Hello,plugging in other fan parameters found in Example 11.1 of Fluid Mechanics, 2nd Edition by Frank M. White, McGraw-Hill, pp. 640 - 641.
I am attempting to estimate the pressure rise generated by a centrifugal fan by first estimating the head (air column) produced by using equations for estimating head of a centrifugal water pump and substituting the density of air for that of water and
The pressure rise was estimated to be 51.53 psi, which seems awfully high to me.
Am I doing this correctly?
The power is computed thus:
P-sub-a = ?*Q*u-sub-2*V-sub-t2 = (0.00237 [slugs/ft^3])*(32.34 [ft^3/s])*(324.63 [ft/s])*(11.84 [ft/s])
= 294.6 [ft-lbf/s]
Where
P-sub-a = power [ft-lbf/s]
? = density of air; std. atm. sea level = (0.00237 [slugs/ft^3])
Q = volume flow rate = (32.34 [ft^3/s])
u-sub-2 = outlet_tip_speed = (324.63 [ft/s])
V-sub-t2 = tangential component at outlet = (11.84 [ft/s])
The head is estimated thus:
H ? P-sub-a/?gQ = 294.6 [ft-lbf/s]/((0.076314 lbf/ft^3)*(32.34 [ft^3/s])) = 119 ft
Where
g = 32.2 ft/s^2
The head can be converted to pressure thus:
p = 0.433 h SG = 0.433 * 119 * 1.000 = 51.53 psi
where
p = pressure (psi)
h = head (ft)
SG = specific gravity of the fluid = 1.000
Source: https://www.engineeringtoolbox.com/pump-head-pressure-d_663.html
Thanks In Advance.
David
Hello,plugging in other fan parameters found in Example 11.1 of Fluid Mechanics, 2nd Edition by Frank M. White, McGraw-Hill, pp. 640 - 641.
I am attempting to estimate the pressure rise generated by a centrifugal fan by first estimating the head (air column) produced by using equations for estimating head of a centrifugal water pump and substituting the density of air for that of water and
The pressure rise was estimated to be 51.53 psi, which seems awfully high to me.
Am I doing this correctly?
The power is computed thus:
P-sub-a = ρ*Q*u-sub-2*V-sub-t2 = (0.00237 [slugs/ft^3])*(32.34 [ft^3/s])*(324.63 [ft/s])*(11.84 [ft/s])
= 294.6 [ft-lbf/s]
Where
P-sub-a = power [ft-lbf/s]
ρ = density of air; std. atm. sea level = (0.00237 [slugs/ft^3])
Q = volume flow rate = (32.34 [ft^3/s])
u-sub-2 = outlet_tip_speed = (324.63 [ft/s])
V-sub-t2 = tangential component at outlet = (11.84 [ft/s])
The head is estimated thus:
H ≈ P-sub-a/ρgQ = 294.6 [ft-lbf/s]/((0.076314 lbf/ft^3)*(32.34 [ft^3/s])) = 119 ft
Where
g = 32.2 ft/s^2
The head can be converted to pressure thus:
p = 0.433 h SG = 0.433 * 119 * 1.000 = 51.53 psi
where
p = pressure (psi)
h = head (ft)
SG = specific gravity of the fluid = 1.000
Source: https://www.engineeringtoolbox.com/pump-head-pressure-d_663.html
Thanks In Advance.
David
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