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An adiabatic closed system is accelerated from 0 m/s to 34 m/s. Determine the specific energy change of this system, in kJ/kg.

Respuesta :

Khoso123

Answer:

Δe=0.578 kJ/kg

Explanation:

Given data

Velocity v₁=0 m/s

Velocity v₂=34 m/s

to find

Specific energy change Δe

Solution

The specific energy change is simply determined from change in velocity

Δe=(v₂²-v₁²)/2

Put the given values to find the specific energy change

[tex]=(\frac{(34)^{2} *10^{-3} }{2} )\\=0.578kJ/kg[/tex]

Δe=0.578 kJ/kg

stigawithfun

Answer:

0.578 kJ/kg

Explanation:

An adiabatic system has no change in heat transferred in or out of it though there may be a change in temperature. And a system is closed if it is isolated from its surrounding and bounded so that no heat is lost or gained.

In an adiabatic closed system, the change in specific energy (ΔE) is related to change in velocity as follows;

ΔE =  (V₂² - V₁²) / 2      ---------------------(i)

Where;

V₂ and V₁ are the final and initial velocities respectively.

From the question, the following are given;

V₁ = 0m/s

V₂ = 34m/s

Substitute these values into equation (i) to give;

ΔE =  (34² - 0²) / 2

ΔE =  (34²) / 2

ΔE =  (1156) / 2

ΔE =  578J/kg

ΔE =  0.578 kJ/kg

Therefore, the specific energy change of this system is 0.578 kJ/kg

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