It is probably best to choose as your base diameter the pipe size of the section of pipe that is responsible for the greatest part of the pressure drop. But you don't know which section that is (unless it is very obvious) until you have done the calculation. Be prepared for a bit of trial and error.
For this type of "single equivalent diameter" calculation it is easier to treat the non-base size pipe as an equivalent length of base sized pipe than as a K value. The pressure drop for turbulent flow in straight pipe varies with the 5th power of the diameter. So you just multiply the length of 2" pipe by (4/2)^5 and add its equivalent length to that of the 4" pipe. In this case you will assume that there is 3200 ft of 4" pipe to represent the 100 ft of 2" pipe.
Remember that a K value is associated with a particular velocity. The K value you have calculated for your reducer (which you have treated as a sudden contraction) is based on the velocity in the smaller diameter. This is what Crane calls K
1. To convert the K
1 to a K
2 i.e. based on the larger diameter, you must divide by (d/D)^4. Compare Crane equations 2-10.1 and 2-11 and my previous post. The K value for the reducer, based on the velocity in the 4" pipe, becomes 0.375 x 16 = 6.0. Or if you want to convert this to an equivalent length of 4" pipe by using the [ƒ] x L/D = K relationship then L = 6 x 4 / 0.02 = 1200" = 100' (This is the method described above by rmw).
This makes the [ƒ] x L/D + [Σ]K terms either
0.02 x 3300 / (4/12) + 6 (leaving the reducer as a K value) or
0.02 x 3400 / (4/12) (treating the reducer as an equivalent length)
In both cases the value is 204
Substituting this value into Eq 3-7 will give you the flowrate.
If you want to calculate the pressure drop across each component you would have to recast Eq 3-7 to calculate P
2 (too difficult for me) or use the Goal Seek function in your spreadsheet to calculate the value of P
2 that would give the known flowrate.
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