Sizing a Steel Penstock for Hydropower

Context: Pressurized HydraulicsBranch of fluid mechanics dealing with confined liquid flow under pressure, typical in pipes. and PenstockA conduit or pipe that carries water from a reservoir to a hydroelectric power plant turbine. Design.

In this exercise, you will act as a hydraulic engineer tasked with designing a steel penstock for a medium-head hydroelectric power plant in Colorado. You must determine the optimal diameter based on velocity constraints, calculate the head losses, determine the net head available at the turbine, and specify the required wall thickness according to US standards (ASME). This task is critical to ensure both the structural integrity of the system and the efficiency of energy generation.

Pedagogical Note: This exercise integrates fluid dynamics concepts (Bernoulli, Darcy-Weisbach) with structural mechanics (Hoop Stress). You will work exclusively with US Customary units (ft, cfs, psi) as is standard in the American industry.

Learning Objectives

Determine the pipe diameter based on allowable velocity criteria.
Calculate major friction losses using the Darcy-Weisbach equation.
Calculate Net Head (\(H_{\text{net}}\)) available for the turbine.
Determine the required pipe wall thickness to withstand hydrostatic pressure.

Study Data

The project involves connecting an upper reservoir to a powerhouse. The conduit (penstock) is made of welded steel.

System Specifications

Characteristic	Value
Material	Welded Steel (ASTM A516 Gr 70)
Design Flow Rate	Steady State
Terrain Profile	Constant slope

Hydraulic Profile Schematic

Parameter	Symbol	Value	Unit
Design Flow Rate	\(Q\)	500	cfs (ft³/s)
Gross Head	\(H_{\text{gross}}\)	300	ft
Penstock Length	\(L\)	1200	ft
Absolute Roughness	\(\epsilon\)	0.00015	ft
Allowable Stress	\(S\)	20,000	psi

Questions to Address

Determine the optimal pipe diameter (\(D\)) assuming a target velocity of 15 ft/s.
Calculate the head loss due to friction (\(h_f\)) using the Darcy-Weisbach equation.
Determine the Net Head (\(H_{\text{net}}\)) available at the turbine inlet.
Calculate the minimum required wall thickness (\(t\)) to withstand the static pressure (Hoop Stress).
Synthesize the results: Determine the theoretical power output assuming 90% efficiency.

Fundamentals of Hydraulics & Mechanics

To solve this problem, we apply the principles of conservation of mass, energy, and basic solid mechanics.

1. Continuity Equation
Relates flow rate (\(Q\)), flow area (\(A\)), and velocity (\(V\)). \[ Q = A \times V = \frac{\pi D^2}{4} \times V \]

2. Darcy-Weisbach Equation
Used to calculate major head losses (\(h_f\)) in a pipe. \[ h_f = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g} \] Where \(g = 32.2 \text{ ft/s}^2\) and \(f\) is the friction factor (Moody).

3. Barlow's Formula / Hoop Stress
Used to determine the thickness (\(t\)) of a thin-walled pressure vessel. \[ t = \frac{P \cdot D}{2 \cdot S \cdot E} \] Where \(P\) is internal pressure, \(S\) is allowable stress, and \(E\) is joint efficiency (assumed 1.0 here).

Solution: Sizing a Steel Penstock for Hydropower

Question 1: Determine Diameter (\(D\))

Principle

We determine the optimal diameter by balancing hydraulic performance (low friction) with economic cost (less steel). A key parameter for this initial sizing is the fluid velocity. For penstocks, a velocity between 10 and 20 ft/s is typical to balance pipe cost (small D) vs. head loss (large D). High velocity leads to smaller, cheaper pipes but higher energy losses.

Mini-Lesson

Continuity Principle: For an incompressible fluid like water, the flow rate is constant throughout the pipe. \(Q = Area \times Velocity\). This fundamental law means that to reduce velocity for a fixed flow, we must increase the cross-sectional area (and thus diameter).

Pedagogical Note

Think of the pipe as a highway: if you want 500 cars (cfs) to pass at a reasonable speed (15 ft/s), you need a specific number of lanes (Diameter). Too narrow, and traffic jams (friction) occur. Too wide, and you waste asphalt (steel) for empty lanes.

