1. Overview & Modes of Heat Transfer
Heat transfer is the movement of thermal energy from high-temperature regions to low-temperature regions. In midstream operations, heat transfer governs process heating/cooling, heat loss prevention, and equipment sizing.
Conduction
Solid materials
Heat transfer through pipe walls, vessel shells, and insulation materials.
Convection
Fluid motion
Heat transfer between fluid and solid surface (forced or natural convection).
Radiation
Electromagnetic waves
Heat transfer via infrared radiation (significant at high temperatures).
Combined modes
Simultaneous transfer
Most real systems involve multiple modes (overall heat transfer coefficient).
Fundamental Heat Transfer Equation
First Law of Thermodynamics (Energy Balance):
Q = ṁ × Cp × ΔT
Where:
Q = Heat transfer rate (Btu/hr or W)
ṁ = Mass flow rate (lb/hr or kg/s)
Cp = Specific heat capacity (Btu/lb·°F or J/kg·K)
ΔT = Temperature change (°F or K)
For constant pressure process:
Q = ṁ × Cp × (T_out - T_in)
Example:
Heat natural gas from 60°F to 120°F at 1000 lb/hr:
Cp ≈ 0.5 Btu/lb·°F (at avg conditions)
Q = 1000 × 0.5 × (120 - 60) = 30,000 Btu/hr
Applications in Midstream Operations
| Application | Heat Transfer Type | Typical Duty |
|---|---|---|
| Gas-gas heat exchanger (feed preheating) | Convection (both sides) | 0.5-5 MMBtu/hr |
| Gas-liquid heat exchanger (reboiler) | Convection + boiling | 1-20 MMBtu/hr |
| Pipeline heat loss (buried) | Conduction to soil | 10-50 Btu/hr·ft |
| Insulated vessel heat loss | Conduction + convection + radiation | 100-1000 Btu/hr |
| Fired heater (process heating) | Radiation + convection | 10-100 MMBtu/hr |
Overall heat transfer coefficient (U): In most equipment, heat must pass through multiple resistances (inside fluid film, pipe wall, outside fluid film, fouling layers). The overall coefficient U combines all resistances into a single value for design calculations. Typical U values: gas-gas 5-50, liquid-liquid 50-300, condensing vapor 100-500 Btu/hr·ft²·°F.
2. Heat Transfer Mechanisms
Conduction (Fourier's Law)
One-Dimensional Steady-State Conduction:
Q = -k × A × (dT/dx)
For constant area and thermal conductivity:
Q = k × A × (T₁ - T₂) / L
Where:
Q = Heat transfer rate (Btu/hr or W)
k = Thermal conductivity (Btu/hr·ft·°F or W/m·K)
A = Cross-sectional area perpendicular to heat flow (ft² or m²)
T₁, T₂ = Temperatures at locations 1 and 2 (°F or K)
L = Distance between locations (ft or m)
Thermal resistance:
R_cond = L / (k × A) (°F·hr/Btu or K/W)
Example - Pipe wall conduction:
For cylindrical coordinates (radial conduction through pipe wall):
Q = 2π × k × L × (T_i - T_o) / ln(r_o / r_i)
Where:
L = Pipe length (ft)
r_i, r_o = Inner and outer radii (ft)
T_i, T_o = Inner and outer surface temperatures (°F)
Thermal Conductivity Values
| Material | k (Btu/hr·ft·°F) | k (W/m·K) | Application |
|---|---|---|---|
| Carbon steel (pipe) | 25-30 | 43-52 | Pipeline, pressure vessels |
| Stainless steel 316 | 8-10 | 14-17 | Process piping, heat exchanger tubes |
| Fiberglass insulation | 0.02-0.025 | 0.035-0.043 | Pipe/vessel insulation |
| Mineral wool | 0.022-0.028 | 0.038-0.048 | High-temperature insulation |
| Polyurethane foam | 0.012-0.016 | 0.021-0.028 | Cryogenic insulation (LNG) |
| Concrete (buried pipe) | 0.4-0.8 | 0.7-1.4 | Pipe coatings, foundations |
| Soil (moist) | 0.5-1.5 | 0.9-2.6 | Buried pipeline heat transfer |
Convection (Newton's Law of Cooling)
Convective Heat Transfer:
Q = h × A × (T_s - T_∞)
Where:
Q = Heat transfer rate (Btu/hr or W)
