Water to 40% glycol: flow, speed and power

Hydronics · glycol derate and the affinity laws

A heating loop serving a loading-dock unit heater runs 120 gpm of water at a 20 °F design ΔT, on a pump sheaved for 1750 rpm and drawing 3.0 kW. The dock keeps freezing the coil, so the loop is being converted to 40% propylene glycol — density 1043 kg/m³, specific heat 3600 J/(kg·K) at operating temperature. The heater still needs the same heat, and the balancer intends to keep the same 20 °F ΔT.

Given

V̇₁ = 120 gpm — Water flow at design
ΔT = 20 F° — Design supply-to-return ΔT
ρ = 1043 kg/m³ — 40% PG density
cₚ = 3600 J/(kg·K) — 40% PG specific heat
N₁ = 1750 rpm — Present pump speed
P₁ = 3 kW — Present pump power

Determine

(a)the heat the water loop delivers at design
(b)the glycol flow needed to carry the same heat at the same ΔT
(c)the pump speed that delivers that flow
(d)the pump power at the new speed

Step 1 of 4(a) · solve for Heat transfer rate

First pin down what the loop actually delivers: 120 gpm across 20 °F is 1.2 million BTU/h by the 500 rule, 351.6 kW by the ρcₚ physics underneath it. This number is the contract — the heater needs it whether the loop runs water, glycol or maple syrup.

Rearranged for Q

\dot{Q} = \rho_w c_w \dot{V} \, \Delta T

V̇

ΔT

Your values, in your units

\dot{Q} = \rho_w c_w \, \left( 120\ \text{gpm} \right) \, \left( 20\ \text{F}^{\circ} \right)

Converted to base units

\dot{Q} = \rho_w c_w \, \left( 454.249\ \text{L/min} \right) \, \left( 11.1111\ \text{C}^{\circ} \right)

Answer

\dot{Q} = 351.6\ \text{kW}

in

Carried onward at full precision, not this rounded figure.

Open the Hydronic Heat Transfer (Water) solver →

Step 2 of 4(b) · solve for Flow rate

Same heat, same ΔT, weaker fluid: glycol's ρcₚ is 3.75 MJ/(m³·K) against water's 4.18, so the flow must rise by that ratio — to 133.6 gpm, 11.3% more. The classic mistake is running the old 500-rule on a glycol loop and reporting a heater 11% bigger than the one installed.

Rearranged for V̇

\dot{V} = \tfrac{\dot{Q}}{\rho c \Delta T}

Q351.6 kWcarried from step 1

ρ

cₚ

ΔT

Your values, in your units

\dot{V} = \tfrac{\left( 351{,}596\ \text{W} \right)}{\left( 1{,}043\ \text{kg/m}^{3} \right) \, \left( 3{,}600\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 20\ \text{F}^{\circ} \right)}

Converted to base units

\dot{V} = \tfrac{\left( 351{,}596\ \text{W} \right)}{\left( 1{,}043\ \text{kg/m}^{3} \right) \, \left( 3{,}600\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 11.1111\ \text{C}^{\circ} \right)}

Answer

\dot{V} = 505.65\ \text{L/min}

in

Carried onward at full precision, not this rounded figure.

Open the Glycol Loop Heat Transfer (Capacity Derate) solver →

Step 3 of 4(c) · solve for Speed 2

Flow follows speed one-for-one on a centrifugal pump, so 11.3% more flow means 11.3% more speed: 1948 rpm, a sheave change or a VFD command. This is the first affinity law doing its day job — the flow target came from heat transfer, the speed comes from the law.

Rearranged for N₂

N_2 = N_1 \cdot \tfrac{Q_2}{Q_1}

Q₁

N₁

Q₂505.65 L/mincarried from step 2

Your values, in your units

N_2 = \left( 1{,}750\ \text{rpm} \right) \cdot \tfrac{\left( 0.00842752\ \text{m}^{3}\text{/s} \right)}{\left( 120\ \text{gpm} \right)}

Converted to base units

N_2 = \left( 1{,}750\ \text{rpm} \right) \cdot \tfrac{\left( 505.651\ \text{L/min} \right)}{\left( 454.249\ \text{L/min} \right)}

Answer

N_2 = 32.467\ \text{Hz}

in

Carried onward at full precision, not this rounded figure.

Open the Pump Affinity Law — Flow vs Speed solver →

Step 4 of 4(d) · solve for Power at speed 2

Power follows the CUBE of speed: an 11.3% speed-up costs 38% more power — 4.14 kW against 3.0. Check the motor before touching the sheave; a 4 kW (5 hp) motor is now at its limit, and the affinity law assumed the fluid stayed the same, so glycol's extra density stacks another ~4% on top of this.

Rearranged for P₂

P_2 = P_1 \left(\tfrac{N_2}{N_1}\right)^{3}

P₁

N₁

N₂32.467 Hzcarried from step 3

Your values, in your units

P_2 = \left( 3\ \text{kW} \right) \left(\tfrac{\left( 32.4671\ \text{Hz} \right)}{\left( 1{,}750\ \text{rpm} \right)}\right)^{3}

Converted to base units

P_2 = \left( 3{,}000\ \text{W} \right) \left(\tfrac{\left( 1{,}948.02\ \text{rpm} \right)}{\left( 1{,}750\ \text{rpm} \right)}\right)^{3}

Answer

P_2 = 4.138\ \text{kW}

in

Carried onward at full precision, not this rounded figure.

Open the Pump Affinity Law — Power vs Speed solver →

Why this order

The chain runs heat → flow → speed → power because each answer is the only bridge to the next. The load is fixed by the building, not the fluid, so it is computed once, on the water side, where the numbers are known good. The glycol flow then falls out of the same Q = ρcₚV̇ΔT relation with the mix's own properties — and the ratio of the two flows, 4.18/3.75 = 1.113, is the whole conversion in one number. The affinity laws then translate that hydraulic requirement into mechanical facts: flow is linear in speed, power is cubic, so the same 1.113 shows up once in part (c) and cubed — 1.379 — in part (d). Cross-check: 1.113³ = 1.379, and 3.0 kW × 1.379 = 4.14 kW, matching the chain.

Glycol conversions go wrong in two places. The first is treating the 500 constant as a law of nature: it is 8.33 lb/gal × 60 × 1 BTU/(lb·°F), water's numbers, and 40% propylene glycol runs nearer 450 — every gpm-based rule on the truck quietly under-delivers heat unless the flow is raised. The second is stopping at the flow answer. The extra 13.6 gpm looks trivial until the cube law prices it: 38% more power, plus the density correction the affinity laws do not include (they assume the fluid is unchanged — power also scales with ρ, so budget ~4% more again, near 4.3 kW). That is how a routine freeze-protection job burns out a motor that ran the water loop for fifteen years: nobody re-did the power arithmetic. And the viscosity penalty on the heater's inside film coefficient means the SAME ΔT may need slightly more surface in deep cold — the flow fix carries the heat to the coil, not necessarily through it.

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