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nikdorinn [45]
2 years ago
9

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

. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Engineering
1 answer:
ludmilkaskok [199]2 years ago
6 0

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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Answer:

I hope this helps you

Explanation:

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Answer:

a) 6.4  mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture= \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}

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b) utlimate BOD at 10,000 m downstream

t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times  \frac{day}{24 hr}

putting Q_r + Q_w = 1+ 10 = 11 m^3/s

t = 0.578  days

we know

L_t = L_o e^{-kt}

L_t = 6.4 \times e^{-0.22 \times 0.578}

L_t = 5.6 mg/l

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Answer:

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Explanation:

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Answer:

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