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Dahasolnce [82]
3 years ago
12

A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf

ace emissivity of 0.95. Heat is lost from the head to the surrounding air at 25°C by convection with a convection coefficient of ???????????????? ???????? ????????????????∙???????? , and by radiation to the surrounding black walls at 15°C. Determine the total rate of heat loss. StefanBoltzmann Constant, ???????? = ????????. ???????????????? × ????????????????−???????? ???????? ????????????????∙???????????????? . (10 points)
Engineering
1 answer:
galina1969 [7]3 years ago
5 0

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

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t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

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Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

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3 0
3 years ago
For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?
juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= 4\sqrt{2} cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as P = \dfrac{\pi ^2EI}{L_e^2}      

If area moment of inertia(I) is high then buckling load will be high.

We know that  area moment of inertia(I)

For circular tube I = \dfrac{\pi }{64}(D_o^4-D_i^4)

For circular bar I = \dfrac{\pi }{64}D^4  

Now by putting the values

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  For circular bar I=50.26 cm^4

So we can say that for same cross sectional area the  area moment of inertia(I) is high for tube as compare to bar.So buckling load  will be higher in tube as compare to bar.

3 0
2 years ago
A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?
stiks02 [169]

Answer:

The tax on this property is 3750 dollars

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Given

Tax on per $100 is $2.50

Tax on every $1 is \frac{2.5}{100} = 0.025 dollars

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The tax on this property is 3750 dollars

7 0
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7 0
3 years ago
What are the three elementary parts of a vibrating system?
zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

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2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

7 0
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