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2 years ago

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In a diesel engine, the fuel is ignited by (a) spark (c) heat resulting from compressing air that is supplied for combustion (d)

ignition (e) combustion chamber. (b) injected fuel

1 answer:

-Dominant- [34]2 years ago

7 0

Answer:

(c) heat resulting from compressing air that is supplied for combustion

Explanation:

In a diesel engine the combustion of the fuel takes place due to the adiabatic compression which leads to elevated temperatures in the cylinder. First the air is compressed by the piston in the cylinder which raises the temperature, then the atomized fuel is put in the cylinder causing the ignition of fuel. Diesel engines have the highest thermal efficiency.

Spark causes ignition in petroleum engine.

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A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0

2 years ago

Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,

aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

System.out.println(birthMonth+"/"+birthYear);

}

}

3 0

3 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

3 years ago

Gaining unauthorized access to a computer's data is called (5 points)

yanalaym [24]

Hacking is correcttttttttt

6 0

3 years ago

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, $T_{a} = 140^{\circ}C$ = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, $T_{i} = 60^{\circ}C$ = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, $C_{p} = 0.949\ kJ/kgK$

At room temperature

Specific heat of iron, $C_{p} = 0.45\ kJ/kgK$

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

$MC_{p}\Delta T = mC_{p}\Delta T$

$28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)$

Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

For Iron:

$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

Thus

$\Delta s =-2.025 + 2.253 = 0.228\ kJ/K$

6 0

3 years ago