Norms

According to the US Bureau of Reclamation (USBR) Engineering Monograph No. 7, economic velocities for steel penstocks typically range from 10 to 15 ft/s for smaller plants and up to 20 ft/s for high-head plants. Exceeding these limits can lead to vibration and excessive liner wear.

Formula(s)

Derived from Continuity Equation \(Q = A \cdot V = \frac{\pi D^2}{4} \cdot V\):

D = \sqrt{\frac{4Q}{\pi V}}

Hypotheses

We assume:

Steady, incompressible flow (constant density).
Full pipe flow (pressurized, no free surface).
Uniform velocity profile across the section.

Data

Parameter	Symbol	Value	Unit
Flow Rate	\(Q\)	500	cfs
Target Velocity	\(V\)	15	ft/s

Tips

Don't forget that the formula outputs the diameter in feet because Q is in cubic feet/sec and V is in feet/sec. No unit conversion is needed here, but always check dimensional consistency.

Schematic (Before Calculations)

Flow Parameters

Calculation(s)

Step 1: Determine Required Flow Area

Using the Continuity Equation \(A = Q/V\), we calculate the area needed to pass 500 cfs at 15 ft/s.

\begin{aligned} A_{\text{required}} &= \frac{Q}{V} \\ &= \frac{500 \text{ ft}^3/\text{s}}{15 \text{ ft/s}} \\ &\approx 33.333 \text{ ft}^2 \end{aligned}

This result indicates that we need a pipe with an internal cross-sectional area of approximately 33.3 square feet.

Step 2: Solve for Diameter

Next, we relate this Area to the Diameter using the geometric formula for a circle, \(A = \frac{\pi \cdot D^2}{4}\). Solving for \(D\), we get \(D = \sqrt{\frac{4A}{\pi}}\).

\begin{aligned} D &= \sqrt{\frac{4 \times A_{\text{required}}}{\pi}} \\ &= \sqrt{\frac{4 \times 33.333}{\pi}} \\ &= \sqrt{\frac{133.333}{3.14159}} \\ &= \sqrt{42.44} \\ &\approx 6.515 \text{ ft} \end{aligned}

The calculated exact diameter is 6.515 ft. In engineering practice, we select the nearest standard size. Here, 6.5 ft (78 inches) is a standard fabrication size that is very close to the optimal value.

Schematic (After Calculations)

Sized Pipe Cross-Section

Analysis

A 6.5 ft diameter pipe is quite large (approx. 78 inches), which is reasonable for such a high flow rate. Reducing it further would cause the velocity to exceed 15 ft/s, which would increase friction losses quadratically (\(h_f \propto V^2\)). Increasing it would increase material costs linearly.

Cautionary Points

In a real project, ensure available standard pipe sizes are checked. 78-inch pipe is custom but standard for penstocks. Avoid specifying non-standard fractions like 6.51 ft.

Key Takeaways

Diameter is proportional to the square root of flow rate.
Velocity is the key constraint for initial sizing to balance cost vs. efficiency.

Did You Know?

The "Economic Diameter" is mathematically defined as the size where the marginal cost of larger pipe equals the marginal value of energy saved from reduced friction over the project life.

FAQ

Final Result

Optimal Diameter \(D \approx 6.5\) ft

Your Turn

If we reduced the target velocity to 10 ft/s, what would be the required diameter in ft?

Memo Card

Q1 Summary:

Concept: Continuity (Conservation of Mass)
Formula: \(D = \sqrt{4Q / \pi V}\)
Result: 6.5 ft

Question 2: Calculate Head Loss (\(h_f\))

Principle

Friction between the flowing water and the fixed steel wall dissipates energy in the form of heat and turbulence. We calculate this "head loss" to quantify exactly how much potential energy is lost before reaching the turbine.

Mini-Lesson

Friction Factor (\(f\)): In the Darcy-Weisbach equation, \(f\) represents the dimensionless resistance coefficient. It depends on the Reynolds number (\(Re\)) and relative roughness (\(\epsilon/D\)). For high \(Re\) (turbulent flow), \(f\) becomes independent of velocity and depends only on roughness.

Pedagogical Note

Head loss is expressed as "energy per unit weight". It is measured in feet, meaning the pressure drop caused by friction is equivalent to losing that specific height of water column from the reservoir.

