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3 years ago

What is engine knock? What cause the engine knock problem?

Engineering

1 answer:

antiseptic1488 [7]3 years ago

3 0

Answer:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

Explanation:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

The engine knock problem can be caused due to the following reason

a) When the octane rating of the fuel used is low.

b) The deposition of the carbon around the cylinder walls takes place.

c) The spark plug used in the vehicle is not correct.

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Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man

strojnjashka [21]

Yes it does the answer is no

5 0

2 years ago

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following

torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches $5\times 10^5$

$(a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\$

$(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e$ $(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}$

7 0

3 years ago

Ann’s Retail, a women’s clothing store, hires female attendants to assist clients in the store’s dressing rooms. Larry, a male,

mojhsa [17]

Answer:

A bona fide occupational qualification defense

Explanation:

Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.

3 0

3 years ago

An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove

Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

$n=Q out / W in$

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

$12000 BTU/hr = 4.72 hp$

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]

6 0

3 years ago

Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with wa

mariarad [96]

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol = 0.8 g/cm³

solution

we get here water content that is express as

water content = $S_s \times \gamma _w$ ....................1

here soil is full saturated so $S_s$ is 100% in both case

so put here value for water

water content = 100 % × 1

water content = 1 g

and

now we get for alcohol that is

water content = 100 % × 0.8

water content = 0.8 g

so here water sample have more water content

5 0

3 years ago