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Ronch [10]
2 years ago
10

Why is oil black and why does oil look black

Engineering
1 answer:
Rina8888 [55]2 years ago
4 0

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

Explanation:

Most people associate soot with diesel engines, but gasoline engines can produce soot as well, particularly modern gasoline-direct-injection engines. ... Any finer filtration and the filter could catch dissolved additives in the motor oil

You might be interested in
A high molecular weight hydrocarbon gas A is fed continuously into a heated mixed flow reactor (0.1liter) where it is thermally
dimulka [17.4K]

Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

5 0
3 years ago
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance
Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

4 0
2 years ago
2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the
Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

  {

  private long[] a;                 // ref to array a

  private int nElems;               // number of data items

  //-----------------------------------------------------------

  public HighArray(int max)         // constructor

     {

     a = new long[max];                 // create the array

     nElems = 0;                        // no items yet

     }

  //-----------------------------------------------------------

  public setValue find(long searchKey)

     {                              // find specified value

     int j;

     for(j=0; j<nElems; j++)            // for each element,

        if(a[j] == searchKey)           // found item?

           break;                       // exit loop before end

     if(j == nElems)                    // gone to end?

        return false;                   // yes, can't find it

     else

        return true;                    // no, found it

     }  // end find()

  //-----------------------------------------------------------

  public void insert(long value)    // put element into array

     {

     a[nElems] = value;             // insert it

     nElems++;                      // increment size

     }

  //-----------------------------------------------------------

  public void display()             // displays array contents

     {

     for(int j=0; j<nElems; j++)       // for each element,

        System.out.print(a[j] + " ");  // display it

     System.out.println("");

     }

  //-----------------------------------------------------------

  }  // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

  {

  public static void main(String[] args)

     {

     int maxSize = 100;            // array size

     HighArray arr;                // reference to array

     arr = new HighArray(maxSize); // create the array

     arr.insert(77);               // insert 10 items

     arr.insert(99);

     arr.insert(44);

     arr.insert(55);

     arr.insert(22);

     arr.insert(88);

     arr.insert(11);

     arr.insert(00);

     arr.insert(66);

     arr.insert(33);

     arr.display();                // display items

     int searchKey = 35;           // search for item

     if( arr.find(searchKey) )

        System.out.println("Found " + searchKey);

     else

        System.out.println("Can't find " + searchKey);

     }  // end main()

  }  // end class HighArrayApp

Explanation:

6 0
2 years ago
Discuss typical advantages and disadvantages of an irrigation system?
trapecia [35]
Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.
4 0
2 years ago
A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ
VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

A=\frac{\pi d^{2}}{4}

A=\frac{\pi\times(10.2)^{2}}{4}

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

D=\frac{A_{i}-A_{f}}{A_{i}}\times100

D=\frac{81.713-52.7}{81.713}\times100

D = 35.5%.

Thus, the percentage ductility is 35.5%.

5 0
2 years ago
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