System curve with a merging stream needs a junction pressure

A gear pump feeds stream 1 into a T-junction. A water-mains stream 2 joins there, and the combined stream 3 runs through a mixer into an atmospheric container. The system curve for this layout is stream 1 losses plus stream 3 losses, referenced to a junction pressure that stream 2 sets. Model the layout as three branches meeting at one node, fix the boundary conditions, and solve for that node pressure at each flow pair. The sections below build the model in order, and each ends with a check.

Branch layout and boundary conditions at the junction

Define p_J as the static pressure at the T-junction node shared by streams 1, 2 and 3. The known boundary is the atmospheric container at the end of stream 3. Stream 3 carries the sum of both inflows. Balance mass first, because the two liquids differ in density by up to a factor of 2:

m3 = m1 + m2
Q3 = m3 / rho_mix = (rho1*Q1 + rho2*Q2) / rho_mix
Q3 = Q1 + Q2 only if the liquids mix with additive volumes (assumption)

Work backwards from the end point. Pick a value of Q2 in the range 1.5 to 20 times Q1 and compute p_J from stream 3 alone:

p_J = p_end + dp_friction3(Q3, rho_mix, mu_mix) + rho_mix*g*dz3 (+ exit velocity head if the outlet is not submerged)

Adding Q2 to the flow in every segment downstream of the junction is correct. Segments upstream of the junction in stream 1 keep Q1 only.

1. Compute p_J for Q2 = 1.5, 5 and 20 times Q1 at fixed Q1. Expected reading: p_J rises monotonically with Q2 and equals p_end plus the static term when Q2 tends to 0.

Pressure as the common currency across different densities

Head is the height of a fluid column, H = Δp / (ρ g). For a single fluid ρ cancels, which is why single-medium system curves are drawn in metres. Across streams of different density, head is not additive and not comparable at the junction. The junction balances pressure, not head. Compute every branch in Pa or bar, then convert the pump requirement to head of the pumped fluid at the end: H_pump = dp_pump / (rho1 * g). For a gear pump, the datasheet usually states pressure directly, so the conversion may be unnecessary.

Density also enters the friction calculation twice. Frictional Δp scales with ρv², and the Reynolds number Re = ρvD/μ scales with ρ. A lower density lowers Re and, in the turbulent regime where the friction factor falls with Re, raises the friction factor. Recompute Re and the friction factor for each segment with the properties of the fluid in that segment.

1. Run the model with both streams set to water. Expected reading: the head-based sum and the pressure-based sum agree exactly. A difference after that swap means density is still being applied twice.

Mixture properties for the segments after the junction

Stream 3 carries a blend, so density, viscosity and temperature for every segment after the junction must describe the blend, not stream 1. With additive volumes:

rho_mix = (rho1*Q1 + rho2*Q2) / (Q1 + Q2)

The stream 2 weight is Q2/(Q1+Q2). That is 60 % at Q2 = 1.5·Q1 and about 95 % at Q2 = 20·Q1. At the top of the range the blend is nearly stream 2 fluid, and stream 1 properties no longer matter downstream. Take viscosity of the blend from laboratory data or a mixing correlation suited to the two fluids, and take temperature from an energy balance across the junction, because both feed Re. The mixer adds its own loss, evaluated at Q3.

1. Recompute rho_mix, mu_mix and Re for each (Q1, Q2) pair instead of once. Expected reading: rho_mix lies between rho1 and rho2 and approaches rho2 as Q2/Q1 grows.

Stream 3 loss at Q1 + Q2 and the marginal-contribution test

Stream 3 loss grows with the square of total flow in the turbulent regime (constant friction factor, assumed for this scaling only) and linearly in the laminar regime. Take a stream 3 pipe sized for stream 1 alone, same density, constant friction factor:

Q2 / Q1	Q3 / Q1	Loss vs. stream 1 only (Q3/Q1)²	Loss vs. Q2 alone (Q3/Q2)²
1.5	2.5	6.25	2.78
5	6	36	1.44
20	21	441	1.10

The last column is the marginal test: compare stream 3 loss at Q1+Q2 with stream 3 loss at Q2 alone. It decides how much modelling stream 2 needs.

Test result	Interpretation	Model
Loss at Q1+Q2 close to loss at Q2 alone	Stream 3 pipe is large relative to stream 1; stream 1 barely moves the header pressure	Model stream 1 only; treat p_J as an assumed header pressure and sweep it across the Q2 range
Loss at Q1+Q2 clearly above loss at Q2 alone	Stream 1 raises the backpressure that stream 2 sees	Model stream 2 and its pressure source explicitly (next section)

1. Evaluate the test at Q2 = 1.5, 5 and 20 times Q1 with your real pipe sizes. Expected reading: the ratio falls toward 1 as Q2 grows. A ratio that stays well above 1 at 20× points to a stream 3 pipe too small for the header assumption.

