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2 years ago

The first law of motion describes the principle of

Engineering

2 answers:

Bumek [7]2 years ago

4 0

Answer:

"First Law of Motion describes the behavior of a massive body at rest or in uniform linear motion, i.e., not accelerating or rotating."

Explanation:

Nadya [2.5K]2 years ago

4 0

It describes the principle of inertia.

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How does Accenture generate value for clients through Agile and DevOps?

Nana76 [90]

By permanently locking in stakeholder requirements during a project's planning phase. -through highly detailed process documentation that is updated following every work cycle.

7 0

1 year ago

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Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v

MAXImum [283]

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = 20.7846 kN / 432 mm^2 = 48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = -36 kN / 648mm^2 = -55.55 MPa

4 0

1 year ago

A double-acting duplex pump with 6.5-in. liners, 2.5-in. rods, and 18-in. strokes was operated at 3,000 psig and 20 cycles/min.

Stella [2.4K]

Answer:

Pump factor = Fp = 7.854 gal/cycle

Ev = 82.00 %

P_H = 183.29 hp

Explanation:

Given data:

Dimension of duplex pump

6.5 inch liner

2.5 inch rod

18 inch strokes

Pressure 3000 psig

Pit dimension

7 ft wide

20 ft long

Ls = 18 inch

Velocity = (18)/10

volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100

we know that flow rate is given as = Area * velocity

Theoritical flow rate $= \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times N$

$Ev = \frac{7\times 12 \times 20\times 12\times 12 \times \frac{18}{10} inch^3/min}{\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 20}$

Ev = 82.00 %

Pump factor $Fp = = \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times Ev$

$Fp =\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 0.82$

Fp = 1814.22 in^3/cyl

Fp = 7.854 gal/cycle

Flow rate $q = NFp = 20 \times 7.854 = 157.08 gal/min$

Power $Ph = \frac{\DeltaP q}{1714} = \frac{3000 \times 157.08}{1714} = 274.93 hp$

6 0

2 years ago

A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses

yaroslaw [1]

Answer:

the temperature of the aluminum at this time is 456.25° C

Explanation:

Given that:

width w of the aluminium slab = 0.05 m

the initial temperature $T_1$ = 25° C

$T{\infty} =600^0C$

h = 100 W/m²

The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;

density ρ = 2702 kg/m³

thermal conductivity k = 231 W/m.K

Specific heat c = 1033 J/Kg.K

Let's first find the Biot Number Bi which can be expressed by the equation:

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}$

$Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}$

$Bi = \dfrac{2.5}{231}$

Bi = 0.0108

The time constant value $\tau_t$ is :

$\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}$

$\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}$

$\tau_t = \dfrac{2702* 0.025*1033}{100}$

$\tau_t = 697.79$

Considering Lumped capacitance analysis since value for Bi is less than 1

Then;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]$

where;

$Q = -\Delta E _{st}$ which correlates with the change in the internal energy of the solid.

So;

$Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}$

The maximum value for the change in the internal energy of the solid is :

$(pVc)\theta_1 = -\Delta E _{st}max$

By equating the two previous equation together ; we have:

$\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}$

Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75

Thus;

$0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}$

So;

$0.75= [1-e^{\dfrac {-t}{ 697.79}}]}$

$1-0.75= [e^{\dfrac {-t}{ 697.79}}]}$

$0.25 = e^{\dfrac {-t}{ 697.79}}$

$In(0.25) = {\dfrac {-t}{ 697.79}}$

$-1.386294361= \dfrac{-t}{697.79}$

t = 1.386294361 × 697.79

t = 967.34 s

Finally; the temperature of Aluminium is determined as follows;

$\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}$

$\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}$

$\dfrac{T - 600}{25-600}= 0.25$

$\dfrac{T - 600}{-575}= 0.25$

T - 600 = -575 × 0.25

T - 600 = -143.75

T = -143.75 + 600

T = 456.25° C

Hence; the temperature of the aluminum at this time is 456.25° C

3 0

2 years ago

Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at 2 MPa pressure. The exhaus

Anastaziya [24]

Answer:

The flowrate of water is 0.03556kg/s

Explanation:

Exhaust gases inlet temperature T1=4000C

Water inlet temperature T3=150C Exit Pressure of water as saturated vapor P4=2MPa

Mass flow rate of exhaust gases Heat lost to the surroundings Qgases=32kg/min

Mass flow rate of exhaust gases is 15 times that of the water

Heat exchangers typically involve no work interactions (w = 0) and negligible...

7 0

2 years ago

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