Solution Found!
Water is to be pumped from a lake to a ranger station on
Chapter 7, Problem 7.55(choose chapter or problem)
Water is to be pumped from a lake to a ranger station on the side of a mountain (see figure). The flow rate is to be 95 gaUmin, and the flow channel is a standard I-in. Schedule 40 steel pipe (ID = 1.049 in.). A pump capable of delivering 8 hp (= - W,) is available. The friction loss F (ft 'Ibf/lbm ) equals 0.04IL, where L(ft) is the length of the pipe. Calculate the maximum elevation, z, of the ranger station above the lake if the pipe rises at an angle of 30.
Questions & Answers
QUESTION:
Water is to be pumped from a lake to a ranger station on the side of a mountain (see figure). The flow rate is to be 95 gaUmin, and the flow channel is a standard I-in. Schedule 40 steel pipe (ID = 1.049 in.). A pump capable of delivering 8 hp (= - W,) is available. The friction loss F (ft 'Ibf/lbm ) equals 0.04IL, where L(ft) is the length of the pipe. Calculate the maximum elevation, z, of the ranger station above the lake if the pipe rises at an angle of 30.
ANSWER:Problem 7.55Water is to be pumped from a lake to a ranger station on the side of a mountain (see figure).The flow rate is to be 95 gal/min, and the flow channel is a standard 1-in. Schedule 40 steel pipe(ID = 1.049 in.). A pump capable of delivering 8 hp (= - Ws) is available. The friction loss F(ft-lbf/lbm) equals 0.041L, where L(ft) is the length of the pipe. Calculate the maximumelevation, z, of the ranger station above the lake if the pipe rises at an angle of 30°. Step-by-step solution Step 1 of 4 ^The relation between the velocity, volumetric flow rate and area of the pipe is as follows:Here, is the volumetric flow rate, ID is the inner diameter of the pipe and u is2he finalvelocity of fluid.Volumetric flow rate, is equal to 95 gal/min.Inner diameter of pipe, ID is equal to 1.049 in.Substitute 95 gal/min for and 1.049 in for ID in the above equation to calculate the value ofu2. 2 3 4 × 95 gal/min (12 in) 1 ft 1 min u2 2 × 2 × 7.4805 gal × 60 s (1.049 in) 1 ft = 35.2 ft/s