Question

E Question-4 1.2 m 04=0 D Pipe 40-mm inside diameter 1.8 m Figure shows a siphon that is used to draw water from a swimming pool. The pipe that makes up the siphon has an inside diameter of 40 mm and terminates with a 25-mm diameter nozzle. Assuming that there are no energy losses in the system, calculate the volume flow rate through the siphon and the pressure at points B-E. 25-mm diameter 1.2 m Pp=0 Question - 5

Answer 1

equation at By using Bernoulli's A points and F :- + At point A :- PA=0 ; VA=0 At point F:- Pf=o PA + VAR +ZA 11 PE + Vf² + ZF Ig ag eg 2 ZA = eft vf 2g Vo = Jag (70-71) from figure! - ZA-Zf= ZN-ZF = 1.8 + 1.2 = 3.0m VF - 249.81) 13.0) = 7.67 m/sec the flow rate By continuity eqh; the siphon is ; through Q = AF VF Vp=7.67 m/sec df=25mm 2 Ap = * (0.025 m)? 4.91 x 10-4 m 4 Q = (4.91 x 10-4) (7067) 3.77 *10-2m%sec

The flow rate ; Q = 3,77 x 10-3 m3/sec Ans for the pressure at points B-E : ean b/w A and B :- points By Bernoulli's PA PB + VB 2 + + ZA (1 ZB Ag sg zg ZA = = PB tuo? +ZB eg zg ZA = ZB PB 39 + VB? = 0 At B :- dB = 40 mm 3.77 X 10 510-3 Q 3.0 m/sec VB = RE AB 5 ( 40x10-3)? VB PB 59 1-2 1000 X 3 2 22 29 PB 4.5 kPa Ans By Bernoulli's eqr blw points Å and ci- 0 11 + P 29 + ZA Pc + ve ² +2c Ig ag P9 2 ZA = + zc Pc + vc fg 29

= 59 [17a-21) – hote] ^ be ZA - Zc = -1.2m ; VA = vc - 3.0 m/sec Pc = (1000)19.81) て 1.2 (3) 219.81) Pe= -16.27 kPa are Since point B and D elevation and UB = UB = UB at equal hence PDF - PB PB - 4.50 kPa Ans By Bernoulli's eqh blw A and -: 3 PZ 19 + va +TA 2 PE 89 + VE วง + LE 2 ZA = PE 19 + VE² + ZE Go PE 89 [(ZA-ZE) - V 29 = 32-42 3.0m VE = VB = 3.0m) sec

L Pe = (1000) (9.81) 3.0 (3)? 219.81) IPE = 24.93 kPa Ans