I think that the book incorrectly solves the problem, but I wanted to see if you guys agree...
Basically, it asks for the directional flow rate, and says it is a 60/40 split (1100 total), so 660.
At the beginning of the solution, they pick data from the tables to correspond with a directional value of 660, which I agree with. However, when they pick the Et value, they use the overall volume of 1100, giving them an Et of 1.9 instead of 1.5.
Then, when they plug the values back into the equation to solve for the Vp, they use the 1100 value and multiply the Vp by .6 at the end after solving for Vp.
This is incorrect, right?
They started the solution by using the directional volume of 660, which is what they should have done because the question asks for the highest directional flow rate for the peak 15 min period. They should have chosen the values from the table that correspond to a directional volume of 660, and then completed the Vp equation like this:
Vp=(660/(.92*.99*.876))... which gives an answer of 827 pcph ©... not 778 pcph (B)
Do you guys agree?
Basically, it asks for the directional flow rate, and says it is a 60/40 split (1100 total), so 660.
At the beginning of the solution, they pick data from the tables to correspond with a directional value of 660, which I agree with. However, when they pick the Et value, they use the overall volume of 1100, giving them an Et of 1.9 instead of 1.5.
Then, when they plug the values back into the equation to solve for the Vp, they use the 1100 value and multiply the Vp by .6 at the end after solving for Vp.
This is incorrect, right?
They started the solution by using the directional volume of 660, which is what they should have done because the question asks for the highest directional flow rate for the peak 15 min period. They should have chosen the values from the table that correspond to a directional volume of 660, and then completed the Vp equation like this:
Vp=(660/(.92*.99*.876))... which gives an answer of 827 pcph ©... not 778 pcph (B)
Do you guys agree?
