Chapter 9

Fluid Handling Equipment

Every flowsheet so far has quietly assumed streams arrive at the pressure a unit op needs. In a real plant, getting them there is its own equipment class: pumps and compressors raise pressure, turbines and valves drop it, and pipe itself eats pressure to friction over distance. None of this changes composition — it's pure mechanical/thermal energy balance — but getting it wrong is a common reason a "correct" flowsheet doesn't match the real plant.

9.1 Pumps: raising liquid pressure

A pump does mechanical work on an (incompressible) liquid. The theoretical work is flow times the pressure rise divided by density; a real pump needs more than that, by the hydraulic efficiency

\eta

W = \frac{F\,(P_{out}-P_{in})}{\rho\,\eta}

$W$: Shaft (pump) work delivered to the fluid, watts (J/s).
$F$: Molar flow through the pump, mol/s.
$P_{out},\; P_{in}$: Outlet and inlet pressure, Pa.
$\rho$: Molar density of the liquid at inlet conditions, mol/m³.
$\eta$: Hydraulic efficiency (0–1) — the fraction of shaft work that goes into raising pressure rather than being lost as heat.

F

is the molar flow,

\rho

the liquid density at the inlet conditions. The shaft work

W

shows up as sensible heating of the liquid too (the inefficiency has to go somewhere) — small for water, but not always negligible for viscous or high-head service. A real pump's head isn't flat with flow either: MaximaLabs' pump can take a

\Delta P = \rho g\,H(Q)

head curve instead of a fixed outlet pressure, so raising throughput on a pinned curve genuinely reduces the achievable head — the same curve a vendor datasheet gives you.

Worked example — same case as §9.6's warm-up exercise

5 mol/s of water (molar density

\rho \approx 55{,}390\ \text{mol/m}^3

) pumped from 1 atm to 10 atm (

\Delta P \approx 9.12\times10^5\ \text{Pa}

) at 70% hydraulic efficiency:

W = \dfrac{5 \times 9.12\times10^5}{55{,}390 \times 0.70} \approx 118\ \text{W}

The same arithmetic §9.6's exercise walks through in full — restated here next to the equation it comes from. 118 W is a modest duty: pumping liquids is cheap, which is why §9.2's compressor example below (compressing a gas, not a liquid) needs so much more power for a comparable pressure ratio.

9.2 Compressors and turbines: isentropic + efficiency

Compressible fluids need the real thermodynamics from Chapter 2, not just a pressure/density ratio. MaximaLabs solves an ideal (isentropic) path first — find the outlet temperature

T_{2s}

that keeps entropy constant across the pressure change, using whatever equation of state or activity model the flowsheet is running:

S(T_{2s}, P_2) = S(T_1, P_1) \quad(\text{solved for } T_{2s})

$T_1,\; P_1$: Known inlet temperature and pressure.
$P_2$: Known outlet pressure (the compression/expansion target).
$T_{2s}$: Ideal (isentropic) outlet temperature — the one unknown, solved so entropy is unchanged across the pressure step.
$S(T,P)$: Molar entropy at (T,P) from the flowsheet's active thermo package — an equation of state or activity model, per Chapter 2, not an ideal-gas shortcut.

then applies the isentropic efficiency

\eta

to get the real outlet state and the actual shaft work — for a compressor the real temperature rise is bigger than the ideal one (efficiency losses show up as extra heat):

T_2 = T_1 + \frac{T_{2s}-T_1}{\eta},\qquad W = F\,\big(H(T_2,P_2) - H(T_1,P_1)\big)

$T_2$: Real (actual) outlet temperature — always hotter than the ideal T_{2s} for a compressor, since inefficiency adds extra heating (see §9.6's core exercise).
$\eta$: Isentropic efficiency (0–1): the ratio of ideal to actual work for a compressor.
$W$: Real shaft work delivered to the gas, watts.
$F$: Molar flow through the machine, mol/s.
$H(T,P)$: Molar enthalpy at (T,P) from the active thermo package.

Worked example — isentropic compression of CO2, PR package

A small illustrative case computed directly from MaximaLabs' Peng-Robinson package (not from a saved showcase example — a standalone solver call, so it's genuinely computed but not the same run as the co2-capture-compression Try-it case below): pure CO₂ at

T_1 = 313.15\ \text{K}

(40 °C),

P_1 = 1\ \text{bar}

, compressed to

P_2 = 8\ \text{bar}

at 78% isentropic efficiency.

S(T_{2s}, P_2) = S(T_1, P_1) \;\Rightarrow\; T_{2s} \approx 476.9\ \text{K}

T_2 = 313.15 + \frac{476.9 - 313.15}{0.78} \approx 523.1\ \text{K}

W/F = H(T_2,P_2) - H(T_1,P_1) \approx 8{,}740\ \text{J/mol} \;\Rightarrow\; W \approx 8.74\ \text{kW per mol/s}

Compare to the pump example above: raising a gas's pressure 8× costs roughly two orders of magnitude more work per mole than raising a liquid's pressure 10× — exactly why compressors, not pumps, dominate a plant's utility bill.

A turbine runs the same isentropic solve in reverse (dropping pressure) and applies

\eta

the other way — the real outlet temperature ends up higher than the ideal expansion, because inefficiency means less energy actually left the fluid as shaft work:

T_2 = T_1 + \eta\,(T_{2s}-T_1),\qquad W_{shaft} = F\,\big(H(T_1,P_1) - H(T_2,P_2)\big)

$T_2$: Real (actual) outlet temperature — always warmer than the ideal T_{2s} for a turbine too, since inefficiency means less energy leaves as work.
$\eta$: Isentropic efficiency (0–1): the ratio of actual to ideal work extracted.
$W_{shaft}$: Real shaft work extracted from the fluid, watts.

