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

What You Need to Know

The big idea

In fluids/pump problems, you constantly bounce between pressure (like $\text{psi}$) and **head** (like $\text{ft}$). Head is often easier because:

• Pump curves are typically in head (e.g., $\text{ft}$ of head vs flow).
• Head “packages” energy per unit weight of fluid, so you can add elevation, pressure, velocity, and loss terms cleanly.

Core conversion you must own

For a fluid with specific gravity $SG$ (relative to water):

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

• $2.31\,\text{ft/psi}$ is the **approximate** head of **water** corresponding to $1\,\text{psi}$.
• For non-water fluids, divide by $SG$.

Reverse conversion:

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

Static vs dynamic head (pump context)

• Static head: the “no-flow” part (elevation and static pressure differences between start and end points).
• Dynamic head: the extra head needed when flowing (primarily friction/minor losses, plus any velocity-head change).
• Total Dynamic Head (TDH): the total head the pump must add at a given flow.

Water horsepower and efficiency

• Water horsepower (WHP) (also called hydraulic horsepower): useful power delivered to the fluid.
• Brake horsepower (BHP): shaft power required at the pump.
• Efficiency ties them together:

$\eta=\frac{\text{WHP}}{\text{BHP}}$

Reminder: You divide by efficiency to go from hydraulic power to required shaft power.

Step-by-Step Breakdown

A) Convert pressure $\leftrightarrow$ head (fast and correct)

1. Confirm what pressure means (gauge vs absolute). In most pump/head problems, you use gauge pressures unless told otherwise.
2. Get the fluid $SG$.
3. Use the conversion:
   • Pressure $\rightarrow$ head:

$h=\frac{P\times 2.31}{SG}$

• Head $\rightarrow$ pressure:

$P=\frac{h\times SG}{2.31}$

4. Sanity check: higher $SG$ means **more pressure per foot**, so for the same $P$ you get less head.

B) Compute TDH (energy equation mindset)

When you’re asked for pump head (TDH), pick a suction reference point (1) and a discharge reference point (2). A common “pump TDH” form is:

$H_p=\left(\frac{P_2-P_1}{\gamma}\right)+(z_2-z_1)+\left(\frac{v_2^2-v_1^2}{2g}\right)+h_L$

Where:

• $\gamma=\rho g$ is specific weight.
• $h_L$ is total head loss (major + minor).

Procedure:

1. Choose points 1 and 2 (often at pressure gauges or free surfaces).
2. Write each term:
   • Pressure head: $\frac{P}{\gamma}$
   • Elevation head: $z$
   • Velocity head: $\frac{v^2}{2g}$
   • Losses: $h_L$
3. Decide what cancels:
   • If both are large tanks: $v_1 \approx v_2 \approx 0$.
   • If both open to atmosphere: $P_1 \approx P_2 \approx 0$ gauge.
4. Add what remains to get $H_p$.

C) Convert TDH to water horsepower and required shaft power

1. Compute hydraulic power (WHP).
2. Divide by efficiency to get BHP.
3. If a motor is involved, account for motor efficiency too.

Key Formulas, Rules & Facts

Pressure-head relationships

The constant $2.31$ comes from water’s specific weight (about $62.4\,\text{lbf/ft}^3$) and $1\,\text{psi}=144\,\text{lbf/ft}^2$.

Head components (what shows up in TDH)

Static vs dynamic head (pump exam language)

• Static head (typical usage):

$H_{static}=\left(\frac{P_2-P_1}{\gamma}\right)+(z_2-z_1)$

• Dynamic head is what grows with flow:

$H_{dynamic}=\left(\frac{v_2^2-v_1^2}{2g}\right)+h_L$

• Total Dynamic Head:

$TDH=H_{static}+H_{dynamic}$

“Static head” definitions vary by industry/source. On exams, it usually means “what remains at $Q=0$.” When in doubt, write the full energy equation and label terms.

Water horsepower (hydraulic horsepower)

Efficiency and power chain

• Pump efficiency:

$\eta_p=\frac{\text{WHP}}{\text{BHP}}$

• Solve for shaft power:

$\text{BHP}=\frac{\text{WHP}}{\eta_p}$

• If motor efficiency $\eta_m$ is included:

$\text{Motor input HP}=\frac{\text{WHP}}{\eta_p\,\eta_m}$

Examples & Applications

Example 1: Convert discharge pressure to head for a non-water fluid

A pump discharge gauge reads $80\,\text{psi}$ pumping a fluid with $SG=0.80$. What head is that?

Setup:

$h=\frac{P\times 2.31}{SG}$

Compute:

$h=\frac{80\times 2.31}{0.80}=231\,\text{ft}$

Key insight: same $\text{psi}$ is **more head** for lighter fluids (lower $SG$).

Example 2: Convert pump head to pressure rise for a heavy liquid

Pump adds $120\,\text{ft}$ of head to a liquid with $SG=1.20$. What pressure rise is that?

