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The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto

lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0

3 years ago

36. What is mass? (4.4)

KatRina [158]

Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter.

Explanation:

Mass is the matter that makes up objects

3 0

3 years ago

Tech A says that the brake pedal uses leverage to multiply foot pressure. Tech B says that when braking hard while moving

Nikolay [14]

Tech- A is correct

Explanation:

Leverage is defined as using a tool to gain mechanical influence. The measure of the benefit gained depends on what kind of lever is used and how it is utilized.
Leverage is designed in such a way that it can reproduce the force from your leg many times before any force is transferred to brake fluid.
The brake pedal size and the measure of leverage received depends on the overall design of the brake system.
The second-order lever is used in the brake pedal. The brake pedal applies leverage to populate the force employed to the master cylinder. The effort needed to drive a load depends on the corresponding distance of the load and the work from the fulcrum. The proportion of load and work is known as mechanical advantage.

7 0

4 years ago

Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original

Liono4ka [1.6K]

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

Aug=[A eye(size(A))]

Above command is tested in MATLAB command window and is attached in figure below

8 0

3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$

$w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um$

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

3 years ago