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Brilliant_brown [7]

3 years ago

8

Starting the vehicle's engine and listening to its operation can

1 answer:

miskamm [114]3 years ago

3 0

I pretty sure it’s true because you can start your car and it sounds different that probably means something is wrong

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The convection heat transfer coefficient for a clothed person standing in moving air is expressed as h 5 14.8V0.69 for 0.15 , V

dolphi86 [110]

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3 years ago

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kakasveta [241]

Answer:

I'm good at drawing and computer-animated design

Explanation:

3 0

2 years ago

A cylinder contains 480 cm3 of loose dry sand which weighs 820 g. Under a static load of 200 kPa the volume is reduced 1%, and t

goblinko [34]

Answer:

a.

b.

c.

Explanation:

a. void ratio is provided by the formula: $e = \frac{V_{p} }{V_{s} }$

where , $V_{p}$ = volume of voids

$V_{s}$ = volume of solid grains

for loose sand, the void space = $\frac{480}{480}$

= 1

b. void ratio after static load = 0.1/(480)/ (480)

= 0.1

c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480

= 0.9

5 0

3 years ago

You are considering purchasing a compact washing machine, and you have the following information: The Energy Guide claims an est

RSB [31]

Answer: $15.34

Explanation: see image below

8 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago