Introduction
Thermodynamics mass‑balance and enthalpy calculations are the backbone of any chemical‑process analysis. Still, this article presents a step‑by‑step guide to solving typical mass‑balance/enthalpy practice problems, explains the underlying concepts, and provides fully worked solutions for several representative examples. In practice, whether you are designing a reactor, sizing a heat exchanger, or troubleshooting a plant upset, the ability to balance mass and apply the first law of thermodynamics to determine enthalpy changes is essential. By the end, you will be comfortable setting up simultaneous mass‑ and energy‑balance equations, selecting the right enthalpy data, and interpreting the results with confidence That's the whole idea..
Easier said than done, but still worth knowing.
1. Core Concepts
1.1 Mass Balance Fundamentals
A mass balance follows the principle of conservation of mass:
[ \sum \dot{m}{\text{in}} - \sum \dot{m}{\text{out}} = \frac{dM_{\text{system}}}{dt} ]
- Steady‑state: (dM_{\text{system}}/dt = 0) → (\sum \dot{m}{\text{in}} = \sum \dot{m}{\text{out}})
- Unsteady‑state: accumulation term must be retained.
Components can be treated individually (component balance) or as a whole (total balance). For reacting systems, the stoichiometric coefficients become part of the component balance That's the part that actually makes a difference. That's the whole idea..
1.2 Enthalpy and the First Law
The first law for a control volume (steady‑state, no shaft work) reads
[ \sum \dot{m}{\text{in}} h{\text{in}} + \dot{Q} = \sum \dot{m}{\text{out}} h{\text{out}} ]
where (h) is the specific enthalpy (kJ kg⁻¹) and (\dot{Q}) is the net heat transfer rate (kW) That alone is useful..
Key points:
- Reference state: Enthalpy values are relative; use the same reference for all streams.
- Ideal gas approximation: (h(T) = C_p(T) , (T - T_{\text{ref}})) if (C_p) is constant, otherwise integrate (C_p(T)).
- Sensible vs. latent: For phase changes, include latent heat (e.g., vaporization enthalpy).
1.3 Linking Mass and Energy Balances
When a reaction occurs, the heat of reaction ((\Delta H_{rxn})) couples the two balances:
[ \dot{Q}{\text{rxn}} = -\Delta H{rxn} , \dot{n}_{\text{reactants}} ]
If the process is adiabatic, (\dot{Q}=0) and the temperature rise is obtained solely from the reaction enthalpy and the sensible heat capacities of the products.
2. Typical Practice Problem Types
| Problem Type | Typical Data | Goal |
|---|---|---|
| Simple non‑reactive mixer | Inlet mass flow rates, temperatures, (C_p) values | Find outlet temperature |
| Adiabatic reactor (single reaction) | Feed composition, conversion, (\Delta H_{rxn}), (C_p) of species | Determine outlet temperature |
| Heat‑exchanger design | Two streams, heat‑transfer coefficient, area | Compute required area or outlet temperatures |
| Batch reactor with accumulation | Initial inventory, reaction rate, heat loss | Predict temperature vs. time |
Below we solve three representative problems covering these categories.
3. Problem 1 – Non‑Reactive Mixing
Statement
Two water streams mix in a perfectly insulated tank.
- Stream A: (\dot{m}_A = 2.0\ \text{kg s}^{-1}), (T_A = 20^{\circ}\text{C})
- Stream B: (\dot{m}_B = 3.
Specific heat of water is constant at (C_p = 4.Still, 18\ \text{kJ kg}^{-1}\text{K}^{-1}). Determine the outlet temperature (T_{out}).
Solution
- Write the total mass balance (steady‑state):
[ \dot{m}_{out} = \dot{m}_A + \dot{m}_B = 5.0\ \text{kg s}^{-1} ]
- Apply the energy balance (no heat loss, no work):
[ \dot{m}A h_A + \dot{m}B h_B = \dot{m}{out} h{out} ]
Because (h = C_p T) (reference cancels), substitute:
[ \dot{m}_A C_p T_A + \dot{m}_B C_p T_B = (\dot{m}_A+\dot{m}B) C_p T{out} ]
- Cancel (C_p) and solve for (T_{out}):
[ T_{out}= \frac{\dot{m}_A T_A + \dot{m}_B T_B}{\dot{m}_A+\dot{m}_B} = \frac{2.0 \times 20 + 3.0 \times 80}{5.
Result: The mixed stream leaves at 56 °C.
Take‑away: When heat capacities are equal, the outlet temperature is the mass‑flow‑weighted average of inlet temperatures.
4. Problem 2 – Adiabatic Plug‑Flow Reactor (PFR)
Statement
A gaseous feed of methane (CH₄) and oxygen (O₂) undergoes complete combustion in an adiabatic PFR. Feed conditions:
- Total molar flow (\dot{n}_{\text{feed}} = 10\ \text{kmol h}^{-1})
- Composition: 20 % CH₄, 80 % O₂ (by moles)
- Inlet temperature (T_{in}= 298\ \text{K})
- Pressure constant, ideal‑gas behavior
Reaction:
[ \text{CH}_4 + 2\ \text{O}_2 \rightarrow \text{CO}_2 + 2\ \text{H}_2\text{O} ]
Standard heat of combustion (\Delta H_{rxn}^{\circ} = -802\ \text{kJ mol}^{-1}) (at 298 K). Assume constant (C_p) values:
- (C_{p,CH_4}=35\ \text{J mol}^{-1}\text{K}^{-1})
- (C_{p,O_2}=29\ \text{J mol}^{-1}\text{K}^{-1})
- (C_{p,CO_2}=37\ \text{J mol}^{-1}\text{K}^{-1})
- (C_{p,H_2O(g)}=34\ \text{J mol}^{-1}\text{K}^{-1})
Determine the outlet temperature (T_{out}).
Solution
- Mole balance – find conversion.
Complete combustion implies all CH₄ reacts.
Moles of CH₄ in feed:
[ \dot{n}_{CH_4}=0.20 \times 10 = 2.0\ \text{kmol h}^{-1} ]
Thus, (\xi = 2.0\ \text{kmol h}^{-1}) (extent of reaction) Turns out it matters..
- Product flow rates:
- CO₂: (\dot{n}_{CO_2}= \xi = 2.0\ \text{kmol h}^{-1})
- H₂O: (\dot{n}_{H_2O}=2\xi = 4.0\ \text{kmol h}^{-1})
- O₂ left: (\dot{n}_{O_2}=0.80 \times 10 - 2\xi = 8.0 - 4.0 = 4.0\ \text{kmol h}^{-1})
Total outlet molar flow:
[ \dot{n}{out}= \dot{n}{CO_2}+\dot{n}{H_2O}+\dot{n}{O_2}=2+4+4=10\ \text{kmol h}^{-1} ]
(Consistent with constant pressure and ideal gas, total moles unchanged.)
- Energy balance (adiabatic, no shaft work, no heat loss):
[ \sum \dot{n}{in} C{p,i} T_{in} + \dot{Q}{rxn}= \sum \dot{n}{out} C_{p,i} T_{out} ]
[ \dot{Q}{rxn}= -\Delta H{rxn}^{\circ} , \xi ]
Convert (\Delta H_{rxn}^{\circ}) to kJ per kmol: (-802\ \text{kJ mol}^{-1} = -802,000\ \text{kJ kmol}^{-1}) Worth keeping that in mind..
[ \dot{Q}_{rxn}= -(-802,000)\times 2.0 = 1.604\times10^{6}\ \text{kJ h}^{-1} ]
- Compute the sensible‑heat term for inlet:
[ \sum \dot{n}{in} C{p,i} = (2.0)(35)+(8.0)(29)=70+232=302\ \text{J K}^{-1}\text{h}^{-1} ]
Convert to kJ: (302\ \text{J K}^{-1}\text{h}^{-1}=0.302\ \text{kJ K}^{-1}\text{h}^{-1}) Less friction, more output..
Sensible heat in:
[ \dot{H}{in}=0.302\ T{in}=0.302\times 298=90.0\ \text{kJ h}^{-1} ]
- Sensible‑heat term for outlet (unknown (T_{out})):
[ \sum \dot{n}{out} C{p,i}= (2)(37)+(4)(34)+(4)(29)=74+136+116=326\ \text{J K}^{-1}\text{h}^{-1}=0.326\ \text{kJ K}^{-1}\text{h}^{-1} ]
Thus
[ \dot{H}{out}=0.326\ T{out} ]
- Set up the energy balance:
[ \dot{H}{in} + \dot{Q}{rxn}= \dot{H}_{out} ]
[ 90.0\ \text{kJ h}^{-1}+1.604\times10^{6}\ \text{kJ h}^{-1}=0.326\ T_{out} ]
[ 0.326\ T_{out}=1.604\times10^{6}+90 \approx 1.60409\times10^{6} ]
[ T_{out}= \frac{1.60409\times10^{6}}{0.326}=4.92\times10^{6}\ \text{K} ]
The absurdly high temperature signals a unit‑consistency error. The mistake stems from mixing per‑hour flow rates (kmol h⁻¹) with heat capacities in J mol⁻¹ K⁻¹ without converting the time basis That alone is useful..
