Question

Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload gas 72 bar. suppose. A- Isothermal Cycle b- Adiabatic cycle, Charge in 25 seconds and discharge in 10 seconds.

Answer 1

1)
`V_o = 10 liters`

2)
`V_o = 12.26 liters`

Step-by-step explanation:

For isothermal process n =1

`V_o =(\Delta V)/(((p_o)/(p_1))^(1/n) -((p_o)/(p_2))^(1/n))`

`V_o  = (5)/([(72)/(80)]^(1/1) -[(72)/(180)]^(1/1))`

`V_o = 10 liters`

calculate pressure ratio to determine correction factor

`(p_2)/(p_1) =(180)/(80) = 2.25`

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

`actual \ volume = c1* 10 = 10.3 liters`

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

`V_o =(\Delta V)/([(p_o)/(p_1)]^(1/n) -[(p_o)/(p_2)]^(1/n))`

`V_o  = (5)/([(72)/(80)]^(1/1.4) -[(72)/(180)]^(1/1.4))`

`V_o = 12.26 liters`

calculate pressure ratio to determine correction factor

`(p_2)/(p_1) =(180)/(80) = 2.25`

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

`actual \volume = c1* 10 = 11.5 liters`