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vlabodo [156]
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.
Physics
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

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A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m
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Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

       = 4.98 × 10¹⁹ protons

       ≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

                                                         = 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

                                                         = 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

     = 30 × 10⁷/14.62

     = 2.05 × 10⁷ °C

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