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3 years ago

List two possible reasons the engine oil could have a strong gasoline smell

1 answer:

8 0

Two reasons why the engine oil could have a strong gasoline smell is- why gas getting into the engine oil, is because your fuel mixture is too rich and If your fuel mixture is too rich, the combustion chamber won't ignite all of the fuel.

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How will you maintain the orderliness of your storage area 2pts?

fiasKO [112]

Explanation:

Label and group products. One would think that a general cleanup would be the first step, but no, it's not. ...Clean up the area. ...Put up demarcation lines. ...Stack properly. ...Keep the aisles, paths and ramps clear. ...Have all the safety signs in place.

4 0

2 years ago

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

2 years ago

What might cause birds in population to immigration To the island

loris [4]

Mass forest destruction

temperatures

7 0

3 years ago

Read 2 more answers

A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible

Zolol [24]

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note: space time (t) =

t = $\frac{A*L}{Vo}$ Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = $\frac{25*3*10^{-2} }{0.03}$ = 25 s

since the reaction is in first order

X = 1 - $e^{-kt}$

$e^{-kt}$ = 1 - X

kt = In $\frac{1}{1-X}$

k = In $\frac{1}{1-X}$ / t

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then

K = 0.156 $s^{-1}$

Calculating Da for a closed vessel

; Da = tk

= 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = $\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})$

therefore

$\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0$

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution

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3 years ago

Jjjjjmmmmmmmmmml.n;nLN/n/kn

GaryK [48]

Jjchdypyepyspgzlbblxlbxljkkvgigirudlh),,$&,‘joyful fhdjdududufjnbd

6 0

3 years ago