Ask question

Ask question

All categories

3 years ago

10

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignit

e the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of 18° C is compressed adiabatically to a temperature of 733° C. Assume the air to be an ideal gas for which
γ = 7/5
Find the compression ratio, which is the ratio of the initial volume to the final volume.

1 answer:

Klio2033 [76]3 years ago

7 0

Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1} = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given: T₁= 18°C we convert to Kelvin(K) by adding 273.

∴ T₁= 18°C + 273 = 291K

T₂ = 733° C also,we convert to Kelvin(K) by adding 273

∴ T₂ = 733° C +273 =1046K

γ = 7/5

∴ Rearranging equation(1), we have

(T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

(V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

V₁/V₂ = 24.42.

∴ Compression ratio = 24.42

You might be interested in

The compressor on an air conditioner draws 35.0 A when it starts up. If the start-up time is 0.59 s, how much charge passes a cr

Elden [556K]

Answer:

is a conditioner es 13.4

6 0

3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

3 0

3 years ago

I need help, please answer

Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0

3 years ago

A 0.15 kg baseball is hit by a baseball bat. Right before it is hit, the ball’s speed is 30 m/s, and right after it is hit, the

Roman55 [17]

Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

Explanation:

We have given mass of the baseball m = 0.15 kg

Ball speed before hit $v_1=30m/sec$

Ball speed after hitting $v_2=-50m/sec$ ( negative direction due to opposite direction )

We have to find the impulse

We know that impulse is equal; to the change in momentum

So change in momentum = $m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec$

So option (b) will be correct option

5 0

2 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0

2 years ago