where VA = VB = PA = PB = 0 — irrigation company must transport 10-2 m3/sec water from reservoir The reservoirs are — Applying energy equation

where VA = VB = PA = PB = 0,

hL = hf + [ΣK](V)2/2g; Ke = 0.5, Kd = 1.0 (exit coef.), hA – hB = 18.4 m, and V = Q/A = 0.0727/D2. Thus, 18.4 m = [f(600m/D) + 1.5]·[(0.0727/D2)2/2g]. Assume D = 0.2 m, thus V = 1.82 m/s, e/D = 0.36/200 = 0.0018; NR = DV/ν = [(0.2)(1.82)]/(1.00x10-6) = 3.64 x 105; and f = 0.023; solving energy e’qn for new D; 18.4 m = [0.023(600m/D) + 1.5]·[(0.0727/D2)2/2g]; D = 0.183 m which differs from the trial size, thus V = 2.17 m/s, e/D = 0.00197; NR = [(0.183)(2.17)]/(1.00x10-6) = 3.97 x 105; and new f = 0.0235; and 18.4 m = [0.0235(600m/D) + 1.5]·[(0.0727/D2)2/2g]; D = 0.184 m; close enough so D = 0.184 m

Without minor losses; 18.4 m = [0.0235(600m/D)]·[(0.0727/D2)2/2g]; D = 0.183 m; essentially the same. The solution process is a little easier without minor losses; solve this first and add minor losses later.