Bernoullis in Parallel pipes

Question

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

$𝑄_𝑆=𝑄𝑎_1+𝑄𝑏_1 =𝑄a_2+𝑄b_2 =𝑄_𝐸$

$𝑣_𝑆.𝐴_𝑆=𝑣_a. 𝐴_a+𝑣_𝑏.𝐴_𝑏=𝑣_𝑎.𝐴_𝑎++𝑣_𝑏 𝐴_𝑏 =𝑣_𝑒.A_𝑒$

$𝑣_𝑆.3=𝑣_a.2+𝑣_𝑏.1=𝑣_𝑎.1++𝑣_𝑏2 =𝑣_𝑒.3𝑒$

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 ?

Answer 1

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 $Q_b$.