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3 years ago

Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!

1 answer:

8 0

Of course they are needed because without them the society wouldn’t be as nice as it is right now and plus there would be no more buildings ! :)

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Oxygen enters an insulated 14.2-cm-diameter pipe with a velocity of 60 m/s. At the pipe entrance, the oxygen is at 240 kPa and 2

MatroZZZ [7]

Answer:

Entropy generation==0.12 KW/K

Explanation:

$s_2-s_1=C_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$

$s_2-s_1=0.891\ln \frac{291}{293}-0.2598\ln \frac{200}{240}$

$s_2-s_1=0.0412\frac{KJ}{kg-K}$

Mass flow rate= $\rho\times\dfrac{\pi}{4}d^2V$

$\rho_1=\dfrac {P_1}{RT_1}$

$\rho_1=\dfrac {240}{0.2598\times 293}$

$\rho_1=3.51\frac{kg}{m^3}$

mass flow rate= $\rho_1A_1V_1$

So by putting the values

Mass flow rate=2.97 kg/s

So entropy generation=(2.97)(0.0412)

=0.12 KW/K

8 0

4 years ago

A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two

vesna_86 [32]

Answer:

One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.

Explanation:

8 0

3 years ago

An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh

Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in

B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

(ii) Average shear stress at the web = 18.289 * 10^5 psi

(iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 * 10^6 * 2.375}{53.54}$ = 0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 *10^6*(8.5-2.375)}{53.54}$ = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = $\frac{wL}{2}$ = $\frac{1000*6*12}{2}$ = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - $\frac{0.5}{2}$ ) * $\frac{(2.375 - (\frac{0.5}{2} ))}{2}$

= 5.878 in^3

t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 ) psi

= 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = $\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}$

= 1.143 *10^5

4 0

4 years ago

Composite beam transformations treat the composite beam as ________________:

dimaraw [331]

Answer:

Steel concrete composite beams consists of a steel beam over which a reinforced concrete slab is cast with shear connectors. In conventional composite construction, concrete slabs are simply rested over steel beams and supported by them. These two components act independently under the action of loads, because there are no connection between the concrete slabs and steel beam.

Hope this helps :) -Mark Brainiest Please :)

4 0

3 years ago

Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l

Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

$m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\$

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

8 0

3 years ago