Size the thermal conductance of a counterflow exchanger from two mass flows and the hot outlet temperature.
1. Calculator Inputs
Select units before entering values.
| Calculator field | Value / setting |
|---|---|
| Arrangement | Counterflow |
| Temperature / mass flow / duty / UA units | °F / lb/h / Btu/h / Btu/(h·°F) |
| Hot inlet / cold inlet | 200 / 80 °F |
| Hot / cold specific heat | 4.18 / 4.18 kJ/(kg·K) |
| Known hot / cold mass flow | 10,000 / 20,000 lb/h |
| Known hot outlet | 140 °F |
| Known boxes | Select only hot mass flow, cold mass flow and hot outlet |
| Cold outlet, duty and UA | Uncheck these three boxes; they are calculated |
| Optional U | Leave blank |
2. Step-by-step Math
Hot-side energy balance
ṁh = 10000 × 0.45359237 / 3600 = 1.259978806 kg/s ΔTh = (200−140)/1.8 ≈ 33.3 K Q = ṁh cp ΔTh ≈ 176 kW Q = 599025.508742 Btu/h
Use temperature differences without the 32-degree offset. Specific heat is entered in kJ/(kg·K).
Cold outlet and counterflow LMTD
Tc,out = 80 + (10000/20000)(200−140) = 110°F ΔT1 = 200−110 = 90°F; ΔT2 = 140−80 = 60°F LMTD = (90−60)/ln(90/60) ≈ 74°F
The terminal differences pair opposite ends of the exchanger.
Required conductance
UA = Q/LMTD = 599025.508742/73.989103871 UA = 8096.131422 Btu/(h·°F) Effectiveness = (200−140)/(200−80) = 0.5
UA is conductance, not area. Area requires a separately supplied overall heat-transfer coefficient.
3. Compare Your Answers
Keep full precision until the final result. Display rounding may differ; compare answers rounded to three significant figures; retain unrounded intermediate values.
| Result | Expected answer | Your answer |
|---|---|---|
| Duty | 599,000 Btu/h | __________ |
| Cold outlet | 110°F | __________ |
| LMTD | 74°F | __________ |
| UA | 8,100 Btu/(h·°F) | __________ |
| Effectiveness / NTU | 50% / 0.811 | __________ |
Second check
Uncheck hot outlet and check UA instead. Enter 8096.131422 Btu/(h·°F), keeping both mass flows known. The calculated hot and cold outlets return to 140°F and 110°F.
4. Assumptions and Method
Steady counterflow, constant specific heats, no phase change or ambient heat loss. This example uses an ideal counterflow arrangement without a multipass correction. Select exactly three independent known quantities.
Equation definitions and implementation: open calculator. See also calculation methodology.