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jenyasd209 [6]
3 years ago
5

Analyze the rate of heat transfer through a wall of an industrial furnace which is constructed from 0.15-m-thick fireclay brick

having a thermal conductivity of 1.7 W/m.K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. The wall dimension is 0.5 m by 1.2 m by side.
Engineering
1 answer:
Simora [160]3 years ago
4 0

Answer:

1700 W

Explanation:

The heat transfer rate P = kA(T - T')/d where k = thermal conductivity of wall = 1.7 W/m-K, A = area of wall = 0.5 m × 1.2 m = 0.6 m², T = temperature of inner surface = 1400 K, T = temperature of outer surface = 1150 K and d = thickness of wall = 0.15 m

So, P = kA(T - T')/d

substituting the values of the variables into the equation, we have

P = 1.7 W/m-K × 0.6 m²(1400 K - 1150 K)/0.15 m

P = 1.7 W/m-K × 0.6 m² × 250 K/0.15 m

P = 255 Wm/0.15 m

P = 1700 W

So, the heat transfer rate through the wall is 1700 W

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state & prove parallelogram law of vector addition &Also determine magnitude &direction of resultant vector.​
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2 years ago
It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec
Serggg [28]

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material remove in 3 min is 16790.4 mm³/s

Explanation:

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length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

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depth = 10 cm

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axial cutting depth = 20 mm

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How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = \frac{length}{advance}

here advance is given as 0.001166 mts / sec

so  time = \frac{800}}{0.001166*1000}

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

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7 0
3 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
2 years ago
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