Norms

We use the Darcy-Weisbach equation as it is universally accepted for pressurized flow. Friction factors are typically derived from the Moody Chart or Colebrook-White equation. For welded steel pipes, \(f\) typically ranges from 0.010 to 0.015 depending on age and coating.

Formula(s)

Darcy-Weisbach Equation:

h_f = f \cdot \frac{L}{D} \cdot \frac{V^2}{2g}

Hypotheses

We assume steady flow. The friction factor will be calculated for fully developed turbulent flow using the Colebrook-White equation.

Data

Parameter	Value
Length (\(L\))	1200 ft
Roughness (\(\epsilon\))	0.00015 ft
Diameter (\(D\))	6.5 ft
Kinematic Viscosity (\(\nu\))	\(1.217 \times 10^{-5} \text{ ft}^2/\text{s}\)
Gravity (\(g\))	32.2 ft/s²

Tips

Always recalculate the actual velocity based on your chosen standard diameter (6.5 ft) before calculating loss. The velocity might differ slightly from the initial target of 15 ft/s.

Schematic (Before Calculations)

Energy Line Concept

Calculation(s)

Step 1: Calculate Actual Flow Area & Velocity

Since we rounded the diameter to 6.5 ft, the area and velocity will shift slightly. First, we calculate the precise area of the 6.5 ft pipe.

\begin{aligned} A_{\text{actual}} &= \frac{\pi \times (6.5)^2}{4} \\ &= \frac{3.14159 \times 42.25}{4} \\ &\approx 33.18 \text{ ft}^2 \end{aligned}

\begin{aligned} V_{\text{actual}} &= \frac{Q}{A_{\text{actual}}} \\ &= \frac{500}{33.18} \\ &\approx 15.07 \text{ ft/s} \end{aligned}

The actual velocity is 15.07 ft/s. We will use this value to find the Reynolds number.

Step 2: Calculate Reynolds Number & Relative Roughness

The Reynolds number determines if the flow is laminar or turbulent. \(Re = \frac{V D}{\nu}\).

\begin{aligned} Re &= \frac{15.07 \times 6.5}{1.217 \times 10^{-5}} \\ &\approx 8.05 \times 10^6 \end{aligned}

Relative roughness \(\frac{\epsilon}{D} = \frac{0.00015}{6.5} \approx 0.000023\).

Step 3: Determine Friction Factor (Colebrook-White Iteration)

The Colebrook-White equation is implicit (cannot isolate \(f\)), so we solve it iteratively: \(\frac{1}{\sqrt{f}} = -2 \log_{10}\left( \frac{\epsilon/D}{3.7} + \frac{2.51}{Re \sqrt{f}} \right)\).

Iteration 1: Assume an initial guess \(f_0 = 0.010\). Calculate the right hand side (RHS).

\begin{aligned} \text{RHS} &= -2 \log_{10}\left( \frac{0.000023}{3.7} + \frac{2.51}{8.05 \times 10^6 \times \sqrt{0.010}} \right) \\ &= -2 \log_{10}(6.22 \times 10^{-6} + 3.12 \times 10^{-6}) \\ &= -2 \log_{10}(9.34 \times 10^{-6}) \\ &\approx 10.059 \end{aligned}

f_1 = (1 / 10.059)^2 \approx 0.00988

Iteration 2: Use \(f_1 = 0.00988\) as the new guess.

\begin{aligned} \text{RHS} &= -2 \log_{10}\left( \frac{0.000023}{3.7} + \frac{2.51}{8.05 \times 10^6 \times \sqrt{0.00988}} \right) \\ &\approx 10.058 \end{aligned}

f_2 = (1 / 10.058)^2 \approx 0.00988

The value has stabilized. We will use \(f = 0.0099\) (rounded slightly up for conservatism).

Step 4: Calculate Friction Head Loss

Now we substitute the converged Friction Factor (\(f=0.0099\)), Length (\(L=1200\) ft), Diameter (\(D=6.5\) ft), Velocity (\(V=15.07\) ft/s), and Gravity (\(g=32.2\) ft/s²) into the Darcy-Weisbach equation.

\begin{aligned} h_f &= f \cdot \frac{L}{D} \cdot \frac{V^2}{2g} \\ &= 0.0099 \times \left( \frac{1200}{6.5} \right) \times \left( \frac{15.07^2}{2 \times 32.2} \right) \\ &= 0.0099 \times 184.61 \times 3.526 \\ &\approx 6.45 \text{ ft} \end{aligned}

The friction implies that for every unit of water traveling down this pipe, 6.45 feet of 'push' (pressure head) is lost to heat and turbulence.

