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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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olasank [31]

A problem that will be handled by a procedure is described by an input-output specification.

<h3>What is input and output specification?</h3>

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The input component outlines the kind of data object that should be provided for each actual argument as well as any presumptions that the process might make.
A technical specification known as an output specification defines the project scope primarily through performance-based requirements.
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11 months ago

Mike is involved in developing the model building codes that various states and local authorities in the United States adopt. He

Lyrx [107]

<h3>Answer:</h3>

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2 years ago

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0

2 years ago

A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125

Gekata [30.6K]

Answer:

a) $\dot m_{3} = 135\,\frac{lbm}{s}$ , b) $h_{3}=168.965\,\frac{BTU}{lbm}$ , c) $T = 200.829\,^{\textdegree}F$

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

$\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

c) After a quick interpolation from data availables on water tables, the final temperature is:

$T = 200.829\,^{\textdegree}F$

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2 years ago

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Rina8888 [55]

Answer:While heat cycles cause oil to darken, soot causes oil to turn black

Explanation:

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2 years ago