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2 years ago

What is the answer of this question please tell me dadada please please

Engineering

1 answer:

charle [14.2K]2 years ago

6 0

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A 1-mm-diameter methanol droplet takes 1 min for complete evaporation at atmospheric condition. What will be the time taken for

Svet_ta [14]

Answer:

Time taken by the $1\mu m$ diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = $1\times 10^{- 6} m$

Diameter of the smaller droplet, R' = $0.5\times 10^{- 6} m$

Now,

Volume of the droplet, V = $\frac{4}{3}\pi R^{3}$

Volume of the smaller droplet, V' = $\frac{4}{3}\pi R'^{3}$

Volume of the droplet ∝ Time taken for complete evaporation

Thus

$\frac{V}{V'} = \frac{t}{t'}$

where

t' = taken taken by smaller droplet

$\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}$

$\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}$

t' = $60\times 10^{- 9} s = 60 ns$

5 0

3 years ago

Nothing. i have nothing to say but that. other than that im good. :))))

ohaa [14]

Answer:

aw good <3

Explanation:

4 0

2 years ago

It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 800 MPa. Calc

Aloiza [94]

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0

2 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

Ratling [72]

Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

$\dfrac{350}{ FOS} =2 \times 88.51$

${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

7 0

2 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

3 0

2 years ago

What is the answer of this question please tell me dadada please please​

What is the answer of this question please tell me dadada please please