riaology
Unit 03 · Lesson 09 of 09

How do we model a pipe system?

The operating point belongs to the pump and the pipe together.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • Build the system curve.
  • Use a pump characteristic.
  • Solve the intersection.
Before you touch the controls

If pipe resistance doubles, can the same pump keep its original flow?

An experiment you can change

Find where a pump meets a piping system

Predict the direction of change, move one control, then explain the response using the equation.

Water passing through a polished pump at 7.071067811865475 liters per minute, driven by shaft workShaft work inInletOutletFollow the water. Account for energy.Find where a pump meets a piping system. Flow (L/s) versus Head (m). Selected result 7.0711 L/s. The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.0012.552510Head (m)Flow (L/s)

Solid red: system head · Dashed blue: pump head · Dot: operating point

q∗=20−50.1+0.2=7.0711 L/s,h∗=15 mq_* =\sqrt{\frac{20-5}{0.1+0.2}}=7.0711\,\mathrm{L/s},\quad h_*=15\,\mathrm m

The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.

Model assumptions and units

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}
01

Build the system curve.

Between reservoir surfaces the required pump head is static lift plus friction and fitting losses. For a teaching model with nearly fixed turbulent loss coefficients, the variable loss scales approximately with flow squared.

hsystem=Δz+kq2h_{system}=\Delta z+kq^2
02

Use a pump characteristic.

A pump at fixed speed supplies a head that varies with flow. Here an illustrative curve falls as flow rises. Its coefficients depend on the chosen flow units and do not represent a specific commercial pump.

hpump=20−0.1q2h_{pump}=20-0.1q^2
03

Solve the intersection.

Steady operation requires pump head to match the system requirement. Equate the two curves and solve for nonnegative flow. If available shutoff head is below static lift, the assumed forward-flow solution does not exist.

20−0.1q2=Δz+kq2⟹q∗=20−Δz0.1+k20-0.1q^2=\Delta z+kq^2\quad\Longrightarrow\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}
04

Verify the complete account.

At the intersection, pump energy addition covers elevation and losses. Continuity converts volume flow to speed in each pipe section; Reynolds number then checks the assumed loss model. Actual design also needs pump efficiency, allowable pressure, and operating constraints.

A worked example

Find the operating flow.

  1. For Δz=5 m\Delta z=5\,\mathrm m and k=0.2 m/(L/s)2k=0.2\,\mathrm{m/(L/s)^2}, solve 20−0.1q2=5+0.2q220-0.1q^2=5+0.2q^2.
  2. Then q2=15/0.3=50q^2=15/0.3=50 and q=7.0711 L/sq=7.0711\,\mathrm{L/s}.
  3. Both curves give h=15 mh=15\,\mathrm m: 5 m static lift plus 10 m loss. Increasing k shifts the system curve upward and lowers flow.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Find an intersection

For hpump=20−0.1hpump = 20 - 0.1q² and hsystem=5+0.2hsystem = 5 + 0.2q², with q in L/s, find operating flow.

Hint

Equate heads and solve for nonnegative q.

Explore the visual solution
Water passing through a polished pump at 7.071067811865475 liters per minute, driven by shaft workShaft work inInletOutletFollow the water. Account for energy.Find where a pump meets a piping system. Flow (L/s) versus Head (m). Selected result 7.0711 L/s. The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.0012.552510Head (m)Flow (L/s)

Solid red: system head · Dashed blue: pump head · Dot: operating point

q∗=20−50.1+0.2=7.0711 L/s,h∗=15 mq_* =\sqrt{\frac{20-5}{0.1+0.2}}=7.0711\,\mathrm{L/s},\quad h_*=15\,\mathrm m

The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.

Model assumptions and units

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}

Controls explore variations. The steps below retain the original question’s values.

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}
2 · Compare

Raise the destination

Raise static lift to 15 m while k stays 0.2. Find operating flow and explain the shift.

Hint

The system curve shifts upward by 10 m.

Explore the visual solution
Water passing through a polished pump at 4.0824829046386295 liters per minute, driven by shaft workShaft work inInletOutletFollow the water. Account for energy.Find where a pump meets a piping system. Flow (L/s) versus Head (m). Selected result 4.0825 L/s. The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.0017.553510Head (m)Flow (L/s)

Solid red: system head · Dashed blue: pump head · Dot: operating point

q∗=20−150.1+0.2=4.0825 L/s,h∗=18.333 mq_* =\sqrt{\frac{20-15}{0.1+0.2}}=4.0825\,\mathrm{L/s},\quad h_*=18.333\,\mathrm m

The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.

Model assumptions and units

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}

Controls explore variations. The steps below retain the original question’s values.

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}
3 · Challenge the model

Reduce piping resistance

Keep static lift 5 m and reduce k to 0.05. Find operating flow. Explain why pump head need not increase.

Hint

The system curve becomes flatter and its intersection moves right.

Explore the visual solution
Water passing through a polished pump at 10 liters per minute, driven by shaft workShaft work inInletOutletFollow the water. Account for energy.Find where a pump meets a piping system. Flow (L/s) versus Head (m). Selected result 10 L/s. The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.001052010Head (m)Flow (L/s)

Solid red: system head · Dashed blue: pump head · Dot: operating point

q∗=20−50.1+0.05=10 L/s,h∗=10 mq_* =\sqrt{\frac{20-5}{0.1+0.05}}=10\,\mathrm{L/s},\quad h_*=10\,\mathrm m

The operating flow is the intersection, not an independently chosen pump rating. Raising static lift or resistance lowers the operating flow.

Model assumptions and units

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}

Controls explore variations. The steps below retain the original question’s values.

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.

hpump=20−0.1q2,hsystem=Δz+kq2,q∗=20−Δz0.1+kh_{pump}=20-0.1q^2,\quad h_{system}=\Delta z+kq^2,\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}
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