Math & Numbers

Bulk density and porosity for granular or porous materials. Three-way solver: pick the unknown (ρ_bulk, m, or V) and supply the other two. ρ_bulk = m/V_bulk. Optionally enter the true particle density ρ_true to compute porosity ε = 1 − ρ_bulk/ρ_true, plus relative density against water. SI units (kg, m³, kg/m³).

Math & Numbers

Convert between Brinell (HB), Vickers (HV), and Rockwell (HRC / HRB) hardness via ASTM E140 tabulated data with linear interpolation. Three material classes: carbon/alloy steel (HV 100–960), austenitic stainless (HV 100–600), and cartridge brass (HV 40–200). Any input scale produces all the others; a scale that does not apply in the current hardness band returns 'out of range'. Indicative only — not for acceptance testing per ASTM E140.

Math & Numbers

Calculate the static safety factor n = σ_limit / σ_applied against yielding (σ_y) or ultimate tensile strength (σ_uts). Built-in typical strengths for carbon/alloy steel, aluminum, copper, austenitic stainless and titanium, plus a custom σ_limit option. Assessment: n<1 failure, 1≤n<1.5 marginal, n≥1.5 safe.

Math & Numbers

Compute engineering tensile strength σ_uts = F_max/A₀, yield strength σ_y = F_y/A₀ (optional), and the yield ratio σ_y/σ_uts from a uniaxial tension test. Forces in kN, area in mm² → stress in MPa. A₀ is the original cross-section. Yield ratio <0.55 high ductility, 0.55–0.85 moderate, >0.85 low ductility.

Math & Numbers

Solve Young's modulus (modulus of elasticity) in the linear-elastic region using E = σ/ε (Hooke's law). Three directions: modulus E = σ/ε, stress σ = E·ε, strain ε = σ/E. Pick the unknown, supply the other two positive values. Result in MPa with a GPa reading (e.g. steel ≈ 200 000 MPa).

Math & Numbers

Compute the air change rate (ACH / n) of a room from the outdoor supply airflow Q and the room volume V: n = Q/V (1/h). Three modes: solve ACH (given Q and V), solve airflow (given n and V), or solve volume (given n and Q). Flow in m³/s/m³/h/CFM, volume in m³/ft³/L. Also reports the well-mixed single-zone purge time to reach a target residual fraction ε (default 1%): t = −ln(ε)/n hours.

Math & Numbers

Split an air-conditioning load into sensible and latent components. Total cooling capacity Qt = ṁ_da·(h1 − h2); sensible capacity Qs = ṁ_da·cp_ma·(T1 − T2) with cp_ma ≈ 1.006 + 1.86·W [kJ/(kg da·K)]; latent capacity Ql = Qt − Qs; Sensible Heat Ratio SHR = Qs / Qt. Entering/leaving states are described by dry-bulb temperature T and humidity ratio W; enthalpy h = 1.006·T + W·(2501 + 1.86·T) [kJ/kg da]. Three modes: full air-state split (T1,W1)→(T2,W2), from Qt & Qs (solve SHR + Ql), or from Qt & SHR (solve Qs, Ql).

Math & Numbers

Compute the effectiveness (ε) of a heat exchanger by the ε-NTU method. C*=C_min/C_max (0..1), NTU=U·A/C_min, q_max=C_min·(T_h,in-T_c,in), and ε=q_actual/q_max. Supports four arrangements: Parallel flow ε=[1-exp(-NTU(1+C*))]/(1+C*); Counter flow ε=[1-exp(-NTU(1-C*))]/[1-C*·exp(-NTU(1-C*))], or NTU/(1+NTU) when C*=1; Shell-and-tube 1-2 ε=2/[1+C*+√(1+C*²)·(1+exp(-NTU√(1+C*²)))/(1-exp(-NTU√(1+C*²)))]; Crossflow (both unmixed) ε=1-exp{(NTU^0.22/C*)·[exp(-C*·NTU^0.78)-1]}. When C*=0 (phase change on one side, boiler/condenser) every arrangement gives ε=1-exp(-NTU). Optionally supply T_h,in and T_c,in to recover the actual heat-transfer rate q=ε·q_max and both outlet temperatures. Temperatures are used only as differences: Δ°C=ΔK and Δ°F×5/9=ΔK.

Math & Numbers

Estimate the heating (heat-loss) load of a room by the simplified steady-state method. Envelope transmission loss Q_trans = Σ(A_i·U_i·ΔT) over surfaces entered one per line as 'area,U' (U in W/(m²·K)). Cold-air infiltration loss Q_inf = 0.018·ACH·V·ΔT (W), where 0.018 W·h/(m³·K) ≈ ρ·c_p/3600. Total load = Q_trans + Q_inf, optionally multiplied by a safety factor (default 1.0). ΔT = indoor - outdoor (heating, >0). Temperature in °C/K/°F (only differences matter), area in m²/ft².

Math & Numbers

Compute 1-D steady-state heat conduction through a flat slab (Fourier's law): heat flux q = k·ΔT/d (W/m²), heat flow rate Q = k·A·ΔT/d (W), and thermal resistance R = d/(k·A) (K/W). k is the thermal conductivity (W/(m·K)); ΔT is the temperature difference (K; a °C difference equals a K difference, a °F difference is converted by ×5/9); d the slab thickness; A the cross-section area. ΔT may be negative (indicating reverse heat flow), but k, d and A must be positive. Thickness in m/cm/mm, area in m²/cm².

Math & Numbers

Compute convective heat transfer (Newton's law of cooling): heat flux q = h·ΔT (W/m²), heat flow rate Q = h·A·ΔT (W), and convective thermal resistance R_conv = 1/(h·A) (K/W). h is the convective heat-transfer coefficient (W/(m²·K)); ΔT is the temperature difference between the surface and the fluid (K; a °C difference equals a K difference, a °F difference is converted by ×5/9); A the heat-transfer area. ΔT may be negative (indicating reverse heat flow), but h and A must be positive. Area in m²/cm².

Math & Numbers

Compute the Log Mean Temperature Difference (LMTD) of a heat exchanger for parallel or counter flow. Parallel flow: ΔT₁ = T_h,in - T_c,in and ΔT₂ = T_h,out - T_c,out; counter flow: ΔT₁ = T_h,in - T_c,out and ΔT₂ = T_h,out - T_c,in. LMTD = (ΔT₁ - ΔT₂)/ln(ΔT₁/ΔT₂), or ΔT₁ when ΔT₁ = ΔT₂. A non-positive terminal difference (temperature cross) is physically impossible and is rejected. Optionally, with the overall heat transfer coefficient U (W/(m²·K)) and the heat transfer area A (m²), the heat transfer rate Q = U·A·LMTD (W) is returned. Temperatures are used only as differences: Δ°C = ΔK and Δ°F ×5/9 = ΔK; the LMTD is reported in K.