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KatRina [158]
4 years ago
6

A diesel engine with an engine volume of 4.0 L and an engine speed of 2500 rpm operates on an air–fuel ratio of 18 kg air/kg fue

l. The engine uses light diesel fuel that contains 800 ppm (parts per million) of sulfur by mass. All of this sulfur is exhausted to the environment, where the sulfur is converted to sulfurous acid (H2SO3). If the rate of the air entering the engine is 336 kg/h, determine the mass flow rate of sulfur in the exhaust. Also, determine the mass flow rate of sulfurous acid added to the environment if for each kmol of sulfur in the exhaust, 1 kmol sulfurous acid will be added to the environment. The molar mass of sulfur is 32 kg/kmol.

Engineering
2 answers:
timama [110]4 years ago
8 0

Answer:

<em>mass flow rate of sulfur = 14.936*10^-3 kg/hr</em>

<em>mass flow rate of sulfurous acid = 38.23 * 10^-3 kg/hr</em>

Explanation:

Please see attached images for detailed explanation.

Feel free to ask anything if it seems confusing. Thank you.

Mars2501 [29]4 years ago
3 0

Answer:

Explanation:

i) (18 X 336) air/h = 6048 air/h

Engine speed = 2500 rpm; Capacity of diesel = 800 ppm

∴ 2500/800 = 3.125 kg/pm

Mass flow rate = (3.125 X 6048) Kg/h/pm =18900 Kg/h/pm

ii) Mass flow rate = 32 X 4/1000 X 2500 = 320 Kg/h/pm

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Create two arrays with 5 elements each: one will hold Strings and the second will hold integers. Write a program to ask the user
MrMuchimi

Answer:

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

int main() {

   string name[5];  

   int age[5];  

   int i,j;  

   for ( i = 0; i<=4; i++ ) {  

       cout << "Please enter student's name:";  

       cin >> name[i];  

       cout << "Please enter student's age:";  

       cin >> age[i];          

   }  

for (i=0;i<=4;i++){

   cout<<"Age of  "<< name[i]<<"  is  "<<age[i]<<endl;  

}

}

Output of above program is displayed in figure attached.

5 0
3 years ago
A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b
mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0
3 years ago
5. (5 points) Select ALL statements that are TRUE A. For flows over a flat plate, in the laminar region, the heat transfer coeff
finlep [7]

Answer:

The following statements are true:

A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction

C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface

E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.

Select ALL statements that are TRUE

B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant

D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

7 0
3 years ago
There are two machines for sale that you are considering purchasing for your sawmill to produce hardwood flooring. You want to f
devlian [24]

Answer:

Machine 2 has a higher process capability index, it would be best considered for purchase.

Explanation:

Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]

For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7

Cpk= [0.952;0.952]= 0.952

For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3

Cpk= [1.111;3.333]= 1.111

It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.

Since machine 2 has a higher process capability index, it would be best considered for purchase.

4 0
3 years ago
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =
oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

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And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0
3 years ago
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