14.5 Mass and Energy Balances on Control Volumes

14.5 Mass and Energy Balances on Control Volumes

NCEES lists "Mass and energy balances" under Thermodynamics and Heat Transfer. It is the most transferable skill in the entire specification: the same bookkeeping structure appears in fluid mechanics continuity, chemistry stoichiometry, safety dilution ventilation, and thermodynamic device analysis. Learn the template once and it covers items in four different areas of the exam.

The General Balance Equation

For any conserved quantity over any control volume:

$\underbrace{\text{IN}}_{\text{across the boundary}} - \underbrace{\text{OUT}}_{\text{across the boundary}} + \underbrace{\text{GENERATION}}_{\text{inside}} - \underbrace{\text{CONSUMPTION}}_{\text{inside}} = \underbrace{\text{ACCUMULATION}}_{\text{rate of change of contents}}$

Two simplifications carry most exam problems:

Total mass is never generated or consumed — even in a chemical reaction. Only species masses and moles can change. Total mass in equals total mass out at steady state, always, which makes it the most reliable equation available for checking an answer.

Steady-State Mass Balance

$\sum \dot{m}_{\text{in}} = \sum \dot{m}_{\text{out}}$

For a single-inlet, single-outlet device, with $\dot{m} = \rho A V$:

$\rho_1A_1V_1 = \rho_2A_2V_2$

and for incompressible flow, where $\rho$ cancels:

$A_1V_1 = A_2V_2 = Q$

Worked Example: Mixing Two Streams

A mixing tee receives 4.0 kg/s of water at 15 °C and 1.5 kg/s at 80 °C. Find the outlet flow and temperature, taking $c_p = 4.18$ kJ/kg·K as constant.

Mass balance: $\dot{m}_3 = 4.0 + 1.5 = 5.5\ \text{kg/s}$

Energy balance (adiabatic, no work, negligible kinetic and potential terms): $\dot{m}_1h_1 + \dot{m}_2h_2 = \dot{m}_3h_3$

With constant $c_p$, enthalpy differences are proportional to temperature differences, so:

$T_3 = \frac{\dot{m}_1T_1 + \dot{m}_2T_2}{\dot{m}_1+\dot{m}_2} = \frac{4.0(15) + 1.5(80)}{5.5} = \frac{60 + 120}{5.5} = \frac{180}{5.5} = \boxed{32.7\ °\text{C}}$

The outlet is a mass-weighted average, not a simple average. The arithmetic mean of 15 and 80 is 47.5 °C — nearly 15 degrees too high — because the cold stream carries 2.7 times the mass. Any mixing problem whose answer sits at the midpoint of the two inlet temperatures has ignored the weighting.

The Steady-Flow Energy Equation

$\dot{Q} - \dot{W} = \dot{m}\left[(h_2-h_1) + \frac{V_2^2-V_1^2}{2} + g(z_2-z_1)\right]$

Sign convention: heat added to the system is positive; work done by the system is positive.

For most devices the kinetic and potential terms are negligible, and the equation collapses to a single useful form per device:

When kinetic energy actually matters. The $V^2/2$ term is in J/kg, while enthalpy changes are typically in kJ/kg. A velocity of 45 m/s contributes $45^2/2 = 1{,}013$ J/kg $= 1.0$ kJ/kg — about 0.3% of a typical turbine's 300 kJ/kg enthalpy drop, so it is rightly neglected. But in a nozzle the entire purpose is velocity change, so it is the dominant term and must never be dropped. Judge by the device, not by habit.

Worked Example: Counter-Flow Heat Exchanger Balance

Hot oil ($c_p = 2.10$ kJ/kg·K) enters at 140 °C and 2.4 kg/s, leaving at 65 °C. It heats water ($c_p = 4.18$ kJ/kg·K) entering at 20 °C at 3.0 kg/s. Find the heat duty and the water outlet temperature.

Heat released by the oil: $\dot{Q} = \dot{m}c_p\Delta T = 2.4(2.10)(140-65) = 2.4(2.10)(75) = \boxed{378\ \text{kW}}$

Energy balance on the water (adiabatic exchanger, so all of it is absorbed): $378 = 3.0(4.18)(T_{\text{out}} - 20) \;\Rightarrow\; T_{\text{out}} - 20 = \frac{378}{12.54} = 30.1$ $T_{\text{out}} = \boxed{50.1\ °\text{C}}$

Sanity check on the temperature approach: the water leaves at 50.1 °C against oil entering at 140 °C, and the water enters at 20 °C against oil leaving at 65 °C. Both approaches are positive (89.9 °C and 45 °C), so the second law is respected and the design is thermodynamically feasible. Any answer that puts a cold-stream outlet above the hot-stream inlet is impossible, and checking both ends of a counter-flow exchanger catches that error immediately.

Transient (Unsteady) Balances

When the contents change, the accumulation term survives:

$\frac{dm_{\text{cv}}}{dt} = \sum\dot{m}_{\text{in}} - \sum\dot{m}_{\text{out}}$

Worked Example: Tank Filling and Draining

A tank holds 800 kg of water. Inflow is 6.0 kg/s and outflow is 2.5 kg/s. How long until the tank holds 3{,}000 kg?

$\frac{dm}{dt} = 6.0 - 2.5 = 3.5\ \text{kg/s}$

$t = \frac{3{,}000 - 800}{3.5} = \frac{2{,}200}{3.5} = \boxed{629\ \text{s} = 10.5\ \text{minutes}}$

Worked Example: Draining with a Concentration Change

A 5{,}000 L tank of brine at 40 g/L is flushed with fresh water at 20 L/min while the well-mixed contents drain at 20 L/min, holding the volume constant. Find the concentration after 90 minutes.

Species mass balance with no generation, outflow at the (well-mixed) tank concentration $C$:

$V\frac{dC}{dt} = 0 - QC \;\Rightarrow\; \frac{dC}{dt} = -\frac{Q}{V}C$

$C(t) = C_0e^{-Qt/V} = 40\,e^{-(20/5{,}000)(90)} = 40\,e^{-0.36} = 40(0.6977) = \boxed{27.9\ \text{g/L}}$

The same exponential decay as radioactive half-life, Newton's law of cooling, and confined-space purge ventilation. The time constant is $V/Q = 250$ minutes, so 90 minutes is only 0.36 time constants — which is why the concentration has fallen just 30% despite flushing 1.8 tank volumes' worth of water through it. Flushing a well-mixed tank is exponential, not linear, and the intuitive "1.8 volumes means it's clean" reasoning is wrong.