Correct approach: Cancel the hour unit by using heat per mole rather than per hour Easy to understand, harder to ignore..
- Compute molar enthalpy for inlet:
[ \bar{h}{in}= \frac{ \sum \dot{n}{in} C_{p,i}}{\dot{n}{total,in}} T{in} ]
Total inlet molar flow = 10 kmol h⁻¹. Average (C_p) inlet:
[ \bar{C}_{p,in}= \frac{302\ \text{J K}^{-1}\text{h}^{-1}}{10\ \text{kmol h}^{-1}} =30.2\ \text{J mol}^{-1}\text{K}^{-1} ]
Thus (\bar{h}_{in}=30.2 \times 298 = 9.0\ \text{kJ mol}^{-1}) Simple, but easy to overlook..
- Similarly, average (C_p) outlet:
[ \bar{C}_{p,out}= \frac{326\ \text{J K}^{-1}\text{h}^{-1}}{10\ \text{kmol h}^{-1}} =32.6\ \text{J mol}^{-1}\text{K}^{-1} ]
Now write the per‑mole energy balance (adiabatic):
[ \bar{h}{in}+ \frac{-\Delta H{rxn}^{\circ}\xi}{\dot{n}{total}} = \bar{C}{p,out}, T_{out} ]
[ \frac{-\Delta H_{rxn}^{\circ}\xi}{\dot{n}_{total}} = \frac{802\ \text{kJ mol}^{-1}\times 2.0\ \text{kmol h}^{-1}}{10\ \text{kmol h}^{-1}} = 160.4\ \text{kJ mol}^{-1} ]
Hence
[ 9.0\ \text{kJ mol}^{-1}+160.4\ \text{kJ mol}^{-1}= \bar{C}{p,out}, T{out} ]
[ 169.4\ \text{kJ mol}^{-1}=32.6\ \text{J mol}^{-1}\text{K}^{-1}, T_{out} ]
Convert (32.6\ \text{J}) to kJ: (0.0326\ \text{kJ mol}^{-1}\text{K}^{-1}).
[ T_{out}= \frac{169.4}{0.0326}=5,197\ \text{K} ]
Even after correction, the temperature is unrealistically high because constant‑(C_p) assumption over such a large temperature range is invalid, and the reaction releases a huge amount of heat. Because of that, in practice, the reactor would be cooled, and the temperature would be limited by material constraints. The exercise, however, illustrates the algebraic steps And it works..
Key lesson: Always keep units consistent and remember that constant‑(C_p) models are only valid for modest temperature excursions.
5. Problem 3 – Heat‑Exchanger Temperature‑Effectiveness
Statement
Two water streams exchange heat in a counter‑current shell‑and‑tube exchanger Worth keeping that in mind. But it adds up..
- Hot stream: (\dot{m}h = 1.5\ \text{kg s}^{-1}), (T{h,in}=120^{\circ}\text{C}), (C_{p}=4.18\ \text{kJ kg}^{-1}\text{K}^{-1})
- Cold stream: (\dot{m}c = 2.0\ \text{kg s}^{-1}), (T{c,in}=20^{\circ}\text{C}), same (C_{p})
Overall heat‑transfer coefficient (U = 500\ \text{W m}^{-2}\text{K}^{-1}). Required heat‑transfer area (A = 15\ \text{m}^2). Determine the outlet temperatures (T_{h,out}) and (T_{c,out}) Worth knowing..
Solution
- Compute heat‑capacity rates:
[ C_h = \dot{m}_h C_p = 1.5 \times 4.18 = 6.
[ C_c = \dot{m}_c C_p = 2.0 \times 4.18 = 8 Simple, but easy to overlook..
- Identify the minimum and maximum heat‑capacity rates:
[ C_{min}=6.27\ \text{kW K}^{-1},\qquad C_{max}=8.36\ \text{kW K}^{-1} ]
Capacity ratio (C_r = C_{min}/C_{max}=0.75).
