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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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What’s another name for a service overcurrent device?

alexandr402 [8]

Overcurrent protective devices, or OCPDs

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3 years ago

What are some common work contexts for Licensing Examiners and Inspectors? Select four options.

Akimi4 [234]

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

Telephone
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O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:

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1. Telephone

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Read more on work contexts here: brainly.com/question/22826220

6 0

3 years ago

Read 2 more answers

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago

What is the smallest variable type I can use to represent the number 27?

oksano4ka [1.4K]

Answer:3

Explanation:

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3 years ago

A 3.0-m wide rectangular asphalt channel discharges 33.84 m^3/s of water with a depth of 2.0 m. What is the head loss in 100 m l

BaLLatris [955]

Answer:

Head loss in 100 m length equals 1.00 m.

Explanation:

The head loss in an open channel is calculated using manning's equation as follows

$Q=\frac{1}{n}\frac{A^{5/3}}{P^{2/3}}S_{f}^{1/2}$

For a asphalt rectangular channel we have

Area of flow = $3.0\times 2.0 = 6m^{2}$

Wetted Perimeter = $3.0+2\times 2.0=7.0m$

manning's roughness coefficient = 0.016

Applying values in the above equation we get

$33.84=\frac{1}{0.016}\times \frac{6^{5/3}}{7^{2/3}}S_{f}^{1/2}\\\\\therefore S^{1/2}_{f}=0.1\\\\\therefore S_{f}=0.01$

Now we know that

$S_{f}=\frac{H_{L}}{L}\\\\\therefore H_{l}=0.01\times 100m=1.00m$

8 0

3 years ago