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2 years ago

What is the braks mean effictive pressure?

Engineering

1 answer:

OverLord2011 [107]2 years ago

7 0

Engine cylinder pressure

<u>Explanation:</u>

Brake mean effective pressure is a method to calculate the engine cylinder pressure which would give the measured brake horsepower. Brake mean effective pressure is used to identify engine efficiency regardless of capacity or engine speed.
It is used to identify engine efficiency.It is measured by means of transducers or pressure gauges.
It is the measure of engine capacity to do work.

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In fully developed laminar flow in a circular pipe the velocity at R/2 (mid-way between the wall surface and the centerline) is

Butoxors [25]

Answer:

$u_{max} = 17.334\,\frac{m}{s}$

Explanation:

Let consider that velocity profile inside the circular pipe is:

$u(r) = 2\cdot U_{avg} \cdot \left(1 - \frac{r^{2}}{R^{2}} \right)$

The average speed at $r = \frac{1}{2} \cdot R$ is:

$U_{avg} = \frac{13\,\frac{m}{s} }{2\cdot \left(1-\frac{1}{4} \right)}$

$U_{avg} = 8.667\,\frac{m}{s}$

The velocity at the center of the pipe is:

$u_{max} = 2\cdot U_{avg}$

$u_{max} = 17.334\,\frac{m}{s}$

6 0

2 years ago

Read 2 more answers

Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase

Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

$\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%$

Where:

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

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

The expression is expanded by using the definition of density and subsequently simplified:

$\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%$

Where:

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

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

$\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%$

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

If $\rho_{gr} = 2.3\,\frac{g}{cm^{3}}$ , $\rho_{fe} = 7.9\,\frac{g}{cm^{3}}$ , $m_{gr} = 3.2\,g$ and $m_{fe} = 96.8\,g$ , the volume percentage of graphite is:

$\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%$

$\%V_{gr} = 10.197\,\%V$

The volume percentage of graphite is 10.197 per cent.

5 0

3 years ago

What is voltage drop?

lord [1]

Answer:

Explanation:

It is the voltage a voltmeter would read when connected across something that has resistance.

___________0_________O______O_______

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|_____________________| |_____________ |

The diagram above is supposed to represent 3 lightbulbs connected in series. The vertical lines in the middle are supposed to be a battery which powers the three light bulbs. If you put a voltmeter across one of the lightbulbs, it will read a voltage that is 1/3 of the voltage of the battery.

Answer

That reading you get across the one light bulb is The Voltage Drop.

6 0

1 year ago

While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

IgorC [24]

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$