h = Convective heat transfer coefficient (Btu/hr·ft²·°F or W/m²·K)
A = Surface area (ft² or m²)
T_s = Surface temperature (°F or K)
T_∞ = Bulk fluid temperature (°F or K)
Thermal resistance:
R_conv = 1 / (h × A) (°F·hr/Btu or K/W)
Convection types:
1. Forced convection: External force (pump, fan) drives fluid motion
2. Natural convection: Buoyancy-driven flow from density gradients
3. Boiling/condensation: Phase change enhances heat transfer
Convection Coefficient Correlations
Convection coefficients depend on fluid properties, flow regime, and geometry. Common correlations:
Dittus-Boelter Equation (Turbulent Flow in Pipes):
Nu = 0.023 × Re^0.8 × Pr^n
Where:
Nu = Nusselt number = h × D / k
Re = Reynolds number = ρ × V × D / μ
Pr = Prandtl number = Cp × μ / k
n = 0.4 for heating, 0.3 for cooling
Valid for:
- Re > 10,000 (turbulent)
- 0.7 < Pr < 160
- L/D > 10 (fully developed)
Example:
Natural gas in 4" pipe, Re = 50,000, Pr = 0.75:
Nu = 0.023 × 50,000^0.8 × 0.75^0.4 = 128
If k = 0.02 Btu/hr·ft·°F, D = 0.333 ft:
h = Nu × k / D = 128 × 0.02 / 0.333 = 7.7 Btu/hr·ft²·°F
Typical Convection Coefficients
| Fluid/Condition | h (Btu/hr·ft²·°F) | h (W/m²·K) |
|---|---|---|
| Natural gas (forced convection) | 5-25 | 28-142 |
| Water (forced convection) | 300-3,000 | 1,700-17,000 |
| Oil/liquid hydrocarbons | 50-500 | 284-2,840 |
| Condensing steam | 1,000-5,000 | 5,680-28,400 |
| Boiling water | 500-5,000 | 2,840-28,400 |
| Air (natural convection) | 1-5 | 6-28 |
| Air (forced, 10 mph wind) | 5-10 | 28-57 |
Radiation (Stefan-Boltzmann Law)
Thermal Radiation:
Q = ε × σ × A × (T₁⁴ - T₂⁴)
Where:
Q = Radiant heat transfer rate (Btu/hr or W)
ε = Emissivity (0-1, dimensionless)
σ = Stefan-Boltzmann constant
= 0.1714×10⁻⁸ Btu/hr·ft²·°R⁴
= 5.67×10⁻⁸ W/m²·K⁴
A = Surface area (ft² or m²)
T₁, T₂ = Absolute temperatures (°R = °F+459.67, or K)
For small temperature differences, linearized form:
Q ≈ h_r × A × (T₁ - T₂)
Where h_r = ε × σ × (T₁² + T₂²) × (T₁ + T₂)
Radiation typically important when:
- High temperatures (> 500°F)
- Low convection coefficients (gas/air)
- View factor close to 1 (large surfaces facing each other)
Emissivity Values
| Surface | Emissivity (ε) | Notes |
|---|---|---|
| Polished aluminum | 0.05-0.10 | Reflective, low radiation |
| Oxidized steel | 0.80-0.90 | Typical piping/vessels |
| Painted surface (any color) | 0.85-0.95 | Paint increases emissivity |
| Black body (theoretical) | 1.00 | Perfect emitter/absorber |
| Insulation jacket (aluminum) | 0.10-0.20 | Reduces radiant heat loss |
Combined heat transfer: In most practical cases, all three modes occur simultaneously. For example, heat loss from an insulated pipe involves: conduction through pipe wall and insulation, convection to ambient air, and radiation to surroundings. Use overall heat transfer coefficient U to combine all resistances in series.
3. LMTD Method (Log Mean Temperature Difference)
The LMTD method is the standard approach for rating (checking performance of) existing heat exchangers or designing new ones when inlet/outlet temperatures are known.
Overall Heat Transfer Equation
Basic Heat Exchanger Equation:
Q = U × A × LMTD × F_t
Where:
Q = Heat duty (Btu/hr or W)
U = Overall heat transfer coefficient (Btu/hr·ft²·°F or W/m²·K)
A = Heat transfer area (ft² or m²)
LMTD = Log mean temperature difference (°F or K)
F_t = Temperature correction factor for exchanger configuration
This equation applies to all heat exchanger types (shell-and-tube,
plate, air-cooled, etc.).
