Similarly

Similarly, branch E must have the same pressure loss as the sum of losses in branches J, H, and F.  From Table 2 this loss is .026+.001+.010 = .037 in. of water.  Branch E has an equivalent length of 17 ft + 35 ft (elbow) = 52 ft.  The loss per 100 ft then is .037(100/52) = .071 in. /100 ft.  Locate the point .071 and 247cfm in an air friction chart.  From this point read the duct velocity to be 650 fpm and the duct size to be 8.5 inches in diameter.  
The remaining branches are sized using the same procedure.  This results in all branches having the same total pressure loss while having the required amount of flow in each branch.
8. Ventilation
The ventilation for this study was assumed to enter through an outside duct into the air handler.  If instead the fresh air is brought directly into the premises, the ventilation contributes to the sensible and latent heat loads of the interior space.  Assuming 320 cfm satisfies code requirements, the sensible load from the ventilation is given by equation 8
                      q = 1.1 (320) (95 – 80) = 5280 Bt/hr
The total internal sensible load is 35274 + 5280 = 40554 Btu/hr and the new flow rate required from the air handler is
© Gary D. Beckfeld  Page 15 of 21
                     w = 40554/ ((1.1) (95-80)) = 2457 ft3 /min
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Similarly, the latent load from the ventilation can be calculated and a new total latent heat load and new sensible heat ratio, SHR, found.  Entering the new SHR on the psychrometric chart locates a new point 1, new coil temperature, and new enthalpy values for resizing the required cooling tonnage.
9. Cooling Load Temperature Difference and Heating Degree Days
In the heat conduction equation 2, the difference between the maximum outside temperature and desired inside temperature was used.   Equation 2 can also be used a