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3 years ago

How does the modern Diesel engine achieve higher power output without the use of higher compression ratio?

Engineering

1 answer:

krok68 [10]3 years ago

3 0

Explanation:

We know that the efficiency of diesel engine given as follows

$\eta=1-\dfrac{\rho^{\gamma}-1}{r^{\gamma-1}\gamma(\rho-1)}$

Where r is the compression ratio ,ρ is the cut off ratio and γ is the heat capacity ratio .

Here given that we want to increase the efficiency with out changing the compression ratio.

So from the efficiency formula we can say that when cut off ratio is decreases then the efficiency of diesel cycle will increases.When efficiency is high it means that the power output of engine is high.

The second method to increase the efficiency of diesel engine is that ,use the anti knocking diesel.Use premium diesel instead of use the normal diesel fuel.

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Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following a

disa [49]

Answer:

See explaination

Explanation:

We can describr Aliasing as a false frequency which one get when ones sampling rate is less than twice the frequency of your measured signal.

please check attachment for the step by step solution of the given problem.

7 0

3 years ago

Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0

bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = 20 MHz

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = 1.025 KW

(d) 33 x10^(-12)F = 33 pico F = 33 pF

8 0

3 years ago

What is this thing on my boat?

MaRussiya [10]

It may be an engine cooler, or it may be for the boats speed

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5 0

2 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

Projectile Motion

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$