LMTD Calculation
For Counterflow (True Countercurrent):
LMTD = (ΔT₁ - ΔT₂) / ln(ΔT₁ / ΔT₂)
Where:
ΔT₁ = T_h,in - T_c,out (hot inlet - cold outlet)
ΔT₂ = T_h,out - T_c,in (hot outlet - cold inlet)
For Parallel Flow (Cocurrent):
ΔT₁ = T_h,in - T_c,in (hot inlet - cold inlet)
ΔT₂ = T_h,out - T_c,out (hot outlet - cold outlet)
LMTD_parallel = (ΔT₁ - ΔT₂) / ln(ΔT₁ / ΔT₂)
Note: Counterflow gives larger LMTD → smaller exchanger for same duty.
Special case - equal temperature differences:
If ΔT₁ = ΔT₂ = ΔT:
LMTD = ΔT (arithmetic mean)
LMTD Example Calculation
Problem:
Heat natural gas from 60°F to 100°F using hot oil cooling from
200°F to 150°F in a countercurrent heat exchanger.
Gas flow: 10,000 lb/hr (Cp = 0.52 Btu/lb·°F)
Calculate LMTD and required area if U = 15 Btu/hr·ft²·°F.
Solution:
Step 1: Calculate duty from gas side
Q = ṁ × Cp × ΔT
Q = 10,000 × 0.52 × (100 - 60) = 208,000 Btu/hr
Step 2: Verify from oil side (energy balance)
Q = ṁ_oil × Cp_oil × (200 - 150)
208,000 = ṁ_oil × 0.5 × 50
ṁ_oil = 8,320 lb/hr ✓
Step 3: Calculate LMTD (counterflow)
ΔT₁ = T_h,in - T_c,out = 200 - 100 = 100°F
ΔT₂ = T_h,out - T_c,in = 150 - 60 = 90°F
LMTD = (100 - 90) / ln(100/90)
LMTD = 10 / ln(1.111)
LMTD = 10 / 0.1054 = 94.9°F
Step 4: Calculate required area (assume F_t = 1.0 for true counterflow)
Q = U × A × LMTD × F_t
208,000 = 15 × A × 94.9 × 1.0
A = 208,000 / (15 × 94.9) = 146 ft²
Result: Require ~150 ft² heat transfer area.
For shell-and-tube with 3/4" OD tubes, 10 ft long:
Area per tube = π × (0.75/12) × 10 = 1.96 ft²
Number of tubes = 150 / 1.96 = 77 tubes
Overall Heat Transfer Coefficient (U)
Resistance Network (Series Resistances):
1/U = 1/h_i + (r_o ln(r_o/r_i))/k_w + R_f,i × (A_o/A_i) + R_f,o + 1/h_o
Where:
h_i = Inside fluid film coefficient (Btu/hr·ft²·°F)
h_o = Outside fluid film coefficient
k_w = Tube wall thermal conductivity
r_i, r_o = Inner and outer tube radii
R_f,i, R_f,o = Fouling resistances (inside and outside)
A_i, A_o = Inside and outside tube areas
For thin-walled tubes, wall resistance negligible:
1/U ≈ 1/h_i + R_f,i + R_f,o + 1/h_o
Note: U based on outside area A_o is common convention.
Fouling factors (from TEMA standards):
- Clean gas: R_f = 0.001 hr·ft²·°F/Btu
- Clean water: R_f = 0.001
- Treated cooling water: R_f = 0.002
- River water: R_f = 0.003-0.005
- Light hydrocarbons: R_f = 0.001-0.002
- Heavy oils: R_f = 0.003-0.005
Typical Overall U Values
| Fluid Combination | U (Btu/hr·ft²·°F) | U (W/m²·K) | Controlling Resistance |
|---|---|---|---|
| Gas to gas | 5-50 | 28-284 | Both film coefficients low |
| Gas to liquid | 10-80 | 57-454 | Gas-side film coefficient |
| Liquid to liquid (light oils) | 50-200 | 284-1,135 | Fouling and film coefficients |
| Liquid to liquid (water/water) | 200-500 | 1,135-2,840 | Mainly fouling |
| Condensing vapor to liquid | 100-500 | 568-2,840 | Liquid-side film + fouling |
| Boiling liquid (reboiler) | 100-300 | 568-1,703 | Heating medium-side film |
Temperature Correction Factor (F_t)
For shell-and-tube exchangers with multiple tube passes, flow is not true counterflow. Correction factor F_t accounts for this:
F_t Calculation:
F_t depends on two dimensionless parameters:
P = (t₂ - t₁) / (T₁ - t₁) (temperature effectiveness)
R = (T₁ - T₂) / (t₂ - t₁) (capacity ratio)
Where:
T₁, T₂ = Shell-side inlet and outlet temperatures
t₁, t₂ = Tube-side inlet and outlet temperatures
F_t is read from charts (TEMA, Kern) based on P, R, and number of shell/tube passes.
