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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)

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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In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o

lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

= $10\times 101325 \ Pa$

= $1013250 \ Pa$

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

= $1 \ m^3$

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ $PV=nRT$

o,

⇒ $n=\frac{PV}{RT}$

By substituting the values, we get

$=\frac{1013250\times 1}{8.3145\times 298}$

$=408.94 \ moles$

As we know,

⇒ $Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}$

or,

⇒ $MW=\frac{m}{n}$

$=\frac{11.5}{408.94}$

$=0.02812 \ Kg/mol$

$=28.12 \ g/mol$

3 0

3 years ago

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0

3 years ago

A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a

fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS ..............1

put here value

H.I = 2156.77 ft + 2.67 ft

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x) = H.I - FS .............2

point (x) = 2159.44 ft - 6.72 ft

point (x) = 2152.72 ft

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3 years ago

3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10

Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

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4 years ago

Explain Flags in ARM Processor

kipiarov [429]

Answer:

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Explanation:

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3 years ago