The power required to pump the oil, P in hp, is determined from
P=4x10-¹3-
w²
p'd³
where
w: oil flow rate in lb/hr
p: density of the oil in lb/ft³
d: pipe diameter in ft
The cost incurred, in dollars, is given by
C = 10000d² +170P
where the first term on the right-hand side represents the capital cost of the pipeline and the
second term represents the cost of the pump and its operation. Power in the cost equation can
be calculated using the equation for power above.
a. Create a user-defined function called pumppower with arguments w, rho, and d that
determines the power requirements (P) to pump oil. Use your function in Excel to determine
the power requirements when the oil flow rate is 107 lb/hr, the density is 50 lb/ft³, and pipe
diameter is 2.0 feet [i.e., calculate "=pumppower(1e7,50,2.0)"].
b. Create a user-defined function called cost with arguments d and P that calculates the cost of
a pipeline. Use your answer from part a along with a diameter of 2.0 ft. to calculate the cost
of a corresponding pipeline [i.e., calculate "=cost(2.0,< result from part a >)].
c. Now, use the Solver tool in Excel along with your pumppower and cost functions to adjust
the diameter to determine the minimum cost for a pipeline pumping 107 lb/hr of oil with a
density of 50 lb/ft³. You may want to generate a quick plot of cost as a function of diameter to
verify your solution.
Fig: 1