Energy balances on non-reactive processes
First law for closed and steady-flow open systems, enthalpy and reference states, simplifications for exchangers, mixers, turbines and valves, with steam-heater and turbine examples.
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Why it matters
Heaters, coolers, condensers, mixers, turbines, pumps and compressors are sized from energy balances. The steam a heater consumes, the cooling water a condenser needs and the power a turbine delivers all come from one equation – the first law written for the system – applied with the right reference states and enthalpy data.
Key ideas
Forms of energy. Internal energy U (molecular motion and interactions), kinetic energy E_k = m·u²/2 and potential energy E_p = m·g·z. Energy crosses a boundary as heat Q (due to a temperature difference) and work W. Sign convention used here: Q positive into the system, W positive when done by the system (shaft work W_s).
Closed systems (batch). No mass crosses the boundary: ΔU + ΔE_k + ΔE_p = Q − W. In most process vessels ΔE_k and ΔE_p are negligible, so Q = ΔU (constant volume) or Q = ΔH (constant pressure, only expansion work).
Open systems at steady state. For flowing streams, the work of pushing fluid in and out (flow work P·V) is combined with U into enthalpy H = U + P·V. The steady-flow energy balance is ΔḢ + ΔĖ_k + ΔĖ_p = Q̇ − Ẇ_s, where Δ means outlet minus inlet summed over all streams.
Simplifications.
- Heat exchangers, mixers, reactors without moving parts: W_s = 0; ΔE_k and ΔE_p usually negligible → Q = ΔH.
- Adiabatic units (well-insulated): Q = 0.
- Turbines, compressors, pumps: W_s ≠ 0; often Q ≈ 0. ΔE_k matters only for large velocity changes (nozzles, high-speed vapour).
- Valves and throttling: Q = 0, W_s = 0 → ΔH = 0 (isenthalpic).
Enthalpy references. Only differences in H matter, so pick a convenient reference state for each species (e.g. liquid water at the triple point, as in steam tables, or the inlet condition). Use steam tables for water, enthalpy tables or charts where available, and otherwise C_p and latent heats (previous topics) along a path from the reference.
Heat losses. Real equipment loses heat to the surroundings; this is a negative Q in the balance or an efficiency factor on the useful duty.
Mechanical energy balance. For incompressible fluids with no heat effects (pumps, piping), the balance reduces to the Bernoulli-type equation with friction losses – covered in fluid mechanics.
Formulas
ΔU + ΔE_k + ΔE_p = Q − W– closed system.ΔḢ + ΔĖ_k + ΔĖ_p = Q̇ − Ẇ_s– open system, steady state; ΔḢ = Σ ṁ_out·Ĥ_out − Σ ṁ_in·Ĥ_in (kW).Ė_k = ṁ·u² / 2– W, with ṁ in kg/s and u in m/s;Ė_p = ṁ·g·z.H = U + P·VQ̇ = ṁ·c_p·(T_out − T_in)– sensible heating, constant c_p.Q̇ = ṁ_steam·λ– condensing steam, λ latent heat at the steam pressure (from steam tables).Σ ṁ_i·c_p,i·(T_i − T_mix) = 0– adiabatic mixing of liquids with constant c_p.ΔḢ = 0– throttling valve.
Worked examples
Example 1 (standard). 5 kg/s of water is heated from 25 °C to 85 °C (c_p = 4.18 kJ/(kg·K)) by saturated steam at 4 bar that condenses and leaves as saturated liquid (λ = 2133.4 kJ/kg from steam tables). Heat losses are negligible. Find the duty and the steam flow.
- Energy balance on the water side (W_s = 0, ΔE_k = ΔE_p = 0): Q̇ = ṁ·c_p·ΔT.
- Q̇ = 5 × 4.18 × (85 − 25) = 1254 kW.
- Steam side: Q̇ = ṁ_s·λ, so ṁ_s = 1254 / 2133.4 = 0.588 kg/s.
Answer: Duty = 1254 kW; steam ≈ 0.588 kg/s (2.12 t/h)
Example 2 (GATE level). 10 kg/s of steam enters a turbine with specific enthalpy 3230 kJ/kg and velocity 50 m/s, and leaves 3 m lower with enthalpy 2560 kJ/kg and velocity 150 m/s. Heat loss to the surroundings is 50 kW. Find the shaft power.
- Steady-flow balance: Ẇ_s = Q̇ − ΔḢ − ΔĖ_k − ΔĖ_p.
- ΔḢ = 10 × (2560 − 3230) = −6700 kW.
- ΔĖ_k = 10 × (150² − 50²) / 2 = 100 000 W = 100 kW.
- ΔĖ_p = 10 × 9.81 × (−3) = −294 W = −0.29 kW.
- Q̇ = −50 kW (lost).
- Ẇ_s = −50 − (−6700) − 100 − (−0.29) = 6550 kW.
Answer: Shaft power ≈ 6.55 MW (kinetic energy change is about 1.5 % of ΔH; potential energy is negligible)
Common mistakes
- Mixing sign conventions for Q and W within one problem.