Typical values:
- True counterflow (1 shell pass, many tube passes): F_t = 1.0
- 1 shell pass, 2 tube passes: F_t = 0.85-0.95
- 1 shell pass, 4+ tube passes: F_t = 0.75-0.90
- 2 shell passes: F_t = 0.90-0.98
Design guideline: F_t should be > 0.75 for economical design.
If F_t < 0.75, consider adding shell passes or switching to true counterflow.
4. Effectiveness-NTU Method
The effectiveness-NTU (Number of Transfer Units) method is preferred when outlet temperatures are unknown, common in preliminary design and performance rating with unknown fouling.
Heat Exchanger Effectiveness
Effectiveness Definition:
ε = Q_actual / Q_max
Where:
Q_actual = Actual heat transfer rate
Q_max = Maximum possible heat transfer rate
Maximum heat transfer occurs when the fluid with minimum heat capacity
rate experiences the largest possible temperature change:
Q_max = C_min × (T_h,in - T_c,in)
Where:
C_min = min(C_h, C_c)
C_h = ṁ_h × Cp_h (hot fluid heat capacity rate, Btu/hr·°F)
C_c = ṁ_c × Cp_c (cold fluid heat capacity rate)
Heat capacity ratio:
C_r = C_min / C_max
Note: 0 ≤ C_r ≤ 1
Number of Transfer Units (NTU)
NTU Definition:
NTU = U × A / C_min
Where:
U = Overall heat transfer coefficient (Btu/hr·ft²·°F)
A = Heat transfer area (ft²)
C_min = Minimum heat capacity rate (Btu/hr·°F)
Physical meaning: NTU represents the "thermal size" of the exchanger.
- Large NTU (> 5): Approaches temperature equilibrium
- Small NTU (< 1): Limited heat transfer
Relationship to effectiveness:
ε = f(NTU, C_r, flow configuration)
The function f depends on exchanger type (counterflow, parallel flow,
cross-flow, shell-and-tube, etc.).
Effectiveness Relations for Common Configurations
Counterflow:
For C_r < 1:
ε = [1 - exp(-NTU × (1 - C_r))] / [1 - C_r × exp(-NTU × (1 - C_r))]
For C_r = 1 (balanced flow):
ε = NTU / (1 + NTU)
Parallel Flow (Cocurrent):
ε = [1 - exp(-NTU × (1 + C_r))] / (1 + C_r)
Condenser or Evaporator (C_r = 0, one fluid phase change):
ε = 1 - exp(-NTU)
Note: This applies when one fluid's temperature is constant (condensing
steam, boiling liquid in pool). NTU = UA / C_min where C_min is the
single-phase fluid.
Cross-Flow (both fluids unmixed):
ε = 1 - exp[(NTU^0.22 / C_r) × (exp(-C_r × NTU^0.78) - 1)]
Note: More complex; use charts or approximations for practical calculations.
NTU Method Example
Problem:
A counterflow heat exchanger with U = 20 Btu/hr·ft²·°F and A = 200 ft²
heats natural gas from 60°F using hot oil.
Gas: 10,000 lb/hr, Cp = 0.52 Btu/lb·°F
Oil: 8,000 lb/hr, Cp = 0.50 Btu/lb·°F, inlet at 180°F
Find: Gas outlet temperature and heat duty.
