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2 years ago

Describe the make-up of an internal combustion engine. Pls answer quickly.

Engineering

1 answer:

Whitepunk [10]2 years ago

8 0

Answer:

The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.

Explanation:

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Consider a cubical furnace with a side length of 3 m. The top surface is maintained at 700 K. The base surface has emissivity of

Ket [755]

Answer:

Check the explanation

Explanation:

Assumptions.

1. The surfaces are diffuse, may and opaque

2. steady operating conditions exist

3. Heat transfer from and to the surfaces is only due to Radiation

Consider the base surface to be surface 2 the top surface to be surface and the side surfaces to surface 3 1. cubical furnace can be considered to be three-surface enclosure. the areas and black body emissive powers of surfaces can be calculated as seen in the attached images below.

8 0

3 years ago

Answer the question faster please

Juliette [100K]

Answer:

Explanation:

3 0

3 years ago

Benzene gas (C6H6) at 25° C and 1 atm, enters a combustion chamber operating at steady state and burns with 95% theoretical air

tiny-mole [99]

Answer:

Explanation:

Benzene gas is burned with 95 percentage theoretical air during a steady â flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined

Assumption

steady operating conditions exit .

Air and combustion gases are gases

Kinetic and potential energies are negligible

The fuel is burned with insufficient amount of air and thus the products will contain some CO as well as CO2, H2O and H2

Combustion equation is

C6H6 + ath (O2 + 3.76N2) â 6CO2 +3H2O +3.76ath N2

Where ath is the stoichiometric coefficient and is determined from the O2 balance

ath = 6+1.5 = 7.5

then the actual combustion equation can be written as

C6H6 + 0.95 * 7.5(O2 + 3.76N2) â xCO2 + (6-x) CO + 3H2O + 26.79N2

O2 balance: 0.95 * 7.5 = x +(6-x)/2 +1.5 â x=5.25

Thus C6H6 + 7.125(O2 +3.76 N2) â 5.25 CO2 + 0.75CO + 3H2O + 26.79N2

The mole fraction of CO in the products is

yCO = N CO/ N total = 0.75/5.25 + 0.75+3+26.79

=0.021 or 2.1% mole fraction of the CO in the products

5 0

3 years ago

Read 2 more answers

In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2

sp2606 [1]

Answer:

$l=24mm$

Explanation:

From the question we are told that:

Plane strain fracture toughness of $T=55 MPa-m1/2$

Y value $Y=1.0$

Stress level of $\sigma =200 MPa$

Generally the equation for length of a surface crack is mathematically given by

$l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2$

$l=\frac{1}{3.142}(\frac{55}{1*200})^2$

$l=0.024m$

Therefore

in mm

$l=24mm$

6 0

3 years ago

A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi

Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

$F_{e}$ = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

= 7.5 x 1000 W

= 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 = F₁ - F₂

750 = 2 F₂ - F₂ ( ∵F₁ = 2 F₂ )

∴F₂ = 750 N

Now F₁ = 2 F₂

F₁ = 2 x F₂

F₁ = 2 x 750

F₁ = 1500 N , this is the maximum force.

Therefore we know,

$F_{max}$ = 3 x $F_{e}$

where $F_{e}$ is centrifugal force

$F_{e}$ = $F_{max}$ / 3

= 1500 / 3

= 500 N

8 0

3 years ago