Answer:
the hurts my brain sorry bud cant help
Explanation:
Here’s some of them
6. J
7. I
10. O
13. F
14. E
15. N
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 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K

Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;

Hence, from;
, we have
5912500 = 90 × A × 341.67

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>
b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>
c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>
c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>
Answer:
i think its A. increasing research to find alternative natural resources for the future