Conservation, inputs, outputs and simple mass or energy balances

A balance starts by saying what is inside

Choose a boundary and one time interval. For ordinary mass accounting with all streams included, final mass = initial mass + mass in − mass out. The boundary may surround the tank and its contents, while the liquid crosses at inlet and outlet. A label such as “tank system” is incomplete until these crossings and any unmeasured routes are clear.

The diagram uses amounts over ten minutes, measured in kg. Initial contents are 20 kg, inflow totals 8 kg and outflow totals 5 kg. Final contents are 23 kg. If someone assumes input must always equal output, they will miss accumulation. Equality of input and output is a special case with no net stored-mass change, not the starting rule for every interval.

A rate carries a time denominator

A mass rate tells how much mass crosses per unit time, such as kg/min. The average rates in this example are 8 ÷ 10 = 0.8 kg/min in and 5 ÷ 10 = 0.5 kg/min out. Their difference, 0.3 kg/min, is the average accumulation rate. Multiplying by ten minutes recovers the 3 kg stored increase.

Keep mass rate distinct from volume rate. A volume in litres becomes a mass only with appropriate density information; different fluids and conditions can have different densities. Do not add kg to kg/min or combine totals covering different intervals. For a time-varying process, shorter interval accounts can show changes that one long average hides.

Use the same reasoning for energy

For an energy account, final stored energy − initial stored energy = energy entering − energy leaving, with every relevant transfer included consistently. Energy can cross by work, by heat transfer or with moving matter. Our simple device example below has no mass crossing; it separates useful mechanical work, heat leaving and energy remaining in the device.

A device warming up can have supplied energy larger than the energy it has already passed onward. A cooling device can release stored energy as well as its current input. Calling either difference an efficiency without accounting for storage can mislead. In steady operation, the stored amount does not keep changing, so the input/output power balance becomes simpler.

Reconcile a device while it warms up

Suppose an enclosed converter receives 500 J during a specified interval. It delivers 300 J mechanical work to the intended output and transfers 100 J as heat to its surroundings. No mass crosses its boundary, and no other energy transfers are included in the constructed record.

Energy accumulation is 500 − 300 − 100 = 100 J. The account is 500 J entering = 300 J useful output + 100 J heat leaving + 100 J stored increase. The last term is not another output crossing the boundary.

If an independent estimate instead reported only 60 J storage increase, there would be a 40 J discrepancy to investigate in measurements, boundary choices or missing transfers. Do not repair the record by relabeling the discrepancy as “energy destroyed.” Conservation supplies a consistency check; it does not identify the missing cause by itself.

Try the reasoning

1. Must the tank’s 8 kg inflow equal its 5 kg outflow?

   Show the answer and reasoning

   Not during an accumulating interval. The 3 kg difference increases the contents from 20 to 23 kg if all streams are accounted for.

2. Can 5 kg be added directly to 0.8 kg/min?

   Show the answer and reasoning

   No. One is an amount and the other a rate. Choose the interval and convert the rate to a corresponding amount first.

3. In the energy example, is the 100 J stored increase the same as heat leaving?

   Show the answer and reasoning

   No. Storage remains within the boundary; heat leaving crosses it. Counting one quantity in both places would double-count energy.

What to carry forward

Declare the boundary and interval, use consistent units, and account for accumulation before equating inputs with outputs.