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KatRina [158]
3 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]3 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]3 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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musickatia [10]

<u>Explanation:</u>

At temperature is 33^{\circ} C and relative humidity is 86% therefore,  the humidity ratio is 0.0223 and the specific volume is 14.289

At temperature is 33^{\circ} C and Relative humidity is 40% therefore, the humidity ratio is  0.0066 and the specific volume is 13.535.

To calculate the mass of air can be calculated as follows:

\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}

Now , we going to calculate the volume,

\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}

The time which is required to fill the cistern can be calculated as follows:

Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)

Now, putting the value in above formula we get,

\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.

Therefore, the hours required to fill the cistern is 4.65 hours.

3 0
2 years ago
A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency?
Oxana [17]

Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

6 0
3 years ago
What is the federal E-Rate program?
Annette [7]

Answer:

c. A program that offers discounts to libraries and schools ensuring they have affordable access to modern telecommunications and information services.

Explanation:

Federal E-Rate program refers to the Schools and Libraries Program of the Universal Service Fund managed by the Universal Service Administrative Company (USAC) and being directed by the Federal Communications Commission (FCC).

The program offers telecommunications and internet access to schools and libraries in the United States at discounts of between 20% and 90% in order to make the services affordable to them.

The discounts received by each of he beneficiary schools receive which is between the rage of 20% and 90% is determined by the degree of poverty and the urban/rural status of the population or students being served.

In the program, connectivity and maintenance services are provided by the Schools and Libraries Program, while school that applied to the program has to provide other items like software, hardware (e.g. computers), and among other items that will make then to use the connectivity provided.

I wish you the best.

8 0
3 years ago
A long corridor has a single light bulb and two doors with light switch at each door.
Zigmanuir [339]

Answer:

The answer is below

Explanation:

Let A represent the first switch, B represent the second switch and C represent the bulb. Also, let 0 mean  turned off and 1 mean turned on. Since when both switches are  in the same position, the light is off. This can be represented by the following truth table:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit can be represented by:

C = A'B + AB'

The output (bulb) is on if the switches are at different positions; if the switches are at the same position, the output (bulb) is off. This is an XOR gate. The gate is represented in the diagram attached below.

6 0
3 years ago
Component of earthing and reasons why each material is being used<br><br>​
timofeeve [1]

Answer:

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