Hydronic, Steam & Piping Systems

7% of exam

Pipe head loss and pump head, hydronic coil capacity, glycol corrections, steam and condensate sizing, and expansion-tank sizing.

5 concepts

Exam areas

C. Fluid Distribution Systems and Piping

Pipe Head Loss, Pump Head and the 500 Water Coefficient

How to size a hydronic loop end to end: friction and fitting head loss, total pump head from the critical circuit, and the q = 500·GPM·ΔT heat-transport rule.

A hydronic system is the circulatory system of a building: a pump pushes water around a closed loop, the water picks up or sheds heat at coils and chillers, and the whole thing lives or dies on two numbers — how much water you move and how much pressure it costs to move it. Get the flow wrong and coils starve; get the head wrong and you buy the wrong pump. Almost every KA02 piping question on the PE Mechanical exam reduces to one of three skills built here: computing friction head loss along a run, adding the minor losses through fittings and equipment, and converting a thermal load into a required GPM with the

q = 500\,\text{GPM}\,\Delta T

relation. The methods come straight from the NCEES PE Mechanical Reference Handbook — §3.4 Fluid Flow and §3.7 Fluid Flow Machinery (the power forms are §3.7.5 Pump Power Equation).

Where the pressure goes

In a closed hydronic loop the water returns to the same elevation it left, so there is no net static lift — the pump only fights friction. Total dynamic head is the sum of distributed pipe friction, the minor (fitting and valve) losses, and the pressure drop through the equipment in the path: the chiller or boiler, the coil, the balancing valve, and the control valve. You do not add up every branch; you find the single worst path — the critical circuit — because the pump must develop enough head to satisfy the most resistant loop, and every other branch is then throttled down to match.

H_{pump} = h_{f,\,pipe} + h_{m,\,fittings} + \sum h_{equipment}

Reynolds number and flow regime

Before you pick a friction model, classify the flow. The Reynolds number

Re = \rho v D/\mu

compares inertia to viscosity. For pipe flow the handbook (§3.4.1) draws the lines at

Re < 2{,}000

laminar, a

2{,}000

–

4{,}000

critical/unstable zone, a

4{,}000

–

12{,}000

transition zone, and

Re > 12{,}000

fully turbulent (the regime tied to the Moody diagram). Building chilled- and hot-water mains run at 4–8 fps with

Re

in the tens of thousands, so they sit in fully turbulent flow regardless, and the friction factor depends on both

Re

and the relative roughness

\varepsilon/D

. (Separately, in heat-transfer work the Dittus–Boelter correlation is taken as valid for

Re > 10{,}000

— a convective-coefficient threshold, not the friction classification above.) A convenient velocity shortcut for water in a round pipe is

v = 0.4085\,Q/d^{2}

with

Q

in gpm and

d

the inside diameter in inches.

Re = \dfrac{\rho v D}{\mu}, \qquad v\,[\text{fps}] = \dfrac{0.4085\,Q\,[\text{gpm}]}{d^{2}\,[\text{in}^{2}]}

Darcy–Weisbach: the universal friction model

The Darcy–Weisbach equation is dimensionally exact for any fluid: head loss equals a friction factor times the length-to-diameter ratio times the velocity head. For laminar flow

f = 64/Re

; for turbulent flow read

f

from the Moody chart or solve the Colebrook relation (the Swamee–Jain explicit fit is exam-friendly). Use Darcy whenever the fluid is not 60°F water — glycol, fuel oil, or hot water where viscosity has shifted — because the friction factor then carries the property change correctly.

h_f = f\,\dfrac{L}{D}\,\dfrac{v^{2}}{2g}, \qquad f = \dfrac{64}{Re}\ (Re<2000)

Hazen–Williams: the water-only shortcut

For ordinary water near room temperature the Hazen–Williams formula skips the friction-factor lookup entirely, folding pipe roughness into a single coefficient

