# Bernoullis in Parallel pipes

Im having trouble with a scenario where a flow 𝑄𝑆𝑡𝑎𝑟𝑡 splits into two parallel pipes 𝑄𝐴 and 𝑄𝐵 then rejoins 𝑄𝐸𝑛𝑑 before exiting the control volume

what makes this scenario difficult is the parallel pipes are of varying diameter

At the diffluence Pipe A has a diameter 2D and Pipe 2 has a diameter of 1D by the time they reach the confluence they have reversed diameter now have Pipe A has a dimeter of 1D and Pipe B has a dimeter of 1D

Bernoullis

Start - assume a flow velocity 5ms static pressure SP 100

$Q_s=A1.V1$

$Q_s=3.5$

$Q_s=15$

$TP=SP+\frac{1}2mv^2$

$TP=100+\frac{1}25^2$

$TP=112.5$

Stream a

$QA=A1.V1=A2.V2$

$A1.V1 =A2.V2$

$2A.5ms = 1A.10ms$

$112.5=SP+\frac{1}2m10^2$

$SP=112.5-50$

$SP=62.5$

Stream b

$A1.V1 =A2.V2$

$1A.5ms^-1 = 2A.2.5ms^-1$

$112.5.5=SP+DP$

$SP=112.5-\frac{1}2m2.5^2$

$SP=112.5-3.125$

$SP=109.375$

Q Check

$Q_s=3.5$=15=1.5+2.5=QA+QB=1.2.5+1.10=3.5=Q_e$Head Loss What we know All elements of flow converging at WILL have the same head loss. The flow will adjust automatically so that the head loss in each branch pipe WILL BE THE SAME$Hl_A=Hl_B$According to resistance coefficient tables the divergent pipe has a K value of 0.46 and the convergent pipe has a K value of 0.1 As these are Losses are proportional to – velocity of flow, this suggests that the expansion pipe will decrease its flow (to decrease its losses) while the convergent pipe must increase its flow to maintain continuity This means that the flow rates have diverged not come together but we know that they must be the same at the exit of the control volume examined Continuity also tells us that the total flow rate must be the same at all points in the pipe$𝑄_𝑆=𝑄𝑎_1+𝑄𝑏_1 =𝑄a_2+𝑄b_2 =𝑄_𝐸𝑣_𝑆.𝐴_𝑆=𝑣_a. 𝐴_a+𝑣_𝑏.𝐴_𝑏=𝑣_𝑎.𝐴_𝑎++𝑣_𝑏 𝐴_𝑏 =𝑣_𝑒.A_𝑒𝑣_𝑆.3=𝑣_a.2+𝑣_𝑏.1=𝑣_𝑎.1++𝑣_𝑏2 =𝑣_𝑒.3𝑒$So on total head/ stagnation value we will have the same value at the convergence as both paths have experienced the same head loss but Bernoullis tells us that we have very different velocities and static pressures at this point . My question is how at the confluence does this follow that we do not require the same value of pressure and velocity at the confluence for both streams? If this can occur we must then have a mechanism to achieve the expected uniform velocity and pressure (Not considering head losses) at the exit$𝑣_𝑆.3 =𝑣_𝑒.3𝑒$What would this mechanism be ? • Did you mean diameters, or did you mean areas? – Chet Miller May 10 '16 at 10:49 • Hi @ChesterMiller Yes it is most like square ducting so area would probably be more appropriate – Tom Chester May 10 '16 at 12:13 • Let's see your two Bernoulli equations for the two parallel ducts. – Chet Miller May 10 '16 at 15:29 • Bernoullis Start - assume a flow velocity 5ms static pressure SP 100$Q_s=A1.V1Q_s=3.5Q_s=15TP=SP+\frac{1}2mv^2TP=100+\frac{1}25^2TP=112.5$Stream a$QA=A1.V1=A2.V2A1.V1 =A2.V22A.5ms = 1A.10ms112.5=SP+\frac{1}2m10^2SP=112.5-50SP=62.5 **Stream b** $A1.V1 =A2.V2$ $1A.5ms^-1 = 2A.2.5ms^-1$ $112.5.5=SP+DP$ $SP=112.5-\frac{1}2m2.5^2$ $SP=112.5-3.125$ $SP=109.375 **Q Check**$Q_s=3.5$=15=1.5+2.5=QA+QB=1.2.5+1.10=3.5=Q_e$ – Tom Chester May 11 '16 at 3:05
• Hi @chestermiller I have added in the breakdown – Tom Chester May 11 '16 at 3:12

