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Plate heat exchanger: duty, LMTD, area

Heat exchangers · stream duty, LMTD and surface area

A brazed-plate exchanger will isolate a boiler plant from a district loop. The boiler side supplies 21,600 kg/h of water entering at 60 °C and leaving at 40 °C; the district side enters at 30 °C and leaves at 50 °C, in true counterflow. The manufacturer rates the plate pack at an overall coefficient of 2500 W/(m²·K), and counterflow means no LMTD correction (F = 1). Take water's specific heat as 4186.8 J/(kg·K).

Given
  • ṁ = 21600 kg/h — Boiler-side mass flow
  • cₚ = 4186.8 J/(kg·K) — Water specific heat
  • T_h,in = 60 °C — Hot side in
  • T_h,out = 40 °C — Hot side out
  • T_c,in = 30 °C — Cold side in
  • T_c,out = 50 °C — Cold side out
  • U = 2500 W/(m²·K) — Overall coefficient
  • F = 1 — — LMTD correction (counterflow)
Determine
  1. (a)the duty the exchanger transfers
  2. (b)the log-mean temperature difference driving it
  3. (c)the plate area the duty requires
Step 1 of 3(a) · solve for Stream duty

Duty comes off ONE stream's energy balance: 6 kg/s dropping 20 K is 502.4 kW. Either stream works — here both flows happen to be equal, and the cold side's ṁcₚΔT must return the same number, which is the first commissioning check on any exchanger: if the two sides disagree, an instrument is lying.

Rearranged for Q
Q˙=m˙ cp ΔT\dot{Q} = \dot{m} \, c_p \, \Delta T
Your values, in your units
Q˙=(21,600 kg/h) (4,186.8 J/(kg⋅K)) (20 C∘)\dot{Q} = \left( 21{,}600\ \text{kg/h} \right) \, \left( 4{,}186.8\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 20\ \text{C}^{\circ} \right)
Answer
Q˙=502.42 kW\dot{Q} = 502.42\ \text{kW}

Carried onward at full precision, not this rounded figure.

Open the Stream Duty from Mass Flow (Q = ṁcΔT) solver →

Step 2 of 3(b) · solve for Log mean temperature difference

Counterflow pairs each inlet with the OTHER stream's outlet: 60 − 50 = 10 at one end, 40 − 30 = 10 at the other. Equal terminal differences make the log-mean exactly 10 K — a balanced exchanger, the special case where the formula's ratio goes to 1 and the mean is just the common value.

Rearranged for ΔT_lm
ΔTlm=(Th,in−Tc,out)−(Th,out−Tc,in)ln⁡ ⁣(Th,in−Tc,outTh,out−Tc,in)\Delta T_{lm} = \frac{(T_{h,in} - T_{c,out}) - (T_{h,out} - T_{c,in})}{\ln\!\left(\frac{T_{h,in} - T_{c,out}}{T_{h,out} - T_{c,in}}\right)}
Your values, in your units
ΔTlm=((60 ∘C)−(50 ∘C))−((40 ∘C)−(30 ∘C))ln⁡ ⁣((60 ∘C)−(50 ∘C)(40 ∘C)−(30 ∘C))\Delta T_{lm} = \frac{(\left( 60\ ^{\circ}\text{C} \right) - \left( 50\ ^{\circ}\text{C} \right)) - (\left( 40\ ^{\circ}\text{C} \right) - \left( 30\ ^{\circ}\text{C} \right))}{\ln\!\left(\frac{\left( 60\ ^{\circ}\text{C} \right) - \left( 50\ ^{\circ}\text{C} \right)}{\left( 40\ ^{\circ}\text{C} \right) - \left( 30\ ^{\circ}\text{C} \right)}\right)}
Answer
ΔTlm=10 K\Delta T_{lm} = 10\ \text{K}

Carried onward at full precision, not this rounded figure.

Open the Log Mean Temperature Difference (Counterflow) solver →

Step 3 of 3(c) · solve for Heat transfer area

A = Q/(U·F·LMTD) = 502,416/25,000 ≈ 20.1 m² of plate. Note what the small LMTD bought: a tight 10 K driving force needs 20 m² where a sloppy 30 K would need 6.7 — close temperature approach is paid for in stainless steel, which is the whole economics of plate exchangers in one line.

Rearranged for A
A=Q˙UF ΔTlmA = \frac{\dot{Q}}{U F \, \Delta T_{lm}}
502.42 kWcarried from step 1
10 Kcarried from step 2
Your values, in your units
A=(502,416 W)(2,500 W/(m2⋅K)) (1) (10 K)A = \frac{\left( 502{,}416\ \text{W} \right)}{\left( 2{,}500\ \text{W/(m}^{2}{\cdot}\text{K)} \right) \, \left( 1 \right) \, \left( 10\ \text{K} \right)}
Converted to base units
A=(502.416 kW)(2,500 W/(m2⋅K)) (1) (10 C∘)A = \frac{\left( 502.416\ \text{kW} \right)}{\left( 2{,}500\ \text{W/(m}^{2}{\cdot}\text{K)} \right) \, \left( 1 \right) \, \left( 10\ \text{C}^{\circ} \right)}
Answer
A=20.097 m2A = 20.097\ \text{m}^{2}

Carried onward at full precision, not this rounded figure.

Open the Heat Exchanger Duty (Q = U·A·F·LMTD) solver →

The exchanger transfers 502.4 kW across a 10 K log-mean temperature difference, which at 2500 W/(m²·K) in true counterflow requires about 20.1 m² of plate area.

Why this order

The three steps are the three questions every exchanger spec has to answer, in the only order they can be answered. Duty first, from a single stream's ṁcₚΔT, because heat transferred is set by the process, not the hardware. The driving force second, and it must be the LOG mean: the temperature difference between the streams varies along the plates, and the log-mean is the exact average for the exponential way it varies — here the counterflow arrangement holds the difference at 10 K along the entire length, the balanced case where arithmetic and log means agree. Area last, because A = Q/(U·F·LMTD) is just the design equation solved for the one thing money buys. The counterflow pairing is the part to internalise: hot-in faces cold-OUT. That is what lets the cold stream leave at 50 °C, hotter than the hot stream leaves at 40 °C — an outcome parallel flow can never produce, where both outlets share an end and the cold side can only ever approach the hot outlet from below.

The two classic wrecks: pairing the temperatures by stream instead of by end — (60−40) and (50−30) gives a '20 K LMTD' and an exchanger half the size it needs to be, the kind of error that surfaces as a district loop that never quite reaches setpoint in January. And borrowing this arithmetic for a shell-and-tube: one shell pass with the same four temperatures — a temperature cross, T_c,out above T_h,out — drives the correction factor F below the workable range, and no F fixes it; the counterflow plate pack (or multiple shells) is not a refinement here, it is the only geometry that can do this duty at all. Last, U = 2500 is a clean-water number: a fouling allowance on both sides routinely takes a third off it, and the honest spec sizes the plate pack for the dirty U, not the brochure one.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.