Schematic (After Calculations)

Visualizing Loss

Analysis

A loss of ~6.45 ft represents about 2.15% of the gross head (300 ft). This is an extremely efficient design; typically, we aim for losses under 5% to maximize revenue. If losses were higher (e.g., 10%), we would need to increase the diameter.

Cautionary Points

Do not forget "minor losses" (bends, valves, entrances) in a real final design. While we only calculate "major" friction losses here, minor losses can add another 0.5-1.0 ft.

Key Takeaways

Loss is proportional to Length and Velocity squared.
Loss is inversely proportional to Diameter.

Did You Know?

Over time, steel pipes can develop "tuberculation" (rust nodules), which effectively increases the roughness \(\epsilon\). This can double friction losses after 20-30 years if not properly coated.

FAQ

Final Result

Head Loss \(h_f \approx 6.45\) ft

Your Turn

If the length was doubled to 2400 ft (keeping everything else same), what would be the head loss?

Memo Card

Q2 Summary:

Concept: Friction Loss
Formula: Darcy-Weisbach
Result: ~6.45 ft

Question 3: Determine Net Head (\(H_{\text{net}}\))

Principle

Net Head is the actual pressure head available to do useful mechanical work at the turbine runner. It is defined as the Gross Head minus all hydraulic losses (friction + minor) in the system.

Mini-Lesson

Bernoulli Principle: Total energy is conserved. \(Energy_{\text{in}} = Energy_{\text{out}} + \text{Losses}\). In our case: \(H_{\text{gross}} = H_{\text{net}} + h_f\). Therefore, Net Head is what remains for the turbine after friction "taxes" the energy.

Pedagogical Note

This is simply an accounting subtraction. Think of Gross Head as your gross salary, Losses as taxes, and Net Head as your take-home pay.

Norms

Standards like IEC 60193 and ASME PTC 18 define exactly how Net Head is measured for performance testing, usually requiring pressure gauges directly at the turbine inlet.

Formula(s)

H_{\text{net}} = H_{\text{gross}} - h_f

Hypotheses

We assume minor losses are negligible for this specific exercise step, focusing only on the major friction loss calculated in Q2.

Data

Parameter	Value
Gross Head (\(H_{\text{gross}}\))	300 ft
Head Loss (\(h_f\))	6.45 ft

Tips

Ensure consistent units. Do not subtract pressure (psi) from Head (ft). Both must be in feet of water column.

Schematic (Before Calculations)

Head Balance Concept

Calculation(s)

This is a direct subtraction of the friction loss calculated in Step 2 from the total elevation difference.

\begin{aligned} H_{\text{net}} &= H_{\text{gross}} - h_f \\ &= 300 \text{ ft} - 6.45 \text{ ft} \\ &= 293.55 \text{ ft} \end{aligned}

This means that out of the original 300 feet of elevation, effectively 293.55 feet of water column pressure is available to spin the turbine.

Schematic (After Calculations)

Energy Budget

Analysis

We retain over 97% of the potential energy. The design is hydraulically efficient. This value will be used to calculate the actual power output of the plant.

Cautionary Points

Ensure you subtract, not add, the losses! A common mistake is to assume \(H_{\text{net}} > H_{\text{gross}}\).

Key Takeaways

Net head is always less than gross head.
Turbine manufacturers rate their machines for Net Head, not Gross Head.

Did You Know?

The type of turbine selected (Pelton, Francis, Kaplan) depends almost entirely on this Net Head value. 300 ft (approx 90m) puts this in the Francis turbine range.

FAQ

Final Result

\(H_{\text{net}} \approx 293.6\) ft

Your Turn

If we ignored friction (\(h_f=0\)), what would \(H_{\text{net}}\) be?

Memo Card

Q3 Summary:

Concept: Energy Budget
Result: 293.6 ft

Question 4: Calculate Wall Thickness (\(t\))

Principle

The pipe must contain the internal water pressure without bursting. The pressure is highest at the bottom of the penstock (at the turbine). We calculate the wall thickness required to keep the material stress below a safe limit.

