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andrey2020 [161]
3 years ago
5

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

Engineering
1 answer:
vovangra [49]3 years ago
6 0

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

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Were women treated as equals to men in early aviation history?
Dmitry_Shevchenko [17]

Answer:

No

Explanation:

In the early Aviation history men have always dominated the world of aviation and women are notoriously under-represented, particularly in technical and leadership roles. Therefore, women was treated unequal.

8 0
2 years ago
Read 2 more answers
What is 29*69+98-78/36=
densk [106]

Answer:

<u>\frac{989}{6}</u>

3 0
3 years ago
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There are 20 forging presses in the forge shop of a small company. The shop produces batches of forgings requiring a setup time
Aleksandr-060686 [28]

Answer:

Considering the guidelines of this exercise.

The pieces produced per month are 504 000

The productivity ratio is 75%

Explanation:

To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21  making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.

4 0
3 years ago
Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count &gt;= 1 then For i = 0 to c
vladimir2022 [97]

Answer:

gg

Explanation:

gg

7 0
3 years ago
Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.012
devlian [24]

The pressure drop of air in the bed is  14.5 kPa.

<u>Explanation:</u>

To calculate Re:

R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}

From the tables air property

\mu_{394 k}=2.27 \times 10^{-5}

Ideal gas law is used to calculate the density:

ρ = \frac{2.2}{2.83 \times 10^{-3} \times 394.3}

ρ = 1.97 Kg / m^{3}

ρ = \frac{P}{RT}

R = \frac{R_{c} }{M} = 8.2 × 10^{-5} / 28.97×10^{-3}

R = 2.83 × 10^{-3} m^{3} atm / K Kg

q is expressed in the unit m/s

q=\frac{2.45}{1.97}

q = 1.24 m/s

Re = \frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}

Re = 2278

The Ergun equation is used when Re > 10,

\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}

\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24 +\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa

4 0
3 years ago
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