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3 years ago

How can I draw this image in 2D form

Engineering

1 answer:

Ket [755]3 years ago

5 0

Answer:

no it is not 2D

Explanation:

it is 3D

ok so follow these steps

- make hole

-make square

-make triangle

ok now your figure is ready

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Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should w

Setler79 [48]

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

$Re_{5}$ = $Re_{12}$

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

$Re_{5}$ = Reynold number of water pipe

$Re_{12}$ = Reynold number of oil pipe

$V_{5}$ = Velocity of water 5 diameter pipe = ?

$V_{12}$ = Velocity of oil 12 diameter pipe = 2.30

$v_{5}$ = Kinetic Viscosity of water = 1 x $10^{-6}$ $m^{2}$ /s

$v_{12}$ = Kinetic Viscosity of oil = 4 x $10^{-6}$ $m^{2}$ /s

$D_{5}$ = Diameter of pipe used for water = 0.05 m

$D_{12}$ = Diameter of pipe used for oil = 0.12 m

Use the formula

$\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}$ = $\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}$

By Removing square rots on both sides

${\frac{V_{5}XD_{5} }{v_{5}}}$ = ${\frac{V_{12}XD_{12} }{v_{12}}}$

${V_{5}$ = ${\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}$ x $v_{5}$

${V_{5}$ = [ (0.23 x 0.12m ) / (4 x $10^{-6}$ $m^{2}$ /s) x 0.05 ] 1 x $10^{-6}$ $m^{2}$ /s

${V_{5}$ = 1.38 m/s

4 0

3 years ago

The point at 0°, 0° is

IgorC [24]

X because it’s at the origin x and y

5 0

3 years ago

Read 2 more answers

Karen just checked her bank balance and the balance was 1111.11. She is kind of freaking out and saved the receipt, thinking tha

Likurg_2 [28]

Answer:

chronic stoner syndrome

Explanation:

"the universe just sends us messages sometimes mannnn, you just have to be ready to listen to them" lol

8 0

3 years ago

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 1.7. If, afte

luda_lava [24]

Answer:

It would take approximately 305 s to go to 99% completion

Explanation:

Given that:

y = 50% = 0.5

n = 1.7

t = 100 s

We need to first find the parameter k from the equation below.

$exp(-kt^n)=1-y$

taking the natural logarithm of both sides:

$-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}$

Substituting values:

$k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}$

Also

$t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}$

Substituting values and y = 99% = 0.99

$t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s$

∴ t ≅ 305 s

It would take approximately 305 s to go to 99% completion

8 0

3 years ago

Read 2 more answers

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 54 percent. Determine the rate of hea

Gennadij [26K]

Answer:

$\dot Q _{L} = 511.111 MW$ . Heat transfer can be higher if themal efficiency is lower.

Explanation:

The heat transfer rate to the river water is calculated by this expression:

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = (\frac{1}{\eta_{th}}-1 )\cdot \dot W\\\dot Q_{L} = (\frac{1}{0.54}-1)\cdot (600 MW)\\\dot Q _{L} = 511.111 MW$

The actual heat transfer can be higher if the steam power plant reports an thermal efficiency lower than expected.

8 0

4 years ago