Stream 2 as a throttled utility supply

A mains stream at high pressure whose flow is set by a control valve behaves as a pressure source in series with a valve. Its head-versus-flow curve is the valve pressure drop, not a pump curve. Measure the mains pressure p_s2 at the tapping point at minimum and maximum site demand, then compute:

dp_valve(Q2) = p_s2 - p_J(Q2) - dp_line2(Q2) - rho2*g*dz2

Valve sizing relation (dp in bar, Q in m3/h, rho_rel = density relative to water):
dp = rho_rel * (Q / Kv)^2

For water, rho_rel = 1. This gives the required Kv at each Q2, and the valve's flow-versus-opening characteristic shows whether it controls across the 1.5× to 20× range. If stream 2 is instead a variable-speed pump, replace p_s2 and the valve with that pump's curve at its speed, together with its own suction conditions.

1. Compute dp_valve at Q2 = 20·Q1, where p_J is highest. Expected reading: positive, with margin. A negative value means the mains cannot deliver that flow against the junction pressure, and the flow target is unreachable without a booster.

Solving the node and handling the check valves

The pressure at the T is one value, p_J, so equal pressure at the junction is the result of the solution, not an input. What needs checking is the direction of flow in each branch. Check valves in streams 1 and 2 upstream of the T block reverse flow and add cracking pressure and loss to each branch:

Stream 1 side:  p_dis = p_J + dp_friction1(Q1) + rho1*g*dz1 + dp_check1(Q1)
                dp_pump = p_dis - p_suction
Stream 2 side:  p_s2 - dp_valve - dp_line2 - dp_check2 - rho2*g*dz2 = p_J

Common failure signatures in a spreadsheet model of this layout:

Symptom in the model or plant	Cause
Pump pressure does not change with Q2	Stream 3 loss evaluated at Q1 only
Curve steps when Q2 changes	Mixture properties held at stream 1 values
Pressure mismatch of about the density ratio	Heads of different fluids added across the junction
Stream 2 flow collapses when stream 1 starts	Required p_J exceeds what the mains supplies at that Q2
Stream 1 check valve chatters or stays shut	p_dis does not exceed p_J plus cracking pressure at low speed

1. For each Q2 point, confirm that p_dis - p_J exceeds the stream 1 check valve cracking pressure plus friction, and that p_s2 minus stream 2 losses exceeds p_J. Expected reading: both differences positive, so both branches carry forward flow.

Gear pump operating point set by speed

A gear pump is positive displacement. Its flow is displacement times rpm minus internal slip, and the pump curve is nearly vertical on a flow-versus-pressure plot. The system curve therefore determines discharge pressure, not flow. Skip the polynomial fit and the curve intersection: set Q1 from rpm and evaluate the required p_dis directly at each Q2 from the previous section. Slip rises with differential pressure and falls with viscosity, so read Q1 from the manufacturer's curve at the computed dp_pump and iterate once if the shift matters.

The Q2 sweep moves p_J, and with it the pump's discharge pressure, far more than it moves Q1. Take the highest p_dis from the sweep and compare it with the pump's rated differential pressure, the relief valve setting, and the drive torque at that pressure. Read all three values from the pump datasheet and the installed relief valve tag.

1. Tabulate dp_pump against Q2/Q1 at fixed rpm. Expected reading: the maximum stays below the relief valve set pressure with margin. A maximum at or above it means the relief valve lifts at high stream 2 flow.

End-to-end verification of the coupled model

1. Mass balance: m1 + m2 - m3 at every sweep point. Expected reading: zero within rounding.
2. Q2 = 0 limit: the model collapses to the original series sum with stream 1 properties. For the worked example of H1 = 3 m and H3 = 4 m at 1 m³/h, the point is 7 m.
3. Large-Q2 limit: p_J approaches the value computed from Q2 alone through stream 3. Expected reading: within the ratio from the marginal-contribution table.
4. Units: every branch in Pa or bar until the final conversion. Expected reading: the same-fluid test from the density section still matches.
5. Branch signs: forward flow in streams 1, 2 and 3 at every point, with check valve cracking included.
6. Field comparison: fit pressure transmitters at the pump discharge, at the junction, and upstream of the stream 2 valve. Compare readings at three Q2 points against the model. Expected reading: residuals explained by roughness and fitting K-value uncertainty. A residual that changes sign between Q2 points indicates wrong mixture density or viscosity in stream 3.

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