Solution:
Step 1: Calculate heat capacity rates
C_gas = 10,000 × 0.52 = 5,200 Btu/hr·°F
C_oil = 8,000 × 0.50 = 4,000 Btu/hr·°F
C_min = 4,000 Btu/hr·°F (oil side)
C_max = 5,200 Btu/hr·°F (gas side)
C_r = 4,000 / 5,200 = 0.769
Step 2: Calculate NTU
NTU = U × A / C_min = 20 × 200 / 4,000 = 1.0
Step 3: Calculate effectiveness (counterflow, C_r = 0.769)
ε = [1 - exp(-1.0 × (1 - 0.769))] / [1 - 0.769 × exp(-1.0 × (1 - 0.769))]
ε = [1 - exp(-0.231)] / [1 - 0.769 × exp(-0.231)]
ε = [1 - 0.794] / [1 - 0.769 × 0.794]
ε = 0.206 / (1 - 0.611)
ε = 0.206 / 0.389 = 0.530
Step 4: Calculate actual heat transfer
Q_max = C_min × (T_h,in - T_c,in) = 4,000 × (180 - 60) = 480,000 Btu/hr
Q_actual = ε × Q_max = 0.530 × 480,000 = 254,400 Btu/hr
Step 5: Find outlet temperatures
From gas side:
Q = C_gas × (T_gas,out - T_gas,in)
254,400 = 5,200 × (T_gas,out - 60)
T_gas,out = 60 + 254,400/5,200 = 109°F
From oil side:
Q = C_oil × (T_oil,in - T_oil,out)
254,400 = 4,000 × (180 - T_oil,out)
T_oil,out = 180 - 254,400/4,000 = 116°F
Results:
Gas outlet: 109°F
Oil outlet: 116°F
Heat duty: 254,400 Btu/hr
NTU vs. LMTD Method Comparison
| Aspect | LMTD Method | NTU Method |
|---|---|---|
| Best use case | Outlet temperatures known (rating) | Outlet temperatures unknown (sizing) |
| Typical application | Performance check, confirm design | Preliminary design, optimization |
| Calculation | Direct (if T known) | Requires charts or equations |
| Iteration required | Sometimes (if T unknown) | No (effectiveness explicit) |
| Industry preference | Shell-and-tube design (TEMA) | Compact exchangers, academic |
Method selection: Use LMTD when all four temperatures are known or specified (rating existing exchangers, verifying vendor proposals). Use NTU when designing new exchangers with unknown outlet temperatures (sizing calculations, sensitivity studies). Both methods give identical results when applied correctly.
5. Applications & Design Considerations
Pipeline Heat Loss Calculation
Insulated Pipe Heat Loss:
Q/L = 2π × (T_fluid - T_amb) / [1/(h_i × r_i) + ln(r_p/r_i)/k_pipe + ln(r_ins/r_p)/k_ins + 1/(h_o × r_ins)]
Where:
Q/L = Heat loss per unit length (Btu/hr·ft)
T_fluid = Fluid temperature (°F)
T_amb = Ambient temperature (°F)
r_i = Inner pipe radius (ft)
r_p = Outer pipe radius (ft)
r_ins = Outer insulation radius (ft)
h_i = Inside convection coefficient (Btu/hr·ft²·°F)
h_o = Outside convection coefficient (includes wind, radiation)
k_pipe = Pipe thermal conductivity
k_ins = Insulation thermal conductivity
Simplified (negligible pipe wall resistance, h_i very large):
Q/L ≈ 2π × (T_fluid - T_amb) / [ln(r_ins/r_p)/k_ins + 1/(h_o × r_ins)]
Example:
6" NPS pipe (OD = 6.625"), 3" insulation, 200°F gas, 50°F ambient
r_p = 6.625/(2×12) = 0.276 ft
r_ins = (6.625 + 2×3)/(2×12) = 0.526 ft
k_ins = 0.025 Btu/hr·ft·°F
h_o = 3 Btu/hr·ft²·°F (still air + radiation)
Q/L = 2π × (200-50) / [ln(0.526/0.276)/0.025 + 1/(3×0.526)]
Q/L = 942 / [25.6 + 0.634] = 942 / 26.2 = 36 Btu/hr·ft
For 1 mile (5280 ft): Q = 36 × 5280 = 190,000 Btu/hr
Insulation Thickness Optimization
Economic insulation thickness balances insulation cost vs. energy savings:
| Service Temperature (°F) | Typical Insulation Thickness (in) | Material |
|---|---|---|
| 50-100 (cold service) | 1-2 | Polyurethane foam, cellular glass |
| 100-200 (warm service) | 2-3 | Fiberglass, mineral wool |
| 200-400 (hot service) | 3-4 | Mineral wool, calcium silicate |
| 400-800 (high temp) | 4-6 | Calcium silicate, ceramic fiber |
| -260 (LNG, cryogenic) | 6-12 | Polyurethane foam, perlite |
