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dimulka [17.4K]

3 years ago

11

4. A 1 m3 rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5 M3 with propane at 250 kPa, 4

00 K. The valve is opened and the two tanks come to a uniform state at 325 K. What is the final pressure?

1 answer:

Bingel [31]3 years ago

5 0

Answer:

139.97 kPa

Explanation:

Assuming propane is an ideal gas

PV = nRT

n₁ ( number of mole f the first gas) of the 1 m³ rigid tank = P₁V₁ / RT₁ R gas constant = 8.314 J/mol

n₁ = 100kPa × 1 m³ / (8.314 J/mol × 300 K) = 0.0401 mol

n₂ = 0.5 m³ × 250 kPa / ( 8.314J/mol × 400 K) = 0.0376 mol where n₂ is the number of mole of the second gas

n total = n₁ + n₂ = 0.0401 mol + 0.0376 mol = 0.0777 mol

PV = nRT

P final pressure = nRT / V where V = V₁ + V₂ = 1 m³ + 0.5 m³ = 1.5 m³

P final pressure = 0.0777 mol × 8.314 J/mol × 325 K / 1.5 m³ = 139.97 kPa

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navik [9.2K]

Answer:

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1 Subsonic

Mach > 1 Supersonic

Ma= Speed of the object/Speed of the sound

$Ma=\frac{260}{330} \\Ma=0.7878$

So, Ma < 1 Flow is Subsonic

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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:

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

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

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Vsevolod [243]

Answer:

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Explanation:

Given:

dc gain $A = 23.6$ dB

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Now convert gain,

$A = 10^{\frac{23.6}{20} } = 15.13$

DC gain at frequency $f = 0$ is given by,

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$V_{out} = 2.72$

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