All categories

2 years ago

Do you get a better performance using premier gasoline (Octane number 93) for your compact car?

Engineering

1 answer:

topjm [15]2 years ago

6 0

Answer:

Yes

Explanation:

As we know that octane number resist the engine from knocking.If knocking can prevent that automatically the performance of engine will increases.If octane number is 100 then it means that knocking tendency in the engine is zero.So higher the octane number better will the performance of the engine.

Generally octane number is 87 but for premier gasoline is 92 or 93.

So we can say that if octane number is 93 then car will give better performance

You might be interested in

Electrical circuits must be locked-out/tagged-out before electricians work on any equipment. Is this true or false?

dybincka [34]

Answer:

true

Explanation:

Equipment that are "locked-out/tagged-out" prevent the electrician from being electrocuted or attaining a serious injury in relation to it. Locking out an equipment prevents it from releasing its energy because such energy can be hazardous to the electrician. There are instances when the equipment accidentally starts up, thus, it is essential that the equipment's source of energy is isolated.

8 0

2 years ago

Drag each item to show if it is an element or not an element.

Flauer [41]

Answer:

CARBON

Explanation:

HOPE THIS HELPS SORRY FOR CAPS

5 0

2 years ago

Which of the following is a possible consequence of poor measurement in construction as stated in the segment?

Kamila [148]

Answer:

Decreased risk of structure failure

6 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

g A primary sedimentation basin is designed for an average flow of 0.3 m3/s. The TSS concentration in the influent is 240 mg/L.

yarga [219]

Answer:

(a) 0.243 m3/day

(b) 96 mg/l

Explanation:

The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240= 9.6 mg/L

Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day

To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3

Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day

Approximately, the volume of sludge is 0.243 m3/day.

(b)

Efficiency of 60 percent is equivalent to 0.6

Efficiency=(influent concentration- flow rate)/influent concentration

0.6=(240-flow rate)/240

Flow rate= 96 mg/l

(c)

Cycle time= 0.243/0.57=0.4263157894736 m3/min

Rounded off, cycle time is 0.426 m3/day

3 0

2 years ago