I wasn't able to figure out what you did, so here is my analysis, without the resistance. Let:

Q = Total volume flow rate

$Q_a$ = Volume flow rate into converging pipe

$Q_b$ = Volume flow rate into diverging pipe

$p_1a$ = static pressure just after entrance to a

$p_2a$ = static pressure just before exit from a

$p_1b$ = static pressure just after entrance to a

$p_2b$ = static pressure just before exit from a

$T_1$ = "total pressure" in channel leading up to diffluence

$T_2$ = "total pressure" in channel after diffluence

$A_{a1}$ = cross sectional area of converging pipe at inlet

$A_{a2}$ = cross sectional area of converging pipe at outlet

$A_{b1}$ = cross sectional area of diverging pipe at inlet

$A_{b2}$ = cross sectional area of diverging pipe at outlet

CASE OF NO FRICTIONAL LOSS

Bernoulli equations relevant to pipe a: $$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$ $$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}$$ $$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$ Adding these three equations together gives $$T_1=T_2$$ Thus, for the case without friction, energy is conserved and the "total pressure" after the split section is equal to the "total pressure" before the split section. This is irrespective of how the flow splits between the two sections. The Bernoulli equations for pipe b will give the same result. Also, the convergence and divergence in the channels doesn't matter, as long as the final outlet pipe has the same cross sectional area as the initial inlet pipe.

CASE WITH FRICTIONAL EFFECTS INCLUDED

Bernoulli equations relevant to pipe a: $$T_1=p_1+\rho \frac{(Q_a/A_{a1})^2}{2}$$ $$p_1+\rho \frac{(Q_a/A_{a1})^2}{2}=p_2+\rho \frac{(Q_a/A_{a2})^2}{2}+k_a\rho \frac{(Q_a/A_{a1})^2}{2}$$ $$p_2+\rho \frac{(Q_a/A_{a2})^2}{2}=T_2$$ Adding these three equations together gives $$T_1=T_2+k_a\rho \frac{(Q_a/A_{a1})^2}{2}\tag{1}$$ Similarly, for channel b:$$T_1=T_2+k_b\rho \frac{(Q_b/A_{b1})^2}{2}\tag{2}$$

Thus, for the case with friction, mechanical energy is not conserved and the "total pressure" after the split section is not equal to the "total pressure" before the split section. Moreover, the split between the two channels is relevant.

Mass balance equation: $$Q_a+Q_b=Q\tag{3}$$

Eqns. 1-3 provide three algebraic equations in the three unknowns $(T_1-T_2)$, $Q_a$, and $Q_b$.

• hi @chestermiller yes that all makes sense with the literature , but only if p2a does not equal p2b is that your expectation in fact no matter how the flow is split I I can't see any way it can (but it does normalise downstream) – Tom Chester May 11 '16 at 13:32
• That's correct. The fluid flow converging from the big pipe into the entrance of pipe a (and the corresponding pressure change) is different from the fluid flow and pressure change from the big pipe into the entrance of pipe b. In the case of pipe a, the flow is slowing down as it approaches the entrance, and, in the case of pipe b, the flow is speeding up as it approaches the entrance. – Chet Miller May 11 '16 at 13:38
• The flow that enters pipe a from the upstream conduit 1 slows down in the immediate vicinity of the entrance to pipe a, with an accompanying increase in pressure. Once inside pipe a, the flow speeds up as it converges toward the exit, with an accompanying decrease in pressure (to a value lower then in either conduit 1 or conduit 2). Once the flow exits pipe a, it slows down to the velocity in conduit 2, with an accompanying increase in pressure, back up to the original value in conduit 1. So the for the flow through a, the pressure first increases, then decreases, then increases again. – Chet Miller May 11 '16 at 22:11
• Of course, the discussion in my previous comment applies specifically to the case without friction. – Chet Miller May 11 '16 at 22:17
• Hi @chestermiller thanks for your explanation , I am interested in why the fluid slows down before it enters pipe a ( and vice versa pipe b) intuitively I know this to be the case but I would love to put a name to the force . Is it due to the orthogonal area presented to the flow by the convergent pipe causes a force impulse . Why I wonder is again this suggests that the Q would deviate to pipe b in an inviscid flow (as you stated) but the k values and head loss suggest that in a viscid flow the Q would deviate to pipe a ... This causes me much confusement – Tom Chester May 12 '16 at 1:09