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3 years ago

Imagine waking up every day and no one is home besides your dog and you how would you feel?

Engineering

2 answers:

Firdavs [7]3 years ago

8 0

Answer:

I would be upset.... are you ok???

Allushta [10]3 years ago

6 0

I think it's better to be alone and not to interrupt you when you're resting. :)

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by

zvonat [6]

Answer:

R= 1.25

Explanation:

As given the local heat transfer,

$Nu_x = 0.035 Re^{0.8}_x Pr^{1/3}$

But we know as well that,

$Nu=\frac{hx}{k}\\h=\frac{Nuk}{x}$

Replacing the values

$h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}$

Reynolds number is define as,

$Re_x = \frac{Vx}{\upsilon}$

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity

Then replacing we have

$h_x=0.035(\frac{Vx}{\upsilon})^{0.8}Pr^{1/3}kx^{-1}$

$h_x=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}kx^{0.8-1}$

$h_x=Ax^{-0.2}$

<em>*Note that A is just a 'summary' of all of that constat there.</em>

<em>That is $A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k$ </em>

Therefore at x=L the local convection heat transfer coefficient is

$h_{x=L}=AL^{-0.2}$

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

$h=\frac{1}{L}\int\limit^L_0 h_x dx\\h=\frac{1}{L}\int\limit^L_0 AL^{-0.2}dx\\h=\frac{A}{0.8L}L^{0.8}\\h=1.25AL^{-0.2}$

The ratio of the average heat transfer coefficient over the entire plate to the local convection heat transfer coefficient is

$R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25$

3 0

3 years ago

A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which

horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0

3 years ago

1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un

ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

Bearing life: $L_1 = L , L_2 = 2L$

Catalog rating: $C_1 = C , C_2 = ? ,$

From given equation bearing life equation,

$F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)$

we Dividing eqn (2) with (1)

$\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C$

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

$R_1 = 0.9 , R_2 = 0.99$

Now calculating life adjustment factor for both value of reliability from Weibull parametres

$a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}$

$= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968$

Similarly

$a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215$

Now calculating bearing life for each value

$L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L$

Now using given ball bearing life equation and dividing each other similar to previous problem

$\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C$

Catalog rating increased by factor of 0.61

6 0

3 years ago

QUESTIONS

igomit [66]

Answer:

According to many of the states' driving rules.

All vehicles of 3,000 pounds or more are required to have a brake system that makes them break as a response to the breaking of the vehicle's tow.

Explanation:

The reason behind this answer is that vehicles of more than 3,000 pounds are extremely dangerous and difficult to control. Therefore, when the tow breaks the automatic brake system is required. Because they are too big to be controlled, and if they are left without a brake system to reduce their damage they can destroy entire houses or other cars if this mechanism is not implemented.

5 0

3 years ago

Every time I take a photo, that photo has to be stored in a file somewhere within "My Files" correct?

Anastaziya [24]

Cut that photo by

1. Left click your mouse on the photo

2. Click cut

Then enter the file where you want to transfer and press

1. ctrl+v

5 0

3 years ago

Read 2 more answers