Introduction & Context

Heat-exchanger duty calculation quantifies the rate at which thermal energy must be transferred from a hot fluid to a cold fluid to achieve a specified temperature change. In process engineering this value is the primary specification used to size exchangers, select utility loads, and verify that product temperatures meet safety or quality targets. Typical applications include pasteurising milk, pre-heating reactor feeds, and recovering energy from hot effluent streams.

Methodology & Formulas

Energy balance on the cold stream
The required heat-transfer rate (duty) equals the enthalpy rise of the fluid being heated: \[ Q = \dot{m}\,C_{p}\,(T_{\text{out}}-T_{\text{in}}) \] where
- \( \dot{m} \) = mass flow rate of the cold fluid
- \( C_{p} \) = specific heat capacity at constant pressure
- \( T_{\text{in}}, T_{\text{out}} \) = inlet and outlet temperatures of the cold stream
Log-mean temperature difference (LMTD)
For counter-current or co-current flow the effective mean driving force is: \[ \Delta T_{\text{lm}} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln(\Delta T_{1}/\Delta T_{2})} \] with \[ \Delta T_{1}=T_{\text{hot}}-T_{\text{cold,in}}, \quad \Delta T_{2}=T_{\text{hot}}-T_{\text{cold,out}} \] To avoid numerical issues when terminal differences are equal, enforce \( \Delta T_{1,2}\geq 10^{-9} \) K.
Overall heat-transfer coefficient
The clean-surface coefficient \( U_{\text{clean}} \) is supplied from literature or vendor data. Fouling resistances are added later during detailed design.
Required heat-transfer area
Combine duty and driving force with the overall coefficient: \[ A = \frac{Q}{U\,\Delta T_{\text{lm}}} \]
Tube length
For plain circular tubes the length follows from the area and inner diameter: \[ L = \frac{A}{\pi\,D_{\text{i}}} \]

Flow regime check (Reynolds number)
\[ Re = \frac{4\,\dot{m}}{\pi\,D_{\text{i}}\,\mu} \]

Regime	Re Range
Laminar	< 3000
Transition	3000 – 20000
Fully turbulent	> 20000

The heat-transfer correlation embedded in \( U \) is valid only in the transition/turbulent window shown above.

Lumped-capacitance check (Biot number)
\[ Bi = \frac{h\,t_{\text{wall}}}{k_{\text{wall}}} \] where

\( h \approx U \) (approximate surface coefficient)
\( t_{\text{wall}} \) = wall thickness
\( k_{\text{wall}} \) = thermal conductivity of the wall material

Limit	Implication
\( Bi \leq 0.1 \)	Temperature gradient inside wall negligible; lumped assumption valid
\( Bi > 0.1 \)	Conduction resistance within wall must be treated explicitly

The fundamental equation is Q = m × C_p × ΔT.

Q = duty (kW or MJ h^-1)
m = mass flow rate of the stream (kg s^-1 or kg h^-1)
C_p = specific heat capacity (kJ kg^-1 °C^-1)
ΔT = true temperature difference (°C)

Always keep units consistent; most process simulators expect SI units internally.

Label the stream that loses heat as hot; the one that gains heat is cold.

If either stream condenses or boils, include latent heat in the duty: Q = m × (C_pΔT + λ).
Check simulator heat curves—duty is positive for the hot side and negative for the cold side by convention.

Fouling factors: vendor ratings include design margin for future fouling.
Heat-loss assumptions: field units lose 1–3% to ambient; calculations often ignore this.
Reference temperatures: ensure C_p and λ are evaluated at the correct bulk temperature.
Impurities: small amounts of non-condensables can shift the effective latent load.

1 kW = 3.412 MMBtu h^-1
1 kW = 859.8 kcal h^-1
1 MMBtu h^-1 = 252 kcal h^-1

Store these in your spreadsheet as named cells to avoid rounding errors.

Only if you know one stream’s flow rate and both C_p values.

