Topic: Fluid Mechanics
Pipes of lengths l1, l2 etc and diameters d1, d2, etc are connected in series. A single pipe of length l1+l2+l3+.... having same pressure loss as all pipes in series will have diameter equal to
- A. ((l1+l2+l3+...) / (l1/d1^4 + l2/d2^4 + ...))^(1/4)
- B. ((l1+l2+l3+...) / (l1/d1^5 + l2/d2^5 + ...))^(1/5)
- C. ((l1+l2+l3+...) / (l1/d1^3 + l2/d2^3 + ...))^(1/3)
- D. ((l1+l2+l3+...) / (l1/d1^2 + l2/d2^2 + ...))^(1/2)
Answer: B. ((l1+l2+l3+...) / (l1/d1^5 + l2/d2^5 + ...))^(1/5)
Explanation: When pipes are connected in series, the equivalent diameter equation is derived from the head loss formula: l/d^5 = l1/d1^5 + l2/d2^5 + l3/d3^5 + ... giving d = ((l1+l2+l3+...) / (l1/d1^5 + l2/d2^5 + ...))^(1/5).