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madam [21]
3 years ago
8

N

Engineering
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

A because i know thats it

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Which of the following are true about the American Wire Gauge?
harina [27]

Answer:

A. smallest wire is No. 12

7 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
Why is it that dislocations play an important role in controlling the mechanical properties of metallic materials, however, they
klio [65]

Answer:

dislocations play an important role in controlling as

Explanation:

As dislocations plays an important role in the ductility, elasticity and plurality of materials

  • The elastic and elastic deflections play a large role in their properties as the metallic materials, because the dislocation of a glass material does not play a major role in their properties.

5 0
3 years ago
If anyone knows manufacturing plz help
sveta [45]

Answer:

I don't know ask my dad he would

Explanation:

but I can't ask him because he went to get milk and forgot to come back

8 0
3 years ago
A pipe of 0.3 m outer diameter at a temperature of 160°C is insulated with a material having a thermal conductivity of k = 0.055
Alekssandra [29.7K]

Answer:

Q=0.95 W/m

Explanation:

Given that

Outer diameter = 0.3 m

Thermal conductivity of material

K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}

So the mean conductivity

K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )

T_m=\dfrac{160+273+40+273}{2}

T_m=373 K

K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )

K_m=0.112 \frac{W}{mK}

So heat conduction through cylinder

Q=kA\dfrac{\Delta T}{L}

Q=0.112\times \pi \times 0.15^2\times 120

Q=0.95 W/m

4 0
3 years ago
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