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2 years ago

9

An engineer has designed and built a prototype to improve the brake system of a car. What is the next step that the engineer sho

uld take in the process

1 answer:

8090 [49]2 years ago

6 0

Answer:

Test the working prototype

Explanation:

After the prototype is built, it is put through basic operational tests and purpose of applicability or performance test

The prototype is also laboratory tested to determine how reliable it is with tests performed to determine the modes of failure. Performance of tests are carried out by subjecting the prototype to stresses beyond its designed stress such as to discover failure modes

The system is however tested in isolation at this stage and the performance and risks of the prototype are quantified after the test, with improvement requirement outlined for modification of the design.

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The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc

Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑ $f_{y}$ = 0, ∑M = 0

If ∑ $M_{A}$ = 0

⇒ $F_{BC}$ ×0.9 - P × 0.6 = 0

⇒ $F_{BC}$ ×3 - P × 2 = 0

⇒ $F_{BC}$ = $\frac{2P}{3}$

If ∑ $M_{B}$ = 0

⇒ $F_{AD}$ ×0.9 = P × 0.3

⇒ $F_{AD}$ ×3 = P

⇒ $F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒ $\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0

2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt

= $\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}$ watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = $\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}$

15999.86 $=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}$

$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

4 0

2 years ago

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

$\frac{p1}{p*}$ = 0.3636

$\frac{T1}{T*}$ = 0.5289

$\frac{T01}{T0*}$ = 0.7934

Isentropic Flow Chart: M1 = 2.0 , $\frac{T01}{T1}$ = 1.8

T1 = $\frac{1}{0.7934}$ (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= $\frac{T02}{T1}T1$ = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = $\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K$ = 135.7* $10^{3}$ J/Kg.

5 0

2 years ago

Read 2 more answers

A paint company produces glow in the dark paint with an advertised glow time of 15 min. A painter is interested in finding out i

andrew11 [14]

Answer:

$p_v = P(z$

Now we can decide based on the significance level $\alpha$ . If $p_v$ we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

$\alpha=0.05$ we see that $p_v< \alpha$ so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

$\alpha=0.01$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

$\alpha=0.001$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

Explanation:

For this case they conduct the following system of hypothesis for the ture mean of interest:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu >15$

The statistic for this hypothesis is:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And on this case the value is given $z = -2.30$

For this case in order to take a decision based on the significance level we need to calculate the p value first.

Since we have a lower tailed test the p value would be:

$p_v = P(z$

Now we can decide based on the significance level $\alpha$ . If $p_v$ we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

$\alpha=0.05$ we see that $p_v< \alpha$ so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

$\alpha=0.01$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

$\alpha=0.001$ we see that $p_v> \alpha$ so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

3 0

2 years ago

For laminar flow of air over a flat plate that has a uniform surface temperature, the curve that most closely describes the vari

Aliun [14]

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = $C$ $(Re_x)^{0.5$ $(Pr)^{1/3$

Nu = $C(\frac{Vx}{v})^{0.5}$ $(Pr)^{1/3$

$\frac{h_xx}{k}$ = $C(\frac{V}{v})^{0.5}$ $(Pr)^{1/3$ $(x)^{0.5$

$h_x$ = $\frac{1}{x^{1/2}}$

Next we write down the expression for the local heat flux from the plate with uniform surface temperature;

q = $h_xA($ T $_s$ - T∞ )

q ∝ $h_x$

∴

q ∝ $\frac{1}{x^{1/2}}$

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

3 0

2 years ago