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Zielflug [23.3K]
3 years ago
12

A strip of AISI 304 stainless steel, 2mm thick by 3cm wide, at 550°C, continuously enters a cooling chamber that removes heat at

a rate of 120 kW. How fast must it move to achieve a temperature of 140 °C at the exit?
Engineering
1 answer:
Zanzabum3 years ago
6 0

Answer:

V = 1.23 m/s

Explanation:

Given data:

AISI 304 steel

thickness of steel = 2 mm

width  =  3 cm

Temperature =  550 degree celcius

wer know that heat is given as Q  

Q = m\times Cp \Delta T

\Delta T = 550 - 140 = 410 \degree\ celcius

for AISI 304 steel Cp is 502 J/kg . K

We know that

\dot mass  =\rho A V

\rho = 7.9\times 10^3 kg/m^3

A = 2\times 10^{-3} \times 3\times 10^{-2} m^2

therefore VELOCITY V  will be

120\times 10^{3} = 7.9\times 10^{3} \times 6\times 10^{-5} \times V\times 502\times 410

solving for V we get

V = 1.23 m/s

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Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

5 0
3 years ago
For the following circuit diagram, if A=010 , B= 101.
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can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

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Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

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6 0
3 years ago
The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?
kodGreya [7K]

Answer:

W = 2695.14 lb

Explanation:

given,

side of cube = 3 ft

density of the cube = 3.1 slugs/ft³

we know,

density = \dfrac{mass}{volume}

mass = density x volume

volume = 3³ = 27 ft³

mass =  3.1  x 27

    m = 83.7 slugs.

weight calculation

converting mass from slug to pound

weight of 1 slug is equal to 32.2 lb

now,

weight of the cube is equal to

  W = 83.7 slugs x 32.2 lb/slug

  W = 2695.14 lb

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4 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
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