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3 years ago

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A car engine operates with a thermal efficiency of 20%. Assume the air-conditioner has a coefficient of performance of β = 2 wor

king as a refrigerator cooling the passenger compartment using engine shaft work to drive it. How much fuel energy per second is spent extra to remove 8 kJ/sec from the inside?

1 answer:

disa [49]3 years ago

5 0

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What are the well-known effects of electricity

Sever21 [200]

Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

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3 years ago

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The components of an electronic system dissipating 90 W are located in a 1-m-long circular horizontal duct of 15-cm diameter. Th

Artyom0805 [142]

Answer:

Given data

electronic system dissipating = 90 W

diameter = 15 cm

The components in the duct are cooled by forced air which enters at 32°C at a rate of 0.65 m3 /min

the duct and the remaining = 15 %

See pictures for solution.

Explanation:

See attached pictures for detailed explanation.

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3 years ago

Explain why it is important to model the context of a system that is being developed. Give two examples of possible errors that

Sedbober [7]

Answer:

Explanation:

System models are created and used in order for all of the individuals involved in the process to be able to see exactly how each module that is being created connects to the rest of the system. This allows every developer to understand each module's/part's function and develop it with the entire system in mind. This also makes it easier to implement newer features. If a model is not made then the developers may miss key features and when it is time to piece all of the components of the system together, the overall process can become a jumbled mess.

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3 years ago

A 12-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

Ira Lisetskai [31]

Answer:

The smallest wire diameter that can be used is 1 cm

Explanation:

First, we find the smallest diameter using the criterion of maximum normal stress:

Max. Stress = 150 x 10^6 Pa = F/A

150 x 10^6 Pa = 12000 N/(πd²/4)

d² = (12000 N)(4)/(150 x 10^6 Pa)(π)

d = √1.0185 x 10^-4 m²

d = 0.010 m = 1 cm

Now, we find the smallest diameter using the criterion of maximum strain:

Max. Strain = Max. Change in Length/Original Length = 0.025 m/50 m

Max. Strain = 5 x 10^-4 mm/mm

Now,

Max. Strain = Stress/E = (F/A)/E = F/AE

using values:

5 x 10^-4 mm/mm = (12000 N)/(200 x 10^9 Pa)(πd²/4)

d =√(12000 N)(4)/(5 x 0^-4)(200 x 10^9 Pa)(π)

d = 0.012 m = 1.2 cm

Now, by comparison in both cases it can be noted that the smallest value of the diameter is <u>1 cm</u>, which is limited by maximum stress.

7 0

3 years ago

An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

Jet001 [13]

Answer:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

Explanation:

Given

$Volume = 9m^3$

Represent the height as h, the length as l and the width as w.

From the question:

$Length = 2 * Width$

$l = 2w$

Volume of a box is calculated as:

$V = l*w*h$

This gives:

$V = 2w *w*h$

$V = 2w^2h$

Substitute 9 for V

$9 = 2w^2h$

Make h the subject:

$h = \frac{9}{2w^2}$

The surface area is calculated as:

$A = 2(lw + lh + hw)$

Recall that: $l = 2w$

$A = 2(2w*w + 2w*h + hw)$

$A = 2(2w^2 + 2wh + hw)$

$A = 2(2w^2 + 3wh)$

$A = 4w^2 + 6wh$

Recall that: $h = \frac{9}{2w^2}$

So:

$A = 4w^2 + 6w * \frac{9}{2w^2}$

$A = 4w^2 + 6* \frac{9}{2w}$

$A = 4w^2 + \frac{6* 9}{2w}$

$A = 4w^2 + \frac{3* 9}{w}$

$A = 4w^2 + \frac{27}{w}$

To minimize the surface area, we have to differentiate with respect to w

$A' = 8w - 27w^{-2}$

Set A' to 0

$0 = 8w - 27w^{-2}$

Add $27w^{-2}$ to both sides

$27w^{-2} = 8w$

Multiply both sides by $w^2$

$27w^{-2}*w^2 = 8w*w^2$

$27 = 8w^3$

Make $w^3$ the subject

$w^3 = \frac{27}{8}$

Solve for w

$w = \sqrt[3]{\frac{27}{8}}$

$w = \frac{3}{2}$

Recall that : $h = \frac{9}{2w^2}$ and $l = 2w$

$h = \frac{9}{2 * \frac{3}{2}^2}$

$h = \frac{9}{2 * \frac{9}{4}}$

$h = \frac{9}{\frac{9}{2}}$

$h = 9/\frac{9}{2}$

$h = 9*\frac{2}{9}$

$h= 2$

$l = 2w$

$l = 2 * \frac{3}{2}$

$l = 3$

Hence, the dimension that minimizes the surface area is:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

6 0

3 years ago