Help with equation

[Guest]

Vmax allowable = 1.2V P {1/EW + D/eER}
where:-
1.2 = air allowance
V = pipe volume in litres
P = pressure drop in kPa
D = pipe internal diameter in metres
e = pipe wall thickness in metres
ER = pipe material modulus kPa (see Table 1)
EW = Bulk modulus of water (see Table 2)


e = 0.017m
V =16.748L
D = 0.146m
ER =@20Celcius = 750
EW =@20Celcius = 2170
P =180 Kpa

[Jonasmathwork]

Solution : Given V max allowable = 1.2V P {1/EW + D/e*ER}

where:-
1.2 = air allowance
V = pipe volume in liter
P = pressure drop in k Pa
D = pipe internal diameter in meter
e = pipe wall thickness in meter 
ER = pipe material modulus k Pa (see Table 1)
EW = Bulk modulus of water (see Table 2)


e = 0.017m
V =16.748L
D = 0.146m
ER =@20Celcius = 750
EW =@20Celcius = 2170
P =180 K pa

then V max =1.2*16.748*180{1/2170+0.146/(0.017*750)}
V max =43.0917