Heat Exchanger Sizing Procedure
Design Steps:
1. Define process requirements
- Fluid flow rates (ṁ_h, ṁ_c)
- Inlet temperatures (T_h,in, T_c,in)
- Required outlet temperatures (T_h,out, T_c,out) or heat duty (Q)
2. Calculate heat duty (if not specified)
Q = ṁ_h × Cp_h × (T_h,in - T_h,out)
Verify: Q = ṁ_c × Cp_c × (T_c,out - T_c,in)
3. Estimate overall heat transfer coefficient (U)
Use typical values or calculate from film coefficients + fouling
4. Calculate LMTD (or use NTU method)
LMTD = (ΔT₁ - ΔT₂) / ln(ΔT₁/ΔT₂)
5. Apply correction factor F_t (for shell-and-tube)
Ensure F_t > 0.75 for economical design
6. Calculate required area
A = Q / (U × LMTD × F_t)
7. Mechanical design
- Select tube size, length, number of tubes
- Determine shell diameter, baffle spacing
- Check for tube-side and shell-side pressure drop
- Verify velocities (erosion, vibration limits)
8. Iterate if needed
- Recalculate U with actual geometry and velocities
- Adjust F_t if changing number of passes
- Re-size if pressure drop excessive or U too low
Common Design Issues
| Issue | Cause | Solution |
|---|---|---|
| Low overall U | Fouling, low film coefficients, gas-gas service | Increase velocity, clean regularly, use extended surfaces (fins) |
| F_t < 0.75 | Temperature cross (T_c,out approaching T_h,in) | Add shell pass, switch to true counterflow, split into 2 exchangers |
| Excessive pressure drop | High velocity, small tubes, long flow path | Increase tube diameter, reduce tube length, add tube passes |
| Tube vibration | High velocity, long unsupported span | Add support baffles, reduce shell-side velocity, thicker tubes |
| Fouling/plugging | Dirty fluids, low velocity, dead zones | Design for cleaning access, maintain velocity > 3 ft/s, install filters |
Temperature Approach and Pinch Point
Minimum Temperature Approach:
ΔT_min = Minimum temperature difference between hot and cold streams
at any point in exchanger
For counterflow:
ΔT_min = min(T_h,out - T_c,in, T_h,in - T_c,out)
Typical minimum approaches:
- Gas-gas: 20-50°F (large ΔT needed due to low U)
- Liquid-liquid: 10-20°F
- Condensing/boiling: 10-30°F (depends on process)
- Refrigeration: 5-10°F (tight approach for efficiency)
Small ΔT_min → large exchanger (expensive)
Large ΔT_min → small exchanger but less heat recovery
Pinch point: Location of minimum ΔT in process heat exchanger network.
Determines maximum heat recovery in multi-exchanger systems.
Heat Exchanger Selection Guide
| Type | Best Application | Advantages | Limitations |
|---|---|---|---|
| Shell & tube | General purpose, high P/T | Robust, repairable, TEMA standardized | Large footprint, expensive |
| Plate & frame | Liquid-liquid, moderate P/T | Compact, high U, easy cleaning | Gasket limited (< 400°F, < 300 psi) |
| Plate-fin (brazed) | Gas service, cryogenic | Very compact, high effectiveness | Not repairable, fouling-sensitive |
| Air-cooled (fin-fan) | Gas/liquid cooling, no water available | No water consumption, low maintenance | Large, ambient-dependent, high power (fans) |
| Double-pipe | Small duty (< 1 MMBtu/hr) | Simple, true counterflow, low cost | Limited area, high pressure drop |
Fouling factor importance: Fouling reduces U by 20-50% over operating life. Always include TEMA-recommended fouling factors in design calculations. Clean exchanger performance will exceed design, but fouled performance must still meet process requirements. Schedule cleaning when measured U drops below 80% of clean design value.
Ready to use the calculator?
→ Launch Calculator