Condensed Problem statement: 40ft of 4in sch. 40 steel pipe containing five long radius 90deg. elbows + one gate valve + one sharp edge. V=16ft/s, f=0.019
Book Solution: K_entrance = 0.5 and all others included in the equivalent length as 40ft + 5(6.4ft) + 2.5ft = 65.5 and total head loss calculated as K*v^2/(2g) + f*(65.5)*v2/(2Dg) and answer = 16.7ft
My method: K_entrance + K_all others = 0.5+5(0.6)+0.19 = 3.69 and total head loss = v^2/(2g)*(3.69+f*(40)/D) = 23ft
I always thought the same answer is achieved whether we take K as a loss or included as an equivalent length, what am I missing here?
Thank you
Book Solution: K_entrance = 0.5 and all others included in the equivalent length as 40ft + 5(6.4ft) + 2.5ft = 65.5 and total head loss calculated as K*v^2/(2g) + f*(65.5)*v2/(2Dg) and answer = 16.7ft
My method: K_entrance + K_all others = 0.5+5(0.6)+0.19 = 3.69 and total head loss = v^2/(2g)*(3.69+f*(40)/D) = 23ft
I always thought the same answer is achieved whether we take K as a loss or included as an equivalent length, what am I missing here?
Thank you