C

(about 150 for plastic, 130 for copper and cast iron, 120 for new welded steel, dropping toward 100 as steel ages). Expressed as head loss in feet per 100 feet of pipe with

Q

in gpm and the inside diameter

d

in inches, it is fast and accurate for sizing. Remember its limits: it is empirical, valid only for water, and degrades badly outside roughly 40–150°F — reach for Darcy the moment a problem mentions glycol or a wide temperature swing.

h_f\,[\text{ft}/100\,\text{ft}] = 0.2083\left(\dfrac{100}{C}\right)^{1.852}\dfrac{Q^{1.852}}{d^{4.8655}}

Minor losses and equivalent length

Every elbow, tee, valve, and reducer dissipates a multiple of the velocity head,

h_m = K\,v^{2}/2g

. Sum the

K

factors along the critical circuit — the handbook tabulates them by fitting type and pipe size — and apply them at the local velocity. An alternative is the equivalent-length method, replacing each fitting with a length of straight pipe that gives the same loss, then running Darcy or Hazen–Williams over the total developed length. Both are correct; do not double-count by applying both to the same fitting.

h_m = K\,\dfrac{v^{2}}{2g}, \qquad L_{eq} = K\,\dfrac{D}{f}

From head to pump power

Once total head is known, the water (hydraulic) horsepower follows directly, and dividing by pump efficiency gives brake horsepower. The handbook's tidy water forms bake in the density of water:

\text{whp} = Q\,h/3960

with

Q

in gpm and head in feet, or

Q\,\Delta P/1714

with

\Delta P

in psi. For any other fluid, multiply by specific gravity. Brake horsepower is what sizes the motor; the operating point itself sits where the rising system curve crosses the falling pump curve.

\text{whp} = \dfrac{Q\,h}{3960}, \qquad \text{BHP} = \dfrac{Q\,h\,(SG)}{3960\,\eta_p}

The 500 coefficient: load to flow

The single most-used hydronic relation converts a sensible thermal load into a water flow rate. It is just

q = \dot m\,c_p\,\Delta T

with water's density and specific heat folded into one constant:

8.33\ \text{lb/gal} \times 60\ \text{min/h} \times 1.0\ \text{Btu/lb·°F} = 500

(precisely

499.8

). So

q\,[\text{Btu/h}] = 500\,Q\,[\text{gpm}]\,\Delta T\,[^\circ\text{F}]

. This is the water-side twin of the air-side

q_s = 1.10\,\text{CFM}\,\Delta T

. The

500

is true only for water near standard conditions — swap in glycol and the coefficient drops because both density and

c_p

change, the subject of the next concept.

q = 500\,Q\,\Delta T \quad\Longleftrightarrow\quad Q = \dfrac{q}{500\,\Delta T}

Exam strategy

Attack piping problems in a fixed order: (1) find the flow — usually from

Q = q/(500\,\Delta T)

; (2) get the velocity from

v = 0.4085\,Q/d^{2}

and confirm it is in the 4–8 fps comfort band; (3) compute distributed friction with Hazen–Williams for plain water or Darcy for anything else; (4) add minor and equipment losses to build total head along the critical circuit only; (5) convert to BHP with

Q\,h\,(SG)/(3960\,\eta)

. Watch the units relentlessly — gpm and inches in the flow formulas, feet of head in the power formula. If a question gives a load in tons, multiply by

12{,}000

to reach Btu/h before touching the 500 rule.

Key equations

Reynolds number

Re = \dfrac{\rho v D}{\mu} = \dfrac{v D}{\nu}

Flow regime for pipe flow (handbook §3.4.1): laminar $Re<2{,}000$ , critical/unstable zone $2{,}000$ – $4{,}000$ , transition zone $4{,}000$ – $12{,}000$ , fully turbulent $Re>12{,}000$ . $D$ in ft, $v$ in fps, $\nu$ kinematic viscosity in ft²/s.

Water velocity shortcut

v = \dfrac{0.4085\,Q}{d^{2}}

$v$ in fps from $Q$ in gpm and inside diameter $d$ in inches; keep mains at 4–8 fps to limit noise and erosion.

