Sizing a steam main from the specific volume

Steam mains · specific volume, velocity and pipe size

A plant is re-routing the steam main that feeds its process building — 5,000 lb/h at 100 psig, run in 3-inch schedule 40 with an inside diameter of 77.93 mm — and the engineer wants the velocity checked before anything is welded. Steam mains are sized on velocity, not on pressure drop alone: too slow and the pipe is oversized and its standing heat loss and warm-up condensate go up with it; too fast and the line becomes noisy, erodes at the elbows where entrained water is thrown against the wall, and hammers on start-up. The band the trade works to for saturated steam mains is 25 to 40 m/s. Turning a mass rate into a velocity needs the specific volume, and that is a table reading: at 100 psig (114.7 psia) the steam tables give v_g = 0.2430 m³/kg at a saturation temperature of 169.9 °C, and at 15 psig (29.7 psia) the same column reads v_g = 0.8665 m³/kg. So the volumetric flow follows straight off the page — 5,000 lb/h is 2,268 kg/h, and 2,268 × 0.2430 = 551 m³/h at 100 psig, against 2,268 × 0.8665 = 1,965 m³/h if the very same duty were carried at 15 psig instead.

100 psig Ø77.93 mm15 psig Ø131.8 mm100 psig · v_g 0.2430 m³/kg32.1 m/s5,000 lb/h · 551 m³/hsmallest bore Ø69.8 mmat 15 psig: 1,965 m³/h · 143 m/s

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • = 5,000 lb/hSteam duty (2,268 kg/h)
  • D = 77.93 mmExisting main, 3 in schedule 40 bore
  • v_g₁ = 0.243 m³/kgSpecific volume at 100 psig, from the tables
  • Q₁ = 551 m³/hVolumetric flow at 100 psig (2,268 × 0.2430)
  • v_g₂ = 0.8665 m³/kgSpecific volume at 15 psig, from the tables
  • Q₂ = 1,965 m³/hVolumetric flow at 15 psig (2,268 × 0.8665)
  • v_max = 40 m/sTop of the trade band for saturated mains
Determine
  1. (a)the velocity 5,000 lb/h makes in the existing 3-inch main at 100 psig
  2. (b)the smallest bore that keeps that duty at or under 40 m/s
  3. (c)the velocity the same duty would reach in that bore if the main ran at 15 psig
  4. (d)the bore a 15 psig main would need to carry the same duty inside the band
Step 1 of 4(a) · solve for Flow velocity

Steam is compressible, so a mass rate says nothing about velocity on its own — the specific volume from the table is what converts one into the other, and it changes by a factor of three and a half across the pressures in this problem. Use the real bore, 77.93 mm, not the nominal 3 inches: velocity goes with the inverse SQUARE of diameter, so a 3 % error in the bore is a 6 % error here, and the difference between schedule 40 and schedule 80 in the same nominal size is far more than 3 %.

Rearranged for v
v=4QπD2v = \tfrac{4Q}{\pi D^{2}}
Your values, in your units
v=4(551 m3/h)π(77.93 mm)2v = \tfrac{4 \cdot \left( 551\ \text{m}^{3}\text{/h} \right)}{\pi \cdot \left( 77.93\ \text{mm} \right)^{2}}
Converted to base units
v=4(9,183.33 L/min)π(77.93 mm)2v = \tfrac{4 \cdot \left( 9{,}183.33\ \text{L/min} \right)}{\pi \cdot \left( 77.93\ \text{mm} \right)^{2}}
Answer
v=32.089 m/sv = 32.089\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 2 of 4(b) · solve for Inside diameter

The same relation read as a sizing rule: hold the velocity at the top of the band and see what bore falls out. The answer, 69.8 mm, is smaller than the 77.93 mm actually installed, which confirms the 3-inch selection with room in hand — 2½-inch schedule 40 at 62.7 mm would be under the bore this calls for, so 3-inch is the honest choice. This is the number the next step needs, so it carries forward as the pipe rather than being rounded to a catalogue size first.