Because both use the flowsheet's real

H(T,P)

S(T,P)

from the active thermo package (Chapter 2) rather than an ideal-gas

\gamma

shortcut, a compressor on a non-ideal gas — dense-phase CO₂, a real natural-gas mix — gets a genuinely different answer than the textbook ideal-gas formula would. A large pressure ratio in one step is unrealistic for real machinery; MaximaLabs' multistage compressorsplits it into

n

stages of equal pressure ratio

r=(P_{out}/P_{in})^{1/n}

, each solved by the same isentropic-efficiency method above, with intercooling between stages — the same reason real plants stage compression instead of doing it in one jump.

9.3 Valves: isenthalpic throttling

A control valve drops pressure with no shaft work and (to a good approximation) no heat loss — a Joule-Thomson throttle. The energy balance collapses to a single statement: enthalpy is conserved across the valve, and the outlet temperature is whatever that enthalpy implies at the new, lower pressure:

h_{out} = h_{in}\;(\text{isenthalpic}) \;\Rightarrow\; T_{out} = T(P_{out}, h_{in})

$h_{in}$: Inlet molar enthalpy — known from the upstream stream's (T,P).
$h_{out}$: Outlet molar enthalpy — equal to h_{in}, since a valve does no shaft work and (ideally) exchanges no heat.
$P_{out}$: Known outlet (downstream) pressure — the valve's set drop.
$T_{out}$: Outlet temperature — the unknown, solved as the (P,H) flash of §3.6 at the fixed (P_{out}, h_{in}).

For an ideal gas that's no temperature change at all; for a real gas or a liquid near its bubble point it can mean significant cooling — flashing part of the stream to vapor, or (for a dense-phase fluid like the CO₂ in §9.5) dropping across a phase boundary entirely. The isenthalpic solve is exactly why a J-T valve shows up as a cheap way to get refrigeration rather than just "friction that wastes pressure."

9.4 Pipelines: friction, not equipment

Pipe itself isn't a piece of "equipment" in the traditional sense, but it drops pressure the same way a valve does — and unlike a valve, that drop is unavoidable and grows with distance. The classic Darcy-Weisbach equation:

\Delta P = f\,\frac{L}{D}\,\frac{\rho v^2}{2} + \rho g L \sin\theta,\qquad f = f(\mathrm{Re},\,\varepsilon/D)

$\Delta P$: Total pressure drop along the pipe segment, Pa.
$f$: Darcy friction factor (dimensionless) — a function of Reynolds number and relative roughness.
$L,\; D$: Pipe length and internal diameter, m.
$\rho,\; v$: Fluid mass density (kg/m³) and mean velocity (m/s).
$g,\; \theta$: Gravitational acceleration and the pipe's inclination angle from horizontal — the elevation term vanishes for a flat run (\theta=0).
$\mathrm{Re},\; \varepsilon/D$: Reynolds number and relative pipe roughness (roughness height over diameter) — the two things f depends on.

splits the loss into friction (Darcy friction factor

f

, a function of Reynolds number and relative pipe roughness

\varepsilon/D

) and elevation (the

\sin\theta

term, zero for a flat run). Over a 150 km trunk line the friction term alone can eat megapascals — which is exactly why long-distance pipeline transport needs intermediate pump/compressor stations to restore pressure, not because the fluid "runs out" of anything.

9.5 Try it

Reproduce it in your browser

1Open the CO₂ capture + compression example and Run it: an absorber recovers 90% of the flue CO₂, a compressor raises it to 8 atm, and an after-cooler brings it back to 313 K.
2Click the COMP node and open the Theory tab — the exact isentropic-efficiency equations from §9.2, with the solved $T_1$ , $T_2$ , and shaft work filled in for this stream.
3Lower the compressor efficiency and re-run — watch the outlet temperature and shaft work both rise for the same pressure ratio, exactly as §9.2 predicts.
4Open the dense-phase CO₂ pipeline example: two 150 km Darcy-Weisbach segments with a pump station (BOOST) in between restoring the pressure the first segment's friction cost.

9.6 Exercises

Practice

Work each problem yourself first, then reveal the solution to check it. Where a problem says so, reproduce it live in MaximaLabs — the solver is the answer key.

1
warm-up
A pump raises 5 mol/s of water (density 997 kg/m³, molar mass 0.018 kg/mol) from 1 atm (101,325 Pa) to 10 atm at 70% hydraulic efficiency. Estimate the shaft work.
2
core
Explain why a real compressor's outlet temperature is higher than the ideal isentropic outlet temperature, while a real turbine's outlet temperature is also higher than its ideal isentropic outlet temperature (not lower, in both cases) — using the $\eta$ equations of §9.2.
3
challenge
In the CO₂ pipeline example, the fluid enters as a liquid-like dense phase at 15 MPa. If the booster pump under-delivers and the pressure after the second 150 km segment drops below CO₂'s critical pressure (7.38 MPa) while still near 305 K, what happens to the fluid, and why does that matter operationally? (Hint: revisit the phase diagram from Chapter 2.)

Open the compression example →← Chapter 8 Chapter 10 →The Handbook