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butalik [34]
3 years ago
15

How to get on your screen on 2k20 in every mode

Engineering
2 answers:
VashaNatasha [74]3 years ago
5 0
D pad or rb or lb hop this helps
nadezda [96]3 years ago
4 0

Answer:

down on the d pad

Explanation:

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Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

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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]

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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.

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