Rearranged for D
D=4QπvD = \sqrt{\tfrac{4Q}{\pi v}}
Your values, in your units
D=4(551 m3/h)π(40 m/s)D = \sqrt{\tfrac{4 \cdot \left( 551\ \text{m}^{3}\text{/h} \right)}{\pi \cdot \left( 40\ \text{m/s} \right)}}
Converted to base units
D=4(9,183.33 L/min)π(40 m/s)D = \sqrt{\tfrac{4 \cdot \left( 9{,}183.33\ \text{L/min} \right)}{\pi \cdot \left( 40\ \text{m/s} \right)}}
Answer
D=69.799 mmD = 69.799\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 3 of 4(c) · solve for Flow velocity

Here is the whole lesson in one line. Same plant, same 5,000 lb/h of steam, same pipe from part (b) — but let down to 15 psig, where each kilogram occupies 0.8665 m³ instead of 0.2430. The volumetric flow rises by that ratio, 3.57, and since the bore has not changed, so does the velocity: 143 m/s, three and a half times the ceiling, a main that would howl and erode its first elbow. Nothing about the STEAM changed. Only the specific volume did.

Rearranged for v
v=4QπD2v = \tfrac{4Q}{\pi D^{2}}
69.799 mmcarried from step 2
Your values, in your units
v=4(1,965 m3/h)π(0.0697991 m)2v = \tfrac{4 \cdot \left( 1{,}965\ \text{m}^{3}\text{/h} \right)}{\pi \cdot \left( 0.0697991\ \text{m} \right)^{2}}
Converted to base units
v=4(32,750 L/min)π(69.7991 mm)2v = \tfrac{4 \cdot \left( 32{,}750\ \text{L/min} \right)}{\pi \cdot \left( 69.7991\ \text{mm} \right)^{2}}
Answer
v=142.65 m/sv = 142.65\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 4 of 4(d) · solve for Inside diameter

So size the low-pressure main honestly: 131.8 mm at the same 40 m/s ceiling. Against part (b)'s 69.8 mm that is the square root of 3.57, or 1.89 times the bore and 3.57 times the cross-section. In catalogue terms it is 6-inch schedule 40 (154.1 mm bore) where the high-pressure main was 3-inch, because 5-inch at 128.2 mm would still run at 42 m/s. Two nominal sizes, for exactly the same pounds per hour.

Rearranged for D
D=4QπvD = \sqrt{\tfrac{4Q}{\pi v}}
Your values, in your units
D=4(1,965 m3/h)π(40 m/s)D = \sqrt{\tfrac{4 \cdot \left( 1{,}965\ \text{m}^{3}\text{/h} \right)}{\pi \cdot \left( 40\ \text{m/s} \right)}}
Converted to base units
D=4(32,750 L/min)π(40 m/s)D = \sqrt{\tfrac{4 \cdot \left( 32{,}750\ \text{L/min} \right)}{\pi \cdot \left( 40\ \text{m/s} \right)}}
Answer
D=131.81 mmD = 131.81\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Answer

Therefore the existing 3-inch main runs at 32.1 m/s, comfortably inside the 25–40 m/s band; 69.8 mm is the smallest bore that duty allows at 100 psig; the same steam at 15 psig would reach 143 m/s in that pipe; and a 15 psig main would have to be opened out to 131.8 mm — a 6-inch line doing the work a 3-inch line does upstairs.

Why this order

The chain is one formula run four times, which is the point: v = 4Q/πD² is not the hard part of sizing a steam main. The hard part is knowing that Q is not given to you. What the plant knows is a mass rate — boilers are rated in pounds per hour, traps are rated in pounds per hour, the load is pounds per hour — and the specific volume column of the steam tables is the only bridge from that to a velocity. Everything surprising about steam distribution lives in that column. It is why the same duty needs a different pipe on either side of a PRV; why a main sized correctly at design pressure screams when the plant runs the header down on a mild day; and why velocity, not pressure drop, is the criterion that governs, since a main can be inside its allowable drop and still be eroding.

The economics run the other way from intuition, and they are worth stating plainly. Distributing at high pressure and letting down close to the load gives smaller mains, less insulation, less standing heat loss, less warm-up condensate and cheaper supports — which is why almost every plant generates higher than it uses and puts pressure-reducing stations at the branches. The cost is on the other side of the ledger: higher pressure means a higher saturation temperature, so the losses that remain are hotter, the flash at every trap is larger (12.7 % on a 150-to-15 psig letdown, as the flash chain works out), and the boiler shell and its fittings are rated for more. As for the band itself, 25 to 40 m/s is not a law but hard-won practice: below 25 the pipe is doing less work than the metal in it justifies, above 40 the noise and the erosion at fittings start to be measurable, and superheated steam — dry, with no droplets to throw — is commonly allowed 50 or more because the erosion mechanism is not there. Rerun this worksheet at half the load and the same 3-inch main drops to 16 m/s, which is why a main sized for a plant's peak spends most of its life oversized, and why the honest question at the design stage is which duty you are sizing for.

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