How would you calculate the time it would take to drain down a cyclinder that is 250mm diameter and 10 meters long with an orifice .025mm. Just using atmospheric pressure. Then I would like to know how much of an increase in flow would i get if I introduce some extra pressure. i.e compressed air of 2 bar. Thank you.
+33658 The corresponding expression for a 2 bar applied pressure is:
τ=√2Aag(√Δpρ+gh0−√Δpρ)
Δp = 2 - 1 = 1 bar = 10^5 N/m^2 assuming 1 atmosphere = 1 bar
Assuming the liquid is water with a density of 1000 kg/m^3 then:
time=(√2×1089.8)×(√1051000+9.8×10−√1051000)(3600×24×365)⇒time=1.8629868806888171
time ≈ 1.9 years
.
+33658 A = pi*(0.25/2)^2 m^2
a = pi*(0.000025/2)^2 m^2
A/a = 10^8
g = 9.8 m/s^2
h0 = 10 m
so
time=108×√2×109.8(3600×24×365)⇒time=4.529970283394941
time ≈ 4.5 years
.
+33658 The corresponding expression for a 2 bar applied pressure is:
τ=√2Aag(√Δpρ+gh0−√Δpρ)
Δp = 2 - 1 = 1 bar = 10^5 N/m^2 assuming 1 atmosphere = 1 bar
Assuming the liquid is water with a density of 1000 kg/m^3 then:
time=(√2×1089.8)×(√1051000+9.8×10−√1051000)(3600×24×365)⇒time=1.8629868806888171
time ≈ 1.9 years
.
