Question

A gas charged accumulator is to provide an hydraulic system with 28 litres of oil from a maximum pressure of 40 MPa to a final pressure of 35 MPa. The accumulator will initially be charged to a pressure of 30 MPa. Determine the total volume of the accumulator. Assume expansion and compression takes place adiabatically and y 1,4

Answer 1

total volume = 0.9125 m³

Step-by-step explanation:

given data

volume = 28 litres

pressure 2 = 40 MPa

pressure 3 = 35 MPa

pressure 1 = 30 MPa

y = 1.4

to find out

total volume of the accumulator

solution

we know here oil deliver 28 litter

so V3 - V2 = 28 litter

so V3 = V2 + 28

V3 = V2 + 0.028

and we know

volume 1 = V2 ×
`((P2)/(P1))^(1/y)` ................1

and volume 2 = V3 ×
`((P3)/(P2))^(1/y)` ..................2

put here value

volume 2 = V3 ×
`((P3)/(P2))^(1/y)`

volume 2 = (V2 + 0.028) ×
`((35)/(40))^(1/1.4)`

volume V2 = 0.274 m³

and from equation 1

volume 1 = V2 ×
`((P2)/(P1))^(1/y)`

volume 1 = 0.274 ×
`((40)/(30))^(1/y)`

volume 1 = 0.3365 m³

and

volume V3 = V2 + 0.028

volume V3 = 0.274 + 0.028

volume V3 = 0.302 m³

so total volume = V1 + V2 + V3

total volume = 0.274 + 0.3365 + 0.302

total volume = 0.9125 m³