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Pachacha [2.7K]

3 years ago

15

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame

ter and is situated in a surrounding ambient temperature of 15°C where the natural convection heat transfer coefficient is 12 W/m2·K. If the efficiency of the electrical heater to transfer heat to the plate is 80%, determine the electric power that the heater needs to keep the surface temperature of the plate at 220°C.

1 answer:

denis23 [38]3 years ago

5 0

Answer:

Q = 125.538 W

Explanation:

Given data:

D = 30 cm

Temperature $T_\infity = 15$ degree celcius

$T_S = 220 + 273 = 473 K$

Heat coefficient = 12 W/m^2 K

Efficiency 80% = 0.8

$Q = hA(T_S - T_{\infty}) \eta$

$= 12(\frac{\pi}{4} 0.3^2) (473 - 288) 0.8$

Q = 125.538 W

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

Where:

$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

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6.1-2. Diffusion of CO, in a Binary Gas Mixture. The gas CO2 is diffusing at stcady state through a tube 0.20 m long having a di

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Answer:

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