Setup:

$P=\frac{h\times SG}{2.31}$

Compute:

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

Key insight: higher $SG$ gives more pressure per foot of head.

Example 3: TDH between two open tanks (classic static + friction)

You pump water from Reservoir A to Reservoir B. Both are open to atmosphere. The free surface in B is $40\,\text{ft}$ above A. Total head loss at the design flow is $18\,\text{ft}$. Find TDH.

Because both tanks are open:

$P_1=P_2=0\,\text{psi (gauge)}$

Because reservoirs are large:

$v_1\approx v_2\approx 0$

So pump head is:

$TDH=(z_2-z_1)+h_L=40+18=58\,\text{ft}$

Key insight: in open-tank transfer, TDH is usually elevation difference + losses.

Example 4: Water horsepower and brake horsepower

A pump delivers $Q=500\,\text{gpm}$ of water at $TDH=120\,\text{ft}$. Pump efficiency is $78\%$. Find WHP and BHP.

Hydraulic power:

$\text{WHP}=\frac{Q\times H\times SG}{3960}=\frac{500\times 120\times 1}{3960}=15.15\,\text{hp}$

Shaft power:

$\text{BHP}=\frac{\text{WHP}}{\eta_p}=\frac{15.15}{0.78}=19.4\,\text{hp}$

Key insight: efficiency increases required shaft power by the factor $\frac{1}{\eta_p}$.

Common Mistakes & Traps

1. Forgetting the specific gravity in head-pressure conversions

   • Wrong move: using $h=P\times 2.31$ for every liquid.
   • Why wrong: $2.31\,\text{ft/psi}$ is for water.
   • Fix: always apply $h=\frac{P\times 2.31}{SG}$.

2. Mixing “ft of water” with “ft of pumped fluid”

   • Wrong move: reporting head as $\text{ft of water}$ when the question implies head of the fluid.
   • Why wrong: “feet of head” is energy per unit weight, but the conversion to pressure depends on $SG$.
   • Fix: state the basis clearly; if converting to $\text{psi}$, include $SG$.

3. Using absolute pressure when the energy equation expects gauge (or vice versa)

   • Wrong move: plugging atmospheric pressure into both sides inconsistently.
   • Why wrong: absolute is fine only if used everywhere; most pump problems use gauge so atmosphere cancels naturally.
   • Fix: choose one system and be consistent. Open tanks usually imply $P=0$ gauge at surfaces.

4. Calling elevation difference “TDH” and ignoring losses

   • Wrong move: $TDH=z_2-z_1$ only.
   • Why wrong: at flow, friction/minor losses can dominate.
   • Fix: $TDH=(z_2-z_1)+\frac{P_2-P_1}{\gamma}+\frac{v_2^2-v_1^2}{2g}+h_L$.

5. Dropping velocity head when points are in pipes (not tanks)

   • Wrong move: assuming $v_1\approx v_2\approx 0$ at gauges in piping.
   • Why wrong: in a pipe, velocity head can be significant.
   • Fix: only neglect velocity head at large reservoirs or when explicitly justified.

6. Efficiency inversion error

   • Wrong move: $\text{BHP}=\text{WHP}\times \eta_p$.
   • Why wrong: efficiency is output/input.
   • Fix: $\text{BHP}=\frac{\text{WHP}}{\eta_p}$.

7. Using the wrong constant (3960 vs 1714) without checking inputs

   • Wrong move: using $3960$ with $\Delta P$ in $\text{psi}$.
   • Why wrong: $3960$ pairs with $H\,(\text{ft})$; $1714$ pairs with $\Delta P\,(\text{psi})$.
   • Fix: match constant to your variables:
     • $Q\,(\text{gpm})$ and $H\,(\text{ft})$: use $3960$ (include $SG$)
     • $Q\,(\text{gpm})$ and $\Delta P\,(\text{psi})$: use $1714$

8. Confusing “static head” definitions across sources

   • Wrong move: memorizing one phrase and misapplying it.
   • Why wrong: some references define static head as elevation only; others include pressure differences.
   • Fix: write the energy equation and identify what remains at $Q=0$.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can convert pressure to head with:

$h=\frac{P\times 2.31}{SG}$

• You can convert head to pressure with:

$P=\frac{h\times SG}{2.31}$

• You can write pump head (TDH) from the energy equation:

$H_p=\left(\frac{P_2-P_1}{\gamma}\right)+(z_2-z_1)+\left(\frac{v_2^2-v_1^2}{2g}\right)+h_L$

• You remember what “static” vs “dynamic” means:

  • Static: exists at $Q=0$ (elevation and possibly pressure differences).
  • Dynamic: grows with flow (losses, velocity terms).

• You can compute water horsepower:

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

• You can compute required shaft horsepower:

$\text{BHP}=\frac{\text{WHP}}{\eta_p}$

• You avoid traps: wrong $SG$ use, gauge vs absolute confusion, and forgetting losses.

You’ve got the key relationships now; do 2–3 conversions and one TDH-to-BHP problem and you’ll be exam-ready.