Measure the unknown stream flow via heat balance: m₂ = (m₁ C_p1 ΔT₁) / (C_p2 ΔT₂).
Without any flow, you must install a temporary flow element or use ultrasonic clamp-on meters.

Worked Example: Sizing a Plate Heat Exchanger for Pasteurised Milk Cooling

A small dairy needs to cool 0.3 kg s^-1 of pasteurised milk from 80 °C to 20 °C using cooling water that enters at 20 °C and leaves at 72 °C. The plant engineer must determine the required heat-transfer area and the corresponding plate length, assuming a clean stainless-steel channel with an internal diameter of 25 mm and a wall thickness of 1 mm.

Mass flow rate of milk, \(\dot{m}\) = 0.3 kg s^-1
Specific heat capacity of milk, \(C_{p,\text{milk}}\) = 3.9 kJ kg^-1 K^-1
Specific heat capacity of water, \(C_{p,\text{water}}\) = 4.18 kJ kg^-1 K^-1
Thermal conductivity of stainless steel, \(k\) = 0.6 W m^-1 K^-1
Overall clean heat-transfer coefficient, \(U_{\text{clean}}\) = 900 W m^-2 K^-1
Channel internal diameter, \(D_i\) = 0.025 m
Wall thickness, \(t_{\text{wall}}\) = 0.001 m
Milk temperatures: \(T_{\text{hot,in}}\) = 80 °C, \(T_{\text{hot,out}}\) = 20 °C
Water temperatures: \(T_{\text{cold,in}}\) = 20 °C, \(T_{\text{cold,out}}\) = 72 °C

Calculate the duty (heat load) \(Q\).
\[ Q = \dot{m}\;C_{p,\text{milk}}\;(T_{\text{hot,in}}-T_{\text{hot,out}}) = 0.3 \times 3.9 \times (80-20) = 70.2\ \text{kW} \]
Convert duty to watts for consistency.
\[ Q = 70.2\ \text{kW} \times 1000 = 70\,200\ \text{W} \]
Determine the two terminal temperature differences.
\[ \Delta T_1 = T_{\text{hot,in}}-T_{\text{cold,out}} = 80-72 = 8\ \text{K} \] \[ \Delta T_2 = T_{\text{hot,out}}-T_{\text{cold,in}} = 20-20 = 0\ \text{K} \] (For the LMTD calculation a small value of 0.1 K is substituted for \(\Delta T_2\) to avoid ln(0).)
Compute the log-mean temperature difference.
\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} = \frac{8 - 0.1}{\ln(8/0.1)} = 1.921\ \text{K} \]
Calculate the required heat-transfer area \(A\).
\[ A = \frac{Q}{U_{\text{clean}}\;\text{LMTD}} = \frac{70\,200}{900 \times 1.921} = 40.605\ \text{m}^2 \]
Relate area to channel length \(L\) for a single 25 mm I.D. tube.
\[ A = \pi D_i L \Rightarrow L = \frac{A}{\pi D_i} = \frac{40.605}{\pi \times 0.025} = 517.1\ \text{m} \]
Check Reynolds number to confirm turbulent flow.
\[ Re = \frac{4\dot{m}}{\pi D_i \mu} = \frac{4 \times 0.3}{\pi \times 0.025 \times 0.002} = 7\,639 \] (3000 < 7639 < 20000 → turbulent regime is satisfied.)
Verify Biot number to ensure wall resistance is negligible.
\[ Bi = \frac{U_{\text{clean}}\;t_{\text{wall}}}{k} = \frac{900 \times 0.001}{0.6} = 1.5 \] Because 1.5 > 0.1, the wall resistance is not negligible; a fouling factor or wall resistance term should be added in detailed design.

Final Answer: The pasteurised milk cooler requires approximately 40.6 m² of heat-transfer area, corresponding to about 517 m of 25 mm internal-diameter channel. The calculated duty is 70.2 kW.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

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