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VikaD [51]
2 years ago
14

What is compression ratio of an Otto cycle? How does it affect the thermal efficiency of the cycle?

Engineering
1 answer:
GuDViN [60]2 years ago
6 0

Answer:

r= \dfrac{V_1}{V_2}

\eta =1-\dfrac{1}{r^{\gamma-1}}

Explanation:

Compression ratio:

   Compression ratio is the ratio of volume in the cylinder.Generally it is denoted by r.

The compression ratio for Otto cycle is about 8 to 12.But the compression ratio of diesel engine is about 14 to 25.

So from the Otto cycle compression ratio can be define as

r= \dfrac{V_1}{V_2}

As we know that efficiency defines as the ratio of work out to the heat input.In the Otto cycle the efficiency is given as in terms of compression ratio r as

\eta =1-\dfrac{1}{r^{\gamma-1}}

Where γ is the heat capacity ratio.

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Witch measuring tool would be used to determine the diameter of a crankshaft journal
djverab [1.8K]
Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

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3 0
2 years ago
A MOSFET differs from a JFET mainly because
Solnce55 [7]

Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.

1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal          Oxide Semiconductor Field Effect Transistor.

  (JFET) stands for junction gate field-effect transistor (JFET)  

2. JFET is a three-terminal semiconductor device, whereas  MOFET a four-terminal semiconductor device.  

3. In terms of areas of application of  JFETs are used in low noise applications while MOSFETs,  are used for high noise applications

5 0
2 years ago
The parts of a feature control frame are the tolerance value, the datum references, and the
Elan Coil [88]

Answer:

d

Explanation:

4 0
3 years ago
Write a Nios II assembly program that reads binary data from the Slider Switches, SW11-0, on the DE2-115 Simulator, and display
Citrus2011 [14]

Algorithm of the Nios II assembly program.

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

and

The decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

<h3>The  Algorithm and decimal equivalent on the seven-segment displays HEX3-0</h3>

Generally,  the program will be written using a  cpulator simulator in order to attain best result.

We are to

  • Attain data for simulation from the  SW11-0, on the DE2-115 Simulator
  • The data will be  read from the switches in loop.
  • The decimal output is displayed using the seven-segment displays and done using the loop.
  • The program is ended by the user operating the SW1 switch

This will be the Algorithm of the Nios II assembly program .

Hence, the decimal equivalent on the seven-segment displays HEX3-0 is

  • DE2-115
  • DE2-115_SW11
  • DE2-115_HEX3
  • DE2-115_HEX4
  • DE2-115_HEX5
  • DE2-115_HEX6
  • DE2-115_HEX7

For more information on Algorithm

brainly.com/question/11623795

6 0
2 years ago
A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th
inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = \int\limits^a_b P \, dV  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 0
2 years ago
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