

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3 years ago

Disconnecting a circuit while in operation can create a voltage blank

zlopas [31]

Answer:

what is the question

Explanation:

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3 years ago

An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo

Vsevolod [243]

Answer:

a) Please see attached copy below

b) 0.39KJ

c) 20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

7 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum

adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress $\sigma$ = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder there exist both the circumferential stress and the longitudinal stress.

Circumferential stress $\sigma = \dfrac{pd}{2t}$

Making thickness t the subject; we have

$t = \dfrac{pd}{2* \sigma}$

$t = \dfrac{560000*3}{2*150000000}$

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

$\sigma = \dfrac{pd}{4t}$

$t= \dfrac{pd}{4*\sigma }$

$t = \dfrac{560000*3}{4*150000000}$

t = 0.0028 mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm

8 0

3 years ago