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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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valkas [14]

Answer:

Increases

Explanation:

By inhibiting the motion of dislocations by impurities in a solid solutions, is a strengthening mechanism. In solid solutions it is atomic level strengthening resulting from resistance to dislocation motion. Hence, the strength of the alloys can differ with respect to the precipitate's property. Example, the precipitate is stronger (ability to an obstacle to the dislocation motion) than the matrix and it shows an improvement of strength.

5 0
3 years ago
A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion
tester [92]

Answer:

mark me as a brainleast

Explanation:

209781

6 0
2 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0
3 years ago
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
Find the resistance of a circuit that draws 4 amperes with 8 volts applied?
vodka [1.7K]

Answer:

2 ohms

Explanation:

V = I * R

8 = 4 * R

8 / 4 = R

R = 2 ohms

5 0
3 years ago
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