Question

Question 1: Calculate the flow in each member of the pipe network shown in Figure 2.4 and also the presure difference in N/m' between C and iE. Data for the individual pipes is given in the table below. 360 Ve 300 V E 600 V 460 Va 100 Vs Figure 2.4 UET Pipe Length (m) Diameter (mm)Friction coefficient (l) 530 0.020 460 BC700 530 0.020 CD 610 600 0.022 DE601 600 0.018 EA 150 450 0.018 Hint: Assume initially that 100 s (Answer: u 92.43 c 257.57 s, c 342.43 Us, Q 242.43 s, 207.57 s Pressure Difference between Cand E 23132 N/m)

Answer 1

Answers:

Now first we have to assume initially the discharges in each pipe.

300 l/s and 600 l/s are entering at A and C

given , initially assume, Qab = 100 l/s

Qcb = 250 l/s , Qbc = - 250 l/s

Qcd = 600 - 250 = 350 l/s ,

Qde = 350 - 100 = 250 l/s

Qae = 300 - 100 = 200 l/s , Qea = - 200 l/s

we assume clockwise flow as positive

we know,

h = K * Q^2

K = 8*f*L / (g*D^5 * $\pi$^2)

$\sum$ k * Q^2 = 0

pipe	Q (m3/s)	f	K	h	h/Q	L
AB	0.100	0.02	18.177	0.182	1.82	460
BC	-0.250	0.02	30.032	-1.877	7.508	760
CD	0.350	0.022	14.26	1.747	5	610
DE	0.250	0.018	11.495	0.718	2.872	601
EA	-0.200	0.018	12.09	-0.484	2.42	150
total				0.286	19.62

$\Delta$Q = - h(total) / (2*(h/Q)total)

= - 0.286 / ( 2* 19.62)

= - 0.007 = - 7 L/s

New discharges:

Qab = 100 - 7 = 93 L/s

Qbc = - 250 - 7 = - 257 L/s

Qcd = 350 - 7 = 343 L/s

Qde = 250 - 7 = 243 L/s

Qea = - 200 - 7 = - 207 L/s

2nd iteration :

all values of f/ L, D, K remaining same

pipe	Q(m3/s)	h	h/Q	K
AB	0.093	0.157	1.688	18.177
BC	- 0.257	-1.984	7.72	30.032
CD	0.343	1.678	4.892	14.26
DE	0.243	0.679	2.794	11.495
EA	-0.207	-0.518	2.502	12.09
total		0.012	19.596

$\Delta$Q = - h(total) / (2*(h/Q)total)

= - 0.012 / ( 2* 19.596)

= - 0.0003 = - 0.3 L/s

New discharges:

Qab = 93 - 0.3 = 92.7 L/s

Qbc = - 257 - 0.3 = - 257.3 L/s

Qcd = 343 - 0.3 = 342.7 L/s

Qde = 243 - 0.3 = 242.7 L/s

Qea = - 207 - 0.3 = - 207.3 L/s

By performing 3rd iteration,

Qab = 92.43 L/s

Qbc = 257.57 L/s

Qcd = 342.43 L/s

Qde = 242.43 L/s

Qea = 207.57 L/s