Darcy–Weisbach head loss

h_f = f\,\dfrac{L}{D}\,\dfrac{v^{2}}{2g}

Universal friction loss (ft) for any fluid; $f$ from Moody/Colebrook, or $64/Re$ if laminar. Use whenever the fluid is not standard water.

Laminar friction factor

f = \dfrac{64}{Re}

Exact for $Re < 2000$ ; independent of roughness.

Swamee–Jain (turbulent f)

f = \dfrac{0.25}{\left[\log_{10}\!\left(\dfrac{\varepsilon/D}{3.7}+\dfrac{5.74}{Re^{0.9}}\right)\right]^{2}}

Explicit fit to the Colebrook equation; $\varepsilon$ is absolute roughness ( $\approx 0.00015$ ft for commercial steel).

Hazen–Williams (water)

h_f = 0.2083\left(\dfrac{100}{C}\right)^{1.852}\dfrac{Q^{1.852}}{d^{4.8655}}

Head loss in ft per 100 ft; $Q$ gpm, $d$ inches. $C\approx$ 150 plastic, 130 copper/cast iron, 120 new welded steel, ~100 old steel (handbook §3.4.2.3). Water only.

Minor (fitting) loss

h_m = K\,\dfrac{v^{2}}{2g}

$K$ is the geometry-dependent loss coefficient (handbook §3.4.2.4, Minor Losses in Pipe Fittings, Contractions, and Expansions); sum the $K$ 's along the critical circuit and apply at local velocity.

Equivalent length

L_{eq} = K\,\dfrac{D}{f}

Straight-pipe length giving the same loss as a fitting; alternative to the $K$ method — do not use both on one fitting.

Total pump head

H_{pump} = h_{f}+ \sum h_{m} + \sum h_{equip}

Closed loop: no static lift, only friction + fittings + equipment drop along the worst (critical) path.

Water horsepower

\text{whp} = \dfrac{Q\,h}{3960} = \dfrac{Q\,\Delta P}{1714}

$Q$ gpm, $h$ ft, $\Delta P$ psi; constants bake in water density (handbook §3.7.5).

Brake horsepower

\text{BHP} = \dfrac{Q\,h\,(SG)}{3960\,\eta_p}

Divide hydraulic power by pump efficiency; multiply by specific gravity for non-water fluids. Sizes the motor.

Sensible water heat transport (500 rule)

q = 500\,Q\,\Delta T

$q$ Btu/h, $Q$ gpm, $\Delta T$ °F. $500 = 8.33\,\text{lb/gal}\times 60\,\text{min/h}\times 1.0\,\text{Btu/lb·°F}$ ; valid for water only.

Worked examples

Required flow for a 100-ton chiller

Problem. A chilled-water plant must reject $100$ tons of cooling using a $44/56^\circ\text{F}$ supply/return schedule (a $12^\circ\text{F}$ range). Find the design GPM.

Solution. Convert tons to Btu/h:

q = 100 \times 12{,}000 = 1{,}200{,}000\,\text{Btu/h}

. Apply the 500 rule:

Q = \dfrac{q}{500\,\Delta T} = \dfrac{1{,}200{,}000}{500 \times 12} = 200\,\text{gpm}

. Sanity check: the classic rule of thumb is

2.4\,\text{gpm per ton}

at a

10^\circ\text{F}

range; at a wider

12^\circ\text{F}

range the flow drops to

2.0\,\text{gpm/ton}

, and

100 \times 2.0 = 200\,\text{gpm}

. ✓

Q = \dfrac{1{,}200{,}000\ \text{Btu/h}}{500 \times 12^\circ\text{F}} = 200\ \text{gpm}

Friction and fitting head loss in the critical circuit

Problem. The $200\,\text{gpm}$ chilled-water main is $4$ -inch Schedule 40 steel (ID $= 4.026\,\text{in}$ , $C = 130$ ), $250\,\text{ft}$ developed length, with fittings totaling $\sum K = 8.0$ . Find the pipe friction and fitting head loss.

