**Head vs Pressure Conversion Guide** - `ft = psi x 2.31 / SG`, static vs dynamic head, water horsepower and efficiency

What You Need to Know

Head is energy per unit weight of fluid, expressed as an equivalent height of fluid column (usually $\text{ft}$). **Pressure** is force per unit area (e.g., $\text{psi}$). You convert between them constantly in pump problems because pump curves are typically in head (ft), while instruments often read pressure (psi).

Core conversion (US customary)

For a liquid with specific gravity $SG$:

$H\,(\text{ft}) = P\,(\text{psi})\,\frac{2.31}{SG}$

And the inverse:

$P\,(\text{psi}) = H\,(\text{ft})\,\frac{SG}{2.31}$

Where $2.31$ comes from water’s specific weight near room temperature:

$\gamma_w \approx 62.4\,\text{lbf}\,\text{ft}^{-3}$

Static vs dynamic head (why you care)

• Static head: elevation and/or pressure difference between start and end points when there is no flow (or ignoring losses). It’s what you get from geometry (and surface pressures).
• Dynamic head / Total Dynamic Head (TDH): what the pump must supply at the operating flow, including static head + friction (major) + minor losses + velocity head changes + pressure head changes.

Water horsepower (hydraulic power) and efficiency

• Water horsepower (WHP) (also called hydraulic horsepower) is the ideal power delivered to the fluid:

$\text{WHP} = \frac{Q\,(\text{gpm})\,H\,(\text{ft})\,SG}{3960}$

• Brake horsepower (BHP) is shaft power required:

$\text{BHP} = \frac{\text{WHP}}{\eta_{pump}}$

• If you need electrical input, include motor/drive:

$\text{Input hp} = \frac{\text{WHP}}{\eta_{pump}\,\eta_{motor}\,\eta_{drive}}$

Critical reminder: Head is energy per unit weight, so for the same pump and same operating point, the pump’s head is largely fluid-independent, but the pressure rise changes with $SG$.

Step-by-Step Breakdown

A) Convert between pressure and head (fast and safe)

1. Identify the fluid and its $SG$.
2. Confirm units: $P$ in $\text{psi}$, head in $\text{ft}$.
3. Use:

$H = P\,\frac{2.31}{SG}$

or

$P = H\,\frac{SG}{2.31}$

4. Sanity check (water):

• $1\,\text{psi} \approx 2.31\,\text{ft}$
• $10\,\text{psi} \approx 23\,\text{ft}$

B) Compute Total Dynamic Head (TDH) across a pump (common exam workflow)

Use an energy balance between suction point $1$ and discharge point $2$.

A practical pump TDH form (heads in $\text{ft}$ of flowing fluid):

$H_{pump} = \left(\frac{P_2}{\gamma} - \frac{P_1}{\gamma}\right) + (z_2 - z_1) + \left(\frac{V_2^2}{2g} - \frac{V_1^2}{2g}\right) + h_L$

Where:

• $\gamma = SG\,\gamma_w$
• $z$ is elevation head
• $V^2/(2g)$ is velocity head
• $h_L$ is total losses (major + minor)

Procedure:

1. Choose points $1$ and $2$ (often suction/discharge gauge taps, or reservoir free surfaces).
2. Convert gauge readings to pressure head:

$\frac{P}{\gamma} = P\,(\text{psi})\,\frac{2.31}{SG}$

3. Add elevation head difference $z_2 - z_1$.
4. Add velocity head change if diameters differ (often small, but don’t assume).
5. Add losses at the operating flow:

• Major (pipe friction): typically via Darcy–Weisbach head loss $h_f$.
• Minor: fittings, valves, entrances/exits, etc.

3. The sum is the required pump head at that flow.

C) Compute WHP and required BHP

1. Compute $\text{WHP}$:

$\text{WHP} = \frac{Q\,(\text{gpm})\,H\,(\text{ft})\,SG}{3960}$

2. Apply pump efficiency:

$\text{BHP} = \frac{\text{WHP}}{\eta_{pump}}$

3. If asked for electrical power, include motor efficiency.

Key Formulas, Rules & Facts

Pressure–head conversions (US customary)

Head components (energy per unit weight)

Static head vs dynamic head (TDH)

Key idea: A pump adds head. The system “spends” it on elevation, pressure differences, velocity changes, and losses.

Water horsepower and efficiency

Examples & Applications

Example 1: Convert pressure rise to head (different fluids)

A pump develops $\Delta P = 50\,\text{psi}$.

• For water $SG = 1.0$:

$H = 50\,\frac{2.31}{1.0} = 115.5\,\text{ft}$

• For a hydrocarbon $SG = 0.85$:

$H = 50\,\frac{2.31}{0.85} = 135.9\,\text{ft}$

Insight: Same pressure rise corresponds to more head in a lighter liquid (smaller $SG$).

Example 2: Convert head to pressure rise (pump curve to gauge reading)

A pump curve indicates $H = 120\,\text{ft}$ at the operating point.

• Water $SG=1.0$:

$\Delta P = 120\,\frac{1.0}{2.31} = 51.9\,\text{psi}$

• Brine $SG=1.20$:

$\Delta P = 120\,\frac{1.20}{2.31} = 62.3\,\text{psi}$

Insight: For the same head, higher $SG$ means higher pressure rise.

