Economic (Least-Cost) Insulation Thickness
The insulation thickness that costs the LEAST over its life, balancing energy saved against insulation bought.
Example
You enter
- Temperature difference (°F) 250
- Bare surface R-value (film, ~0.5) 0.5
- Insulation k (BTU-in/hr-ft²-F) 0.27
- Operating hours (h/yr) 8000
- Energy cost ($/MMBtu) 12
- System efficiency (0-1) 0.8
- Installed cost ($ per in per ft²) 3
- Service life (years) 10
- Discount rate (0.08 = 8%, 0 = straight line) 0.08
You get
- Capital recovery factor CRF 0.149029
- Economic thickness 4.12143
- Annual cost at the optimum 3.74565
- Versus bare 60
- Heat loss reduction (%) 96.828
Details, formula, and sources
Energy falls as 1/R while insulation cost rises linearly, so the total has a single minimum with a closed form: t = k(sqrt(C/(k x price x CRF)) - R0). A 250 F line running 8,000 h/yr wants 4.12 in, cutting heat loss 96.8% and paying back in 0.21 years. Deliberately THINNER than a surface-temperature or condensation limit - those are separate limits and they win where they apply.
Annual energy $ = dT/(R0 + t/k) x hours x $/MMBtu / (1e6 x efficiency); annual capital $ = price_per_in_sf x t x CRF with CRF = i/(1-(1+i)^-n); the minimum of the sum is t_opt = k(sqrt(C/(k x price x CRF)) - R0), C = dT x hours x $/MMBtu / (1e6 x efficiency).
Economic-thickness (least-life-cost) analysis built from the standard conduction relation and a capital recovery factor - closed form, no table reproduced.
Both the conduction relation and the capital recovery factor are public engineering and finance formulas.
Estimate. AHJ and licensed professional govern.
Field names used by the API: delta_t_f, bare_r_value, k_btu_in, operating_hours, energy_cost_per_mmbtu, system_efficiency, installed_cost_per_in_sf, life_years, discount_rate, crf, optimum_thickness_in, total_annual_cost, bare_annual_cost, heat_loss_reduction_pct
- Closed-form optimum t_opt from setting the derivative of total annual cost to zero; checked against a numerical scancalculus on the cost model
- Capital recovery factor CRF = i/(1-(1+i)^-n) annualizes installed cost; 0 rate gives straight-line 1/nstandard engineering economics
- Flat surface plane-wall conduction; a pipe optimum runs thicker because curvature adds area per inchstated scope limit