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3 years ago

What is the purpose of O-ring and valve seals in a cylinder head?

Engineering

1 answer:

Andrews [41]3 years ago

8 0

Answer:

its to show the shape is flat and only flat at the botom and top and you can set it up ther way and it wlll still look the same.

Explanation:

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Explain how you would solve for total resistance in a parallel circuit versus a series circuit. How would you apply and solve fo

Mekhanik [1.2K]

Answer:

No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.

Explanation:

8 0

3 years ago

function summedValue = SummationWithLoop(userNum) % Summation of all values from 1 to userNum summedValue = 0; i = 1; % Write a

Alexeev081 [22]

Answer:

function summedValue = SummationWithLoop(userNum)

% Summation of all values from 1 to userNum

summedValue = 0;

i = 0;

% use a while loop that assigns summedValue with the

% sum of all values from 1 to userNum

while(i <= userNum)

summedValue = summedValue + i;

i = i + 1;

end

8 0

3 years ago

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:

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