Solution. Velocity:

v = 0.4085(200)/4.026^{2} = 5.04\,\text{fps}

(in the 4–8 fps band). Velocity head:

v^{2}/2g = 5.04^{2}/64.4 = 0.395\,\text{ft}

. Hazen–Williams:

h_f = 0.2083\,(100/130)^{1.852}\,200^{1.852}/4.026^{4.8655} = 2.67\,\text{ft per 100 ft}

, so over

250\,\text{ft}

h_f = 2.67 \times 2.5 = 6.67\,\text{ft}

. Minor losses:

h_m = \sum K\,(v^{2}/2g) = 8.0 \times 0.395 = 3.16\,\text{ft}

. Piping + fittings total

= 6.67 + 3.16 = 9.83\,\text{ft}

(to 3 sig figs,

9.83\,\text{ft}

). Sanity check:

\approx 4\,\text{ft}/100\,\text{ft}

of developed length is a healthy design value for chilled-water mains. ✓

h_{f}+h_{m} = 6.67 + 3.16 = 9.83\ \text{ft}

Total head and pump brake horsepower

Problem. Add the equipment drops to the previous circuit: chiller $18\,\text{ft}$ , cooling coil $12\,\text{ft}$ , control valve $10\,\text{ft}$ . With pump efficiency $\eta_p = 0.75$ , size the pump (total head and BHP).

Solution. Total dynamic head:

H = 9.83 + 18 + 12 + 10 = 49.8 \approx 50\,\text{ft}

(closed loop, no static lift). Water horsepower:

\text{whp} = Q\,h/3960 = 200(50)/3960 = 2.53\,\text{hp}

. Brake horsepower:

\text{BHP} = \dfrac{Q\,h\,(SG)}{3960\,\eta_p} = \dfrac{200(50)(1.0)}{3960(0.75)} = 3.37\,\text{hp}

. Select a

5\,\text{hp}

motor (next standard size up). Sanity check:

\text{whp}/\eta = 2.53/0.75 = 3.37\,\text{hp}

— consistent. ✓

\text{BHP} = \dfrac{200(50)(1.0)}{3960(0.75)} = 3.37\ \text{hp}

Common pitfalls

•Using Hazen–Williams for glycol or hot water. It is calibrated for room-temperature water only; viscosity shifts make it wrong — switch to Darcy–Weisbach the moment the fluid or temperature changes.
•Adding static lift in a closed loop. The water returns to its starting elevation, so net static head is zero; the pump fights friction only. Static head matters only in open systems (cooling towers, drain-down).
•Sizing the pump for the sum of all branch flows but the head of only one branch — or vice versa. Pump head is set by the single most resistant path (the critical circuit), at the total system flow.
•Double-counting minor losses by applying both a K factor and an equivalent length to the same fitting. Pick one method per fitting.
•Forgetting the 500 in $q = 500\,Q\,\Delta T$ is water-specific. For glycol the coefficient falls (lower density and $c_p$ ), so the same load needs more GPM.
•Confusing water horsepower with brake horsepower. whp is the fluid power; BHP = whp/η is what the motor must deliver — always size the motor on BHP, then round up.
•Reading a load in tons straight into the 500 rule. Convert tons to Btu/h first ( $\times 12{,}000$ ); a ton is not a Btu/h.

References

NCEES PE Mechanical Reference Handbook — §3 Hydraulics, Fluids, and Pipe Flow (§3.4 Fluid Flow, §3.7 Fluid Flow Machinery — §3.7.5 Pump Power Equation)
ASHRAE Handbook—Fundamentals, Pipe Sizing (Ch. 22) — Hazen–Williams C values and friction charts for hydronic pipe.
ASHRAE Handbook—HVAC Systems and Equipment, Hydronic Heating and Cooling (Ch. 13) — Critical-circuit method and pump selection practice.

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