Mini-Lesson

Hoop Stress: In a cylindrical pipe under pressure, the force acts to separate the two halves of the pipe. This stress acts perpendicular to the radius and length. We treat the pipe as a "thin-walled pressure vessel", assuming stress is uniform across the thickness.

Pedagogical Note

Imagine the pipe trying to burst open like a sausage casing. The steel wall thickness provides the resisting area to hold that pressure back.

Norms

ASME Boiler and Pressure Vessel Code Section VIII or B31.3 is typically used. We apply a joint efficiency factor (\(E\)), usually 1.0 for fully X-rayed welds, and 0.85 for spot-checked welds.

Formula(s)

Barlow's Formula (Hoop Stress):

t = \frac{P \cdot D}{2 \cdot S \cdot E}

Hypotheses

Thin-walled assumption is valid (\(t/D < 0.1\)). Static pressure only (surge pressure handled in 'Your Turn'). Joint efficiency \(E=1.0\).

Data

Gross Head \(H = 300\) ft. Allowable Stress \(S = 20,000\) psi. Diameter \(D = 6.5 \text{ ft} = 78 \text{ in}\).

Tips

The most common error is units mismatch. Allowable stress is in PSI (lbs/in²), so Diameter must be in Inches, and Pressure must be converted from feet to PSI.

Schematic (Before Calculations)

Forces on Pipe Wall

Calculation(s)

Step 1: Convert Head to Pressure (psi)

First, we must quantify the hydrostatic pressure at the bottom of the pipe. We convert the Gross Head (\(H=300\) ft) into pressure (psi). Note that 1 ft of water \(\approx\) 0.433 psi (62.4 / 144).

\begin{aligned} P &= \frac{H \times 62.4}{144} \\ &= \frac{300 \times 62.4}{144} \\ &= 300 \times 0.4333 \\ &\approx 130 \text{ psi} \end{aligned}

The internal pressure pushing outwards on the pipe walls is approximately 130 psi.

Step 2: Convert Diameter to Inches

Barlow's formula requires consistent units. Since stress \(S\) is in psi (pounds per square *inch*), our diameter must also be in inches.

\begin{aligned} D &= 6.5 \text{ ft} \times 12 \text{ in/ft} \\ &= 78 \text{ inches} \end{aligned}

Step 3: Calculate Thickness

Now we substitute the pressure (\(P=130\) psi), diameter (\(D=78\) in), allowable stress (\(S=20,000\) psi), and joint efficiency (\(E=1.0\)) into the Hoop Stress formula.

\begin{aligned} t &= \frac{P \cdot D}{2 \cdot S \cdot E} \\ &= \frac{130 \times 78}{2 \times 20,000 \times 1.0} \\ &= \frac{10,140}{40,000} \\ &\approx 0.254 \text{ inches} \end{aligned}

The calculation shows a minimum structural thickness of 0.254 inches. Any thickness less than this would risk bursting the pipe under static load.

Schematic (After Calculations)

Wall Thickness Detail

Analysis

The theoretical minimum thickness is about 1/4 inch. However, in practice, we add a "corrosion allowance" (e.g., 1/16") and consider "water hammer" (pressure surges), so we might specify 0.375 (3/8") or 0.5 inch plate.

Cautionary Points

Always convert Diameter to Inches when working with PSI (Pounds per Square Inch). Using feet would give a result 12 times too small!

Key Takeaways

Thickness increases linearly with Pressure and Diameter.
Higher strength steel (ASTM A517) allows for thinner walls but is harder to weld.

Did You Know?

Early penstocks were made of wood staves held together by steel hoops, similar to a barrel! The wood sealed when wet due to swelling.

FAQ

Final Result

Minimum thickness \(t \approx 0.254\) inches

Your Turn

If the pressure surged to 200 psi (due to water hammer), what would the required thickness be?

Memo Card

Q4 Summary:

Concept: Hoop Stress
Formula: Barlow's
Result: ~0.254 in

Question 5: Theoretical Power Output

Principle

Power depends on flow, net head, and efficiency. This calculation tells us the potential revenue of the plant. It combines hydraulic energy (Q, H) with mechanical conversion.

Mini-Lesson

Power Formula: Power is Work over Time. \(P = \rho \cdot g \cdot Q \cdot H \cdot \eta\). The constants depend on the units used (kW, HP, MW). In US units, to get kW from cfs and ft, we divide by 11.8.