- Maximum possible heat transfer (NTU method):
[ q_{max}= C_{min},(T_{h,in}-T_{c,in}) = 6.27,(120-20)=6.27\times100=627\ \text{kW} ]
- Calculate NTU (Number of Transfer Units):
[ \text{NTU}= \frac{U A}{C_{min}} = \frac{500\ \text{W m}^{-2}\text{K}^{-1}\times15\ \text{m}^2}{6.27\ \text{kW K}^{-1}} = \frac{7,500\ \text{W K}^{-1}}{6,270\ \text{W K}^{-1}} \approx 1.20 ]
- Effectiveness for counter‑current exchanger (C_r ≠ 1):
[ \varepsilon = \frac{1 - \exp[-\text{NTU}(1-C_r)]}{1 - C_r\exp[-\text{NTU}(1-C_r)]} ]
Compute exponent:
[ \text{NTU}(1-C_r)=1.20\times(1-0.75)=1.20\times0.25=0.30 ]
[ \exp(-0.30)=0.741 ]
Now
[ \varepsilon = \frac{1-0.741}{1-0.75\times0.741}= \frac{0.259}{1-0.556}= \frac{0.259}{0.444}=0.583 ]
- Actual heat transfer:
[ q = \varepsilon , q_{max}=0.583 \times 627 = 365\ \text{kW} ]
- Determine outlet temperatures:
Hot side:
[ q = C_h (T_{h,in} - T_{h,out}) ;\Rightarrow; T_{h,out}= T_{h,in} - \frac{q}{C_h}=120 - \frac{365}{6.Day to day, 27}=120 - 58. 2 \approx 61 Not complicated — just consistent. Which is the point..
Cold side:
[ q = C_c (T_{c,out} - T_{c,in}) ;\Rightarrow; T_{c,out}= T_{c,in} + \frac{q}{C_c}=20 + \frac{365}{8.36}=20 + 43.6 \approx 63 Still holds up..
Result: The hot water leaves at ≈ 62 °C, the cold water at ≈ 64 °C. The small temperature cross‑over (hot outlet slightly cooler than cold outlet) confirms the counter‑current operation.
6. Frequently Asked Questions
6.1 Why do we often assume constant (C_p) in practice problems?
Constant (C_p) simplifies the integration of (h(T)=\int C_p,dT). For small temperature ranges (≤ 50 K) the error is typically < 2 %. Also, when temperature swings are large, use temperature‑dependent correlations (e. g., NASA polynomials) or tabulated data.
6.2 How do I decide whether to use a molar or mass basis?
- Molar basis is natural for gas‑phase reactions, equilibrium calculations, and when using standard enthalpies of formation (kJ mol⁻¹).
- Mass basis is convenient for liquid‑phase streams, especially when densities and mass flow rates are given.
Always keep the chosen basis consistent throughout a single problem.
6.3 What is the difference between adiabatic and isothermal reactors?
- Adiabatic: No heat exchange with surroundings ((\dot{Q}=0)). Temperature changes are driven solely by reaction enthalpy and sensible heating/cooling of the mixture.
- Isothermal: Temperature is held constant, typically by external heating/cooling. The required (\dot{Q}) is calculated from the energy balance.
6.4 Can I ignore the pressure‑volume work term ((P\Delta V))?
For incompressible liquids, (P\Delta V) is negligible. For gases, the first law for an open system already incorporates flow work via the enthalpy definition ((h = u + Pv)). Which means, you do not need a separate (P\Delta V) term when using enthalpy.
6.5 How do I handle phase changes in a mass‑balance/enthalpy problem?
Treat the phase change as a reaction with a known latent heat ((\Delta H_{vap}) or (\Delta H_{fus})). Include the latent term in the energy balance:
[ q_{phase}= \dot{m}{phase}, \Delta H{phase} ]
Make sure the mass balance accounts for the disappearance of the liquid phase and the appearance of the vapor phase (or vice versa).
7. Tips for Mastering Mass‑Balance & Enthalpy Problems
- List all known data in a table (flow rates, compositions, temperatures, (C_p), (\Delta H_{rxn})).
- Choose a basis (per hour, per mole, per kilogram) and stick to it.
- Write separate balances for each component before summing to a total balance.
- Identify the controlling equation: is the problem energy‑limited (adiabatic) or mass‑limited (conversion)?
- Check units at every step; convert kJ ↔ J, kmol ↔ mol, etc.
- Validate the result: temperatures should be within realistic limits, and mass flow out should equal mass flow in for steady‑state non‑accumulating systems.
- Use the NTU‑effectiveness method for heat exchangers; it avoids iterative calculations.
8. Conclusion
Mass‑balance and enthalpy calculations are fundamental tools that turn raw process data into actionable engineering insight. The worked examples above demonstrate the logical flow from problem statement to solution, while the FAQ and tip sections address common stumbling blocks. By methodically applying the conservation laws, selecting appropriate thermodynamic data, and maintaining strict unit consistency, you can solve a wide variety of practice problems—from simple mixers to adiabatic reactors and heat exchangers. Master these techniques, and you’ll be equipped to tackle real‑world process simulations, design tasks, and optimization studies with confidence.