Example 3: TDH from suction/discharge gauges (include elevation + velocity)

Given:

• Fluid: water $SG=1$
• Suction gauge at point $1$: $P_1 = -3\,\text{psi}$ (vacuum relative to atmosphere)
• Discharge gauge at point $2$: $P_2 = 40\,\text{psi}$
• Discharge gauge is $20\,\text{ft}$ higher: $z_2 - z_1 = 20\,\text{ft}$
• Same pipe diameter at gauges, so $V_2 \approx V_1$
• Losses between gauges: $h_L = 12\,\text{ft}$

Compute pump head:

Pressure head rise:

$\left(\frac{P_2}{\gamma} - \frac{P_1}{\gamma}\right) = (40 - (-3))\,2.31 = 99.3\,\text{ft}$

Total:

$H_{pump} = 99.3 + 20 + 0 + 12 = 131.3\,\text{ft}$

Insight: Negative suction gauge pressure increases required pump head because the pump must lift from a lower suction pressure level.

Example 4: Water horsepower and required brake horsepower

Given:

• $Q = 500\,\text{gpm}$
• $H = 130\,\text{ft}$
• $SG=1.0$
• Pump efficiency $\eta_{pump} = 0.78$

Water horsepower:

$\text{WHP} = \frac{500\times 130\times 1.0}{3960} = 16.4\,\text{hp}$

Brake horsepower:

$\text{BHP} = \frac{16.4}{0.78} = 21.0\,\text{hp}$

Insight: Efficiency penalties can add a lot to required shaft power. Always compute WHP first, then divide by $\eta$.

Common Mistakes & Traps

1. Forgetting specific gravity in the head–pressure conversion

   • Wrong: using $H = 2.31\,P$ for any liquid.
   • Why wrong: $2.31$ is for water; other liquids scale by $1/SG$.
   • Fix: always write $H = P\,\frac{2.31}{SG}$ before plugging numbers.

2. Mixing gauge and absolute pressure inconsistently

   • Wrong: using suction gauge (relative) with discharge absolute (or vice versa).
   • Why wrong: only pressure differences matter, but they must be on the same reference.
   • Fix: use gauge pressures for both points unless the problem explicitly uses absolute.

3. Confusing “static head” with “TDH”

   • Wrong: selecting a pump based only on elevation difference.
   • Why wrong: friction/minor losses at flow can dominate.
   • Fix: compute $H_{pump}$ including $h_L$ at the operating $Q$.

4. Double-counting elevation head in a closed loop

   • Wrong: adding all upward elevation changes as static head without subtracting downward legs.
   • Why wrong: in a loop with the same start/end elevation, net $\Delta z = 0$.
   • Fix: for a closed recirculating loop, static lift cancels; pump head mainly covers losses (plus any equipment pressure drops).

5. Ignoring velocity head changes when diameters differ

   • Wrong: assuming $V_1 = V_2$ automatically.
   • Why wrong: if suction and discharge diameters differ, $\Delta (V^2/2g)$ can be several feet.
   • Fix: check diameters; include velocity head if asked for precision.

6. Using the $3960$ constant with the wrong flow units

   • Wrong: plugging $Q$ in $\text{ft}^3\,\text{s}^{-1}$ into $\text{WHP} = QHSG/3960$.
   • Why wrong: $3960$ assumes $Q$ in $\text{gpm}$.
   • Fix: either convert $Q$ to $\text{gpm}$ or use a consistent base formula.

7. Sign mistakes with suction vacuum

   • Wrong: treating $P_1 = -5\,\text{psi}$ as if it reduces required head.
   • Why wrong: suction vacuum increases the pressure head rise needed across the pump.
   • Fix: compute $\Delta P = P_2 - P_1$ carefully; subtracting a negative increases $\Delta P$.

8. Thinking “more psi” always means “more head” across fluids

   • Wrong intuition: comparing pumps by psi when fluids differ.
   • Why wrong: pumps fundamentally impart head; pressure rise depends on $SG$.
   • Fix: compare pumps by head for hydraulic capability, then convert to psi for the fluid.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can convert confidently:
  • $H\,(\text{ft}) = P\,(\text{psi})\,\frac{2.31}{SG}$
  • $P\,(\text{psi}) = H\,(\text{ft})\,\frac{SG}{2.31}$
• You can explain the difference:
  • Static head: geometry/level/pressure differences without losses.
  • TDH: static + losses + velocity head changes at operating flow.
• You remember what to include in TDH:
  • Pressure head change, elevation change, velocity head change, and all losses.
• You can compute hydraulic power:
  • $\text{WHP} = \frac{Q\,(\text{gpm})\,H\,(\text{ft})\,SG}{3960}$
• You can get shaft power:
  • $\text{BHP} = \frac{\text{WHP}}{\eta_{pump}}$
• You avoid the classic traps:
  • Don’t forget $SG$, don’t mix gauge/absolute, don’t double-count elevation in loops, and don’t misuse $3960$ with wrong units.

You’ve got this: if you keep units and sign conventions clean, these problems become plug-and-play.