Pedagogical Note

Remember: Head is height (push), Flow is volume (amount). Combining them gives power. High head with low flow (Swiss Alps) can produce as much power as low head with high flow (Mississippi River).

Norms

NEMA MG 1 standards govern generator ratings in the US. Efficiency \(\eta\) typically ranges from 0.85 to 0.93 for modern Francis turbines.

Formula

P (\text{kW}) = \frac{Q \cdot H_{\text{net}} \cdot \eta}{11.8}

Note: The constant 11.8 comes from: \(\frac{62.4 \text{ lb/ft}^3}{550 \text{ ft-lb/s per HP}} \times 0.746 \text{ kW/HP} \approx \frac{1}{11.81}\).

Hypotheses

Water density assumed standard (62.4 lb/ft³). Overall efficiency (turbine + generator) is assumed to be 90%.

Data

\(Q=500\), \(H_{\text{net}}=293.55\), Efficiency \(\eta = 0.90\).

Tips

If calculating in Horsepower (HP), divide by 8.82 instead of 11.8. kW is generally preferred for electrical output.

Schematic (Before Calculations)

Power Equation Inputs

Calculation

Step 1: Apply Power Formula

We calculate the power delivered to the generator shaft. We use the Net Head (\(H_{\text{net}}\)) because it accounts for losses, and the Flow Rate (\(Q\)). The factor 11.8 converts these specific US units into Kilowatts.

\begin{aligned} P (\text{kW}) &= \frac{Q \cdot H_{\text{net}} \cdot \eta}{11.8} \\ &= \frac{500 \times 293.55 \times 0.90}{11.8} \end{aligned}

Step 2: Compute Numerator and Divide

Substituting the values:

\begin{aligned} P (\text{kW}) &= \frac{132,097}{11.8} \\ &\approx 11,195 \text{ kW} \end{aligned}

\begin{aligned} P (\text{MW}) &= \frac{11,195}{1000} \\ &\approx 11.2 \text{ MW} \end{aligned}

The result is approximately 11,195 kW. Converting to Megawatts by dividing by 1000 gives us the final plant capacity rating.

Schematic (After Calculations)

Final Output Visualization

Analysis

This is a significant output (11.2 MW), suitable for supplying roughly 9,000 homes (assuming avg US usage). It classifies as a "Small Hydro" project.

Cautionary Points

Do not confuse MW (Megawatts) with MWh (Megawatt-hours). MW is power (rate), MWh is energy (quantity).

Key Takeaways

Power is linear with Head and Flow.
Efficiency losses directly reduce revenue, making high-efficiency turbines worth the cost.

Did You Know?

Grand Coulee Dam produces over 6,800 MW, which is about 600 times larger than this project!

FAQ

Final Result

Power Output \(\approx 11.2\) MW

Your Turn

What is 11.2 MW in Horsepower? (Multiply kW by 1.341)

Memo Card

Q5 Summary:

Concept: Power Generation
Result: 11.2 MW

Interactive Tool: Optimization Simulator

Use this tool to see how changing the pipe diameter affects the head loss and the required wall thickness. Find the "sweet spot"!

Input Parameters

Flow Rate \(Q\) (cfs) 500 cfs

Diameter \(D\) (ft) 6.5 ft

Key Indicators

Velocity (ft/s) -

Head Loss \(h_f\) (ft) -

Final Quiz: Test Your Knowledge

1. Why do we limit the maximum velocity in the penstock?

To increase the water temperature.
To minimize head loss and prevent water hammer issues.
To make the pipe look smaller.

2. If we double the diameter of the pipe (keeping \(Q\) constant), what happens to the velocity?

It is reduced by a factor of 4.
It is cut in half.
It doubles.

(Because Area depends on \(D\) squared!)

3. Which unit is used for pressure in the US Customary system?

Pascal (Pa)
Newton (N)
Pounds per Square Inch (psi)

Glossary

cfs: Cubic Feet per Second. The standard unit of flow rate in the US.
Head Loss (\(h_f\)): Energy loss due to friction and turbulence in the pipe, measured in feet of water.
Water Hammer: A pressure surge or wave caused when a fluid in motion is forced to stop or change direction suddenly.
Hoop Stress: The mechanical stress defined for rotationally-symmetric objects (like pipes) being the force perpendicular to the radius.