riaology
Unit 02 · Lesson 07 of 09

What enters, leaves, and accumulates?

A steady inlet does not guarantee a steady tank.

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

What you’ll be able to do
  • Conservation starts with accumulation.
  • Convert a port measurement.
  • Steady flow constrains the sum.
Before you touch the controls

Which condition makes a tank steady: constant inflow, or equal mass inflow and outflow?

An experiment you can change

Watch a tank accumulate mass

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

Tank: 4 L/s in, 1 L/s out, 74 percent full4 L/s1 L/sFixed tank boundaryWatch a tank accumulate mass. Time (s) versus Liquid level (m). Selected result 0.74 m. Density cancels in the volume form; mass accumulation is still density times net volume flow.000.5501100Liquid level (m)Time (s)
h=0.5+(4−1)10−30.5(40)=0.74 mh=0.5+\frac{(4-1)10^{-3}}{0.5}(40)=0.74\,\mathrm m

Density cancels in the volume form; mass accumulation is still density times net volume flow.

Model assumptions and units

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t
01

Conservation starts with accumulation.

The rate of change of mass within a control volume equals mass entering minus mass leaving. Multiple ports are summed. A fixed control volume is a region, so its contents can change even when its geometry does not.

dmCVdt=∑inm˙−∑outm˙\frac{dm_{CV}}{dt}=\sum_{in}\dot m-\sum_{out}\dot m
02

Convert a port measurement.

With approximately uniform density and mean normal speed at a port, mass flow is density times volume flow. The mean speed is the area-averaged normal component, not necessarily the speed at every point.

m˙=ρV˙=ρAV\dot m=\rho\dot V=\rho AV
03

Steady flow constrains the sum.

At steady state no mass accumulates. In a one-inlet/one-outlet device the mass rates are equal; volume rates need not be equal if density changes. For incompressible flow with equal density, A1V1=A2V2A_1V_1=A_2V_2.

m˙1=m˙2,ρ1A1V1=ρ2A2V2\dot m_1=\dot m_2,\quad\rho_1A_1V_1=\rho_2A_2V_2
04

Unsteady tanks need integration.

Constant density and vertical walls give m=ρAhm=\rho Ah. With prescribed constant flows the height varies linearly until emptying or overflow changes the physical model. A gravity-fed outlet instead needs its depth-dependent discharge law.

h(t)=h0+V˙in−V˙outAth(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t
A worked example

Unequal volume flow in a steady device.

  1. Air enters at density 1.2 kg/m31.2\,\mathrm{kg/m^3} and volume rate 0.10 m3/s0.10\,\mathrm{m^3/s}.
  2. The steady mass rate is m˙=1.2(0.10)=0.12 kg/s\dot m=1.2(0.10)=0.12\,\mathrm{kg/s}.
  3. If exit density is 0.8 kg/m30.8\,\mathrm{kg/m^3}, exit volume rate is 0.12/0.8=0.15 m3/s0.12/0.8=0.15\,\mathrm{m^3/s}. Mass is conserved; volume rates differ.
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

A rising liquid level

A 0.5 m² tank starts 0.5 m deep. Inflow is 4 L/s and outflow 1 L/s. Find height after 40 s.

Hint

Convert liters to cubic meters before dividing net volume by area.

Explore the visual solution
Tank: 4 L/s in, 1 L/s out, 74 percent full4 L/s1 L/sFixed tank boundaryWatch a tank accumulate mass. Time (s) versus Liquid level (m). Selected result 0.74 m. Density cancels in the volume form; mass accumulation is still density times net volume flow.000.5501100Liquid level (m)Time (s)
h=0.5+(4−1)10−30.5(40)=0.74 mh=0.5+\frac{(4-1)10^{-3}}{0.5}(40)=0.74\,\mathrm m

Density cancels in the volume form; mass accumulation is still density times net volume flow.

Model assumptions and units

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t

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

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t
2 · Compare

Constant flows, steady level

The same tank has inlet and outlet both 3 L/s. Find level at 100 s and explain what is steady.

Hint

The accumulation term is zero when mass rates balance.

Explore the visual solution
Tank: 3 L/s in, 3 L/s out, 50 percent full3 L/s3 L/sFixed tank boundaryWatch a tank accumulate mass. Time (s) versus Liquid level (m). Selected result 0.5 m. Density cancels in the volume form; mass accumulation is still density times net volume flow.000.25500.5100Liquid level (m)Time (s)
h=0.5+(3−3)10−30.5(100)=0.5 mh=0.5+\frac{(3-3)10^{-3}}{0.5}(100)=0.5\,\mathrm m

Density cancels in the volume form; mass accumulation is still density times net volume flow.

Model assumptions and units

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t

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

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t
3 · Challenge the model

Find a model boundary

With inlet 6 L/s and no outlet, find the model state at 100 s. Use the graph to identify when the original equation stops applying.

Hint

Freeboard volume is 0.5 × 0.5 = 0.25 m³; divide by 0.006 m³/s.

Explore the visual solution
Tank: 6 L/s in, 0 L/s out, 100 percent full6 L/s0 L/sFixed tank boundaryWatch a tank accumulate mass. Time (s) versus Liquid level (m). Selected result 1 m. The unbounded balance has crossed empty/full. The tank display clips; a new outflow/empty-tank model is required beyond that event.000.5501100Liquid level (m)Time (s)
h=0.5+(6−0)10−30.5(100)=1.7 mh=0.5+\frac{(6-0)10^{-3}}{0.5}(100)=1.7\,\mathrm m

The unbounded balance has crossed empty/full. The tank display clips; a new outflow/empty-tank model is required beyond that event.

Model assumptions and units

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t

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

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

dmCVdt=∑m˙in−∑m˙out,h(t)=h0+V˙in−V˙outAt\frac{dm_{CV}}{dt}=\sum\dot m_{in}-\sum\dot m_{out},\quad h(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t
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