- Forgetting to convert m²/s² (J/kg) to kJ/kg when adding E_k to enthalpy – a factor of 1000.
- Using U instead of H for flowing streams.
- Using different reference states for the same species in inlet and outlet.
- Forgetting the latent heat when steam is used for heating (counting only sensible heat).
- Treating a throttling valve as if it cools a liquid or ideal gas by itself – for an ideal gas ΔH = 0 means ΔT = 0.
For GATE CH
Expect NAT questions on steam or cooling-water requirements, adiabatic mixing temperatures, turbine or compressor work with kinetic-energy terms, and closed-system heating at constant volume or pressure. Practise writing the general balance first and striking out terms with a stated reason.
Quick check
- What is the energy balance for an adiabatic throttling valve?
- Adiabatic mixing of 3 kg/s water at 80 °C with 2 kg/s at 20 °C gives what outlet temperature?
- 2 kg/s of water heated by 60 K needs how much steam with λ = 2114 kJ/kg?
- What is the kinetic energy of 1 kg/s flowing at 100 m/s, in kW?
- When can Q = ΔH be used for an open system?
Answers: 1. ΔH = 0; 2. 56 °C; 3. 0.237 kg/s; 4. 5 kW; 5. when W_s = 0 and ΔE_k, ΔE_p are negligible.
Interview questions
All Process Calculations interview questionsTry answering each one aloud before you open it.
1.What is an energy balance in the context of non-reactive processes?Concept
An energy balance in non-reactive processes involves accounting for all forms of energy entering and leaving a system. It ensures that the energy input equals the energy output plus any changes in the energy stored within the system. This principle is based on the first law of thermodynamics, which states that energy cannot be created or destroyed.
2.Explain the significance of the first law of thermodynamics in energy balances for non-reactive processes.Concept
The first law of thermodynamics, also known as the law of energy conservation, is crucial for energy balances in non-reactive processes. It states that the total energy of an isolated system remains constant. This means that any energy entering a system must either be stored, used to do work, or leave the system, ensuring that energy is conserved throughout the process.
3.How do you differentiate between sensible and latent heat in energy balances?Concept
Sensible heat refers to the heat exchanged by a system that results in a temperature change without a phase change. Latent heat, on the other hand, is the heat exchanged during a phase change at constant temperature. In energy balances, it's important to account for both types of heat to accurately determine the energy changes in a system.
4.Why is it important to consider heat losses in energy balances for non-reactive processes?Application
Considering heat losses is important because they represent energy that is not used for the intended purpose within the process. Ignoring heat losses can lead to inaccurate energy balances, resulting in inefficient process design and operation. By accounting for heat losses, engineers can optimize processes to minimize energy waste and improve efficiency.
5.What happens if you neglect work interactions in an energy balance for a non-reactive process?Application
Neglecting work interactions can lead to an incomplete energy balance, as work interactions represent energy transferred across the system boundary due to forces acting over a distance. This can result in errors in calculating the energy input or output, potentially leading to incorrect conclusions about the system's efficiency and performance.
6.In what scenarios would you use an enthalpy balance instead of an internal energy balance?Application
An enthalpy balance is often used when dealing with open systems where mass crosses the system boundary, such as in flow processes. Enthalpy accounts for both the internal energy and the flow work associated with the mass entering or leaving the system. In contrast, an internal energy balance is more suitable for closed systems where no mass crosses the boundary.
7.How does the choice of system boundary affect the energy balance in a non-reactive process?Application
The choice of system boundary determines what energy interactions are considered in the balance. A well-defined boundary ensures that all relevant energy inputs and outputs are accounted for, leading to an accurate energy balance. If the boundary is poorly defined, it may exclude significant energy interactions, resulting in errors and inefficiencies in the process analysis.
8.Calculate the energy change for a system where 500 kg of water is heated from 20°C to 80°C. Assume the specific heat capacity of water is 4.18 kJ/kg·°C.Numerical
To calculate the energy change, use the formula: Q = m·c·ΔT. Here, m = 500 kg, c = 4.18 kJ/kg·°C, and ΔT = 80°C - 20°C = 60°C. Therefore, Q = 500 kg × 4.18 kJ/kg·°C × 60°C = 125,400 kJ.
9.A heat exchanger transfers 200 kJ of energy to a fluid. If the fluid's mass is 10 kg and its specific heat capacity is 2 kJ/kg·°C, what is the temperature change of the fluid?Numerical
Use the formula: Q = m·c·ΔT. Rearrange to find ΔT: ΔT = Q / (m·c). Here, Q = 200 kJ, m = 10 kg, and c = 2 kJ/kg·°C. Therefore, ΔT = 200 kJ / (10 kg × 2 kJ/kg·°C) = 10°C.
10.Explain why energy balances are crucial in the design of industrial processes.Application
Energy balances are crucial because they ensure that all energy inputs, outputs, and losses are accounted for, leading to efficient process design. They help identify areas where energy can be conserved or recovered, reducing operational costs and environmental impact. Accurate energy balances also ensure that processes meet safety and performance standards.
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