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Elenna [48]
3 years ago
11

For heat transfer purposes, a standing man can be mod-eled as a 30-cm-diameter, 170-cm-long vertical cylinderwith both the top a

nd bottom surfaces insulated and with theside surface at an average temperature of 34°C. For a con-vection heat transfer coefficient of 15 W/m2·K, determinethe rate of heat loss from this man by convection in still airat 20°C. What would your answer
Engineering
1 answer:
jeyben [28]3 years ago
8 0

Answer:

Rate of Heat Loss = 336 W

Explanation:

First, we will find the surface area of the cylinder that is modelled as the man:

Area = A = (2\pi r)(l)

where,

r = radius of cylinder = 30 cm/2 = 15 cm = 0.15 m

l = length of cylinder = 170 cm = 1.7 m

Therefore,

A = 2\pi(0.15\ m)(1.7\ m)\\A = 1.6\ m^2

Now, we will calculate the rate of heat loss:

Rate\ of\ Heat\ Loss = hA\Delta T

where,

h = convective heat tranfer  coefficient = 15 W/m²K

ΔT = Temperature difference = 34°C - 20°C = 14°C

Therefore,

Rate\ of\ Heat\ Loss = (15\ W/m^2K)(1.6\ m^2)(14\ K)\\

<u>Rate of Heat Loss = 336 W</u>

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Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
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Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1:  ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2:  ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3:  ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4:  ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5:  ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

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6 0
3 years ago
The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space
Snezhnost [94]

Answer:

answer is C. his seat belt

5 0
2 years ago
What are the two main what are the two main concerns in the research of fluid power efficiency?
Galina-37 [17]

Answer:

The correct option is;

Materials and Components

Explanation:

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3) Making use of the other externally available technological advantages

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3 0
2 years ago
Read 2 more answers
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
Can some help me with this !!! Is 26 points!!
Aleonysh [2.5K]
Third one
15,000,000 ohms because M=10^6
8 0
3 years ago
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