The Bernoulli equation, with friction, has the following form:
$$
p_1+\frac{1}{2}{\rho}v_1^2+{\rho}gz_1=p_2+\frac{1}{2}{\rho}v_2^2+{\rho}gz_2+\left(f\frac{L}{D}+{\sum}K\right)\frac{1}{2}{\rho}v_2^2
$$
where $p$ is pressure, $\rho$ desity, $v$ velocity, $z$ height, $g$ the acceleration due to gravity, $f$ the friction factor, $L$ the length of the tube, $D$ the hydraulic diameter and $K$ additional sources of local friction.
A tap falls under the category of additional sources of local friction. Adjusting the tap (its $K$ value) will change the flow-rate, because this factor gets multiplied by the velocity squared.
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 $Q_b$.
Best Answer
This is a great question, and the answer relates intimately to why turbofan engines equipped with afterburners require variable geometry exhaust nozzles. Without increasing the throat area to accommodate the larger volumetric flow rate, lighting the afterburner would back-pressure the fan and very possibly lead to a compressor stall. Similarly, mechanically reducing the downstream area (all other things being equal) will require the flow to have a higher upstream stagnation pressure, which means the fan/pump will be required to work harder.
Now, as to your question about why the flowrate decreases when the exit area is closed, we need to expound a bit on how fans and compressors operate. The fan speed, massflow rate, and pressure ratio are related in a complex way and are usually represented graphically by a fan map.
For a given rotational speed, there is a single steady-state characteristic relating pressure ratio and massflow rate. The shape of this curve can vary (e.g. compare the 40% Nf line with the 100% Nf line above), but generally speaking the higher the pressure ratio, the lower the massflow rate for a given engine RPM. This makes some intuitive sense because the faster the bladetip velocity compared to the axial velocity, the higher will be the flow turning within the bladerow. Work done and pressure rise are proportional to flow turning within the rotor, so higher pressure ratios are positively correlated with lower massflow rates/axial velocities (up to a point).
To truly understand the causal relationship between massflow rate and back-pressure requires that we abandon steady-state thinking altogether. If the exit area is reduced, unsteady compression waves propagate upstream at the speed of sound, incrementally increasing the static pressure at the exit of the fan. This increased back-pressure means that the entering flow now encounters an adverse streamwise pressure gradient and slows down. This slower flow is then worked harder by the spinning bladerow, which results in larger stagnation pressure and temperature rises.
Remember that the flow always exits the device at atmospheric pressure so long as it is subsonic, precisely because of the information propagated upstream by unsteady pressure waves. Thus, if reducing the exit area means a higher exit Mach number is required to conserve mass, the total-to-static pressure ratio must increase. A fixed exit static pressure and increased total-to-static pressure ratio demands that the upstream stagnation pressure increase, and so the upstream turbomachinery will be affected.
If you are looking to put numbers on things, the isentropic flow function is a useful and straightforward way to determine the massflow rate of a compressible fluid if other of the fluid's basic properties are known. In general, the massflow rate of a fluid through a cross-sectional area $A$ is equal to
$\dot{m}=\rho VA$.
Now, if the fluid is compressible and the Ideal Gas Law applies, then
$\dot{m}=\rho VA=\left(\frac{P}{RT}\right)(M\sqrt{\gamma RT})A=PAM\sqrt{\frac{\gamma}{RT}}$.
Both the stagnation temperature and stagnation pressure are preferred flow variables to their static counterparts, so the above equation can be rewritten as
$\dot{m}=P_0 \left(\frac{P}{P_0}\right)AM\sqrt{\frac{\gamma (T_0/T)}{R(T_0)}}$,
and the stagnation properties (as well as the through-flow area) can be moved to the LHS of the equation:
$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=\left(\frac{P}{P_0}\right)M\sqrt{\frac{\gamma}{R}\left(\frac{T_0}{T}\right)}$
If the flow is isentropic (as we are assuming), we know that
$\frac{P}{P_0}=\left(\frac{P_0}{P}\right)^{-1}=\left(\frac{T_0}{T}\right)^\frac{\gamma}{1-\gamma}$,
which gives us
$\frac{\dot{m}\sqrt{T_0}}{P_0 A}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1}{2}+\frac{\gamma}{1-\gamma}}=M\sqrt{\frac{\gamma}{R}}\left(\frac{T_0}{T}\right)^{\frac{1+\gamma}{2(1-\gamma)}}$.
Again invoking our assumption of isentropic flow, we know that the stagnation temperature ratio is related to the local Mach number by the following equation:
$\frac{T_0}{T}=1+\frac{\gamma-1}{2}M^2$
which, when plugged into the previously derived expression gives us the isentropic flow function $FF_T$:
$FF_T=\frac{\dot{m}\sqrt{T_0}}{P_0 A}=M\sqrt{\frac{\gamma}{R}}\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{1+\gamma}{2(1-\gamma)}}$
To compute the massflow rate we simply rearrange the isentropic flow function relation...
$\boxed{\dot{m}=P_0 AM\sqrt{\frac{\gamma}{RT_0}}\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{1+\gamma}{2(1-\gamma)}}}$.
**Note: The above equation is true at any given section within a compressible flow, but the stagnation properties may change from location to location (or over time) based on the specifics of the exact flow the equation is being applied to.