Calculating Cooling Water Velocity

What matters is the volumetric flow rate because it is cooling water temperature and mass that are the major factos at work.

trying to calculate the diameter of the shroud to have the correct flow velocity in 35 degree water to keep the motor below 85 degrees.

Determining the water velocity is more than just a single simple equation. For a simple analysis, needed inputs would be the heat rejection of the motor, the motor surface area exposed to the cooling water stream, the target surface temperature of motor surface, the flow area and perimeter of the cooling water passage(s), and any limitations on the outlet water temperature. From this information, pertinent equations follow:

Q = m*cp*(Tout-Tin) ~ h*Asurf*(0.5*(Tout+Tin)-Tsurf)

h ~ 0.023*kw/Dh*(m/Aflow*Dh/mu)

Dh ~ 4*Aflow/p

V = m/(rho*Aflow)

where

Q = heat rejection rate (BTU/s)

m = water flow rate (lbm/s)

cp = specific heat of water (BTU/lbm-F)

Tin = water inlet temperature (deg F)

Tout = water outlet temperature (deg F)

h = water cooling heat transfer coefficient (BTU/s-ft^2-F)

Asurf = surface area exposed to the cooling water (ft^2)

Tsurf = surface temperature of the motor (deg F)

kw = thermal conductivity of water (BTU/s-ft-F)

Dh = hydraulic diameter (ft)

Aflow = water passage flow area (ft^2)

mu = water absolute viscosity (lbm/ft-s)

p = water passage perimeter (ft)

V = velocity in ft/s

rho = density of water (lbm/ft^3)

These equations are generally good for a coarse estimate of the needed cooling flow. The provided units for the variables are an example of consistent units, but any desired units can be used as long as consistent. For a more refined answer that takes into account variations of water properties, setup multiple calculation stations along the cooling flow passage, and use the "outlet" of a preceding station as the "inlet" to the next station.

From the equations, note that the velocity is function of the mass flow rate and the flow area. So the mass flow rate and the flow area are the design variables to adjust to meet the design needs. Although keep in mind that higher velocity also means needing a higher pressure drop in the cooling flow in order to flow the water, i.e.

m = K*Aflow*sqrt(2*gc*rho*(Pin - Pout))

where

K = effective flow resistance

gc = conversion factor (32.2 lbm/slug)

Pin = inlet pressure (lbf/ft^2)

Pout = outlet pressure (lbf/ft^2)

Thus the overall solution is a balancing act between available pressure drop, the motor temperature limits, and the needed heat rejection rate.

Hopefully this helps. Depending on your application, there may be approaches that can enhance the heat rejection rate in order to keep the motor cooler. And if adequate cooling is critical, a more detailed analysis could be done.

what si the rho value...........?