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aliina [53]
3 years ago
12

A spherical, stainless steel (k 16 W m1 K-1) tank has a wall thickness of 0.2 cm and an inside diameter of 10 cm. The inside sur

face of the tank wall is held constant at 25 oC and the outside surface heat transfer coefficient is 6 W m2 K. Calculate the rate of heat loss from the tank when the outside air temperature is 15°C.
Engineering
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

the rate of heat loss is 2.037152 W

Explanation:

Given data

stainless steel K = 16 W m^{-1}K^{-1}

diameter (d1) = 10 cm

so radius (r1)  = 10 /2 = 5 cm = 5 × 10^{-2}

radius (r2)  = 0.2 + 5 = 5.2 cm = 5.2 × 10^{-2}

temperature = 25°C

surface heat transfer coefficient = 6 6 W m^{-2}K^{-1}

outside air temperature = 15°C

To find out

the rate of heat loss

Solution

we know current is pass in series from temperature = 25°C to  15°C

first pass through through resistance R1  i.e.

R1  = ( r2 -  r1 ) / 4\pi  × r1 × r2 × K

R1  = ( 5.2 - 5 ) 10^{-2}  / 4\pi  × 5 × 5.2 × 16 × 10^{-4}

R1  = 3.825 ×  10^{-3}

same like we calculate for resistance R2 we know   i.e.

R2 = 1 / ( h × area )

here area = 4 \pi r2²

area = 4 \pi (5.2 × 10^{-2})²  =  0.033979

so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979  )

R2 = 4.90499

now we calculate the heat flex rate by the initial and final temp and R1 and R2

i.e.

heat loss = T1 -T2 / R1 + R2

heat loss = 25 -15 / 3.825 ×  10^{-3} + 4.90499

heat loss =  2.037152 W

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Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

  • Voltage amplifier needs high input and low output  resistance.
  • Current amplifier needs Low Input and High Output  resistance.
  • Trans-conductance amplifier Low Input and High Output resistance.
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Therefore, the correct answer is "None of these "

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3 years ago
A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

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The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 106 Btu/h. The comb
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The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
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Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = \frac{100}{100 - P} M kg  per day

put her value

mass of sludge produced = \frac{100}{100 - 78} 7240 kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge =  \frac{\rho sludge}{\rho water }

and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}

S sludge = 1.152

so that

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so that

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volume of biological sludge = \frac{32909.09}{1152}

volume of biological sludge = 28.566 m³ per day

6 0
3 years ago
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