Respuesta :
Answer:
1) [tex]V_o = 10 liters[/tex]
2) [tex]V_o = 12.26 liters[/tex]
Explanation:
For isothermal process n =1
[tex]V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}[/tex]
[tex]V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}[/tex]
[tex]V_o = 10 liters[/tex]
calculate pressure ratio to determine correction factor
[tex]\frac{p_2}{p_1} =\frac{180}{80} = 2.25[/tex]
correction factor for calculate dpressure ration  for isothermal process is
c1 = 1.03
[tex]actual \ volume = c1\times 10 = 10.3 liters[/tex]
b) for adiabatic process
n =1.4
volume of hydraulic accumulator is given as
[tex]V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}[/tex]
[tex]V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}[/tex]
[tex]V_o = 12.26 liters[/tex]
calculate pressure ratio to determine correction factor
[tex]\frac{p_2}{p_1} =\frac{180}{80} = 2.25[/tex]
correction factor for calculate dpressure ration  for isothermal process is
c1 = 1.15
[tex]actual \volume = c1\times 10 = 11.5 liters[/tex]
