a non-flow quasi-equilibrium process occurs for which p=â’1.5v+20 bar , where v is volume in m3. what is work done when v changes from 1 to 5 m3?
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Answered by
0
The formula is :
Wd = ∫pdv
p = 101325 pa
= 101325 × ₁∫⁵1.5V + 20
= ₁⁵[1.5/2V² + 20V]
= (1.5/50 + 100) - (1.5/2 + 20)
= 79.28
= 79.28 × 101325 = 803.3046kJ
Answered by
0
Formula for word done Wd = integral of pdv,
p = 101325 pa, = 101325 × integral of 1 to 51.5V + 20,
= integral of 1 to 5 [1.5/2V² + 20V], = (1.5/50 + 100) - (1.5/2 + 20),
= 79.28, = 79.28 × 101325 = 803.3046kJ.
So v changes from 1 to 5 m3 then word done is 803.3046kJ
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