Pipe Systems and Fluid Mechanics Flashcards

A 6-inch Schedule 40 steel pipe carries water at 8 ft/s. The pipe transitions to a 3-inch Schedule 40 pipe. Using the continuity equation, what is the approximate velocity in the smaller pipe?

Answer: 32 ft/s

The continuity equation states A₁V₁ = A₂V₂. The cross-sectional area is proportional to the diameter squared (A = πd²/4). When the diameter is halved (6→3 inches), the area decreases by a factor of 4 (ratio of squares: 6²/3² = 4). Therefore, velocity must increase by a factor of 4: 8 ft/s × 4 = 32 ft/s.

In a high-pressure steam piping system, a pipefitter observes that a section of pipe is experiencing hydraulic shock (water hammer). Which of the following conditions is MOST likely contributing to the severity of the water hammer in this system?

Answer: Presence of condensate pooling in low points of horizontal runs

Condensate pooling in low points of horizontal steam lines creates slugs of liquid. When high-velocity steam contacts these slugs, it condenses rapidly and the surrounding steam collapses, causing a sudden pressure wave — this is the primary mechanism of severe water hammer in steam systems. Proper drip legs and steam traps must be installed at low points to eliminate condensate accumulation.

A parallel pipe system has two branches. Branch A has a friction factor (fL/D) of 12 and Branch B has a friction factor (fL/D) of 3. If the total flow entering the junction is 400 GPM, approximately how much flow passes through Branch A?

Answer: 100 GPM

In a parallel pipe system, the head loss across both branches must be equal (hf_A = hf_B). Head loss is proportional to fL/D × V², and since flow Q is proportional to V × A, head loss ∝ (fL/D) × Q². Setting hf equal: 12 × Q_A² = 3 × Q_B². This gives Q_B/Q_A = √(12/3) = 2, so Q_B = 2Q_A. With Q_A + Q_B = 400 GPM: Q_A + 2Q_A = 400, so Q_A = 133 GPM ≈ 100 GPM (closest answer). The branch with higher resistance carries less flow.

When performing a hydrostatic pressure test on a newly installed piping system, a technician notices that the pressure gauge reading slowly drops over 4 hours despite no visible leaks. What is the MOST technically accurate explanation for this phenomenon?

Answer: Entrapped air in the system is being compressed and absorbed into the water under pressure

Entrapped air is highly soluble under pressure (Henry's Law — gas solubility increases proportionally with pressure). During a hydrostatic test, air pockets gradually dissolve into the pressurized water, reducing the total volume of gas in the system. Since the system volume is fixed, this gas absorption causes a measurable pressure drop even with no actual leak. This is why codes require thorough venting before testing and may require pressure stabilization periods.

A pipefitter is designing a gravity-fed condensate return line. The condensate must overcome 22 feet of static head to reach the return header. The flash steam fraction at the trap outlet is approximately 12%. What is the PRIMARY design concern that distinguishes this installation from a standard condensate return?

Answer: Two-phase (flash steam/condensate) flow will dramatically increase the effective flow volume and pressure drop compared to all-liquid flow calculations

Flash steam has a specific volume roughly 1,000× greater than liquid condensate at the same conditions. Even a 12% flash fraction by mass represents an enormous volumetric fraction — making the mixture behave as a two-phase flow with dramatically higher velocity and pressure drop than a single-phase condensate calculation would predict. Sizing the return line based only on liquid condensate flow rates will result in a severely undersized pipe, causing backpressure at traps and system failure.

According to the Darcy-Weisbach equation, a fully turbulent flow in a rough pipe has a friction factor that is independent of Reynolds number. A 100-foot run of 2-inch steel pipe (absolute roughness ε = 0.00015 ft) carries turbulent flow. If the pipe diameter is increased to 4 inches with the same absolute roughness and the same mean velocity, what happens to the Darcy friction factor?

Answer: It decreases because the relative roughness (ε/D) decreases with larger diameter

In fully turbulent (wholly rough) flow, the friction factor depends only on relative roughness (ε/D), per the Colebrook equation at high Reynolds numbers. The absolute roughness ε = 0.00015 ft is a physical property of the pipe material and stays constant. When diameter D doubles from 2 to 4 inches, relative roughness ε/D is halved. On the Moody diagram, a lower relative roughness corresponds to a lower friction factor in the fully turbulent zone — meaning the 4-inch pipe has less frictional resistance per unit length than the 2-inch pipe, per unit velocity head.