Viscous flow of fluids is a fundamental concept in mechanical production, especially for technicians involved in systems where fluid movement affects machine performance and product quality. Understanding how viscous fluids behave in various configurations such as pipes and bearings enables technicians to diagnose flow problems, design efficient lubrication systems, and optimize machinery operation. In Kenya, industries ranging from food processing to pharmaceutical manufacturing rely on precise control of viscous fluid flow for consistent output and equipment longevity. This chapter focuses on applying theoretical principles to practical scenarios encountered by mechanical production technicians.
A viscous fluid resists being deformed. When one layer of fluid slides over another, a shear force acts between the layers. Newton's law of viscosity states that the shear stress is proportional to the rate at which the velocity changes across the flow:
τ = μ × du ÷ dy
where τ is the shear stress in Pa (N/m²), μ is the dynamic viscosity in Pa·s and du ÷ dy is the velocity gradient in 1/s. The kinematic viscosity is ν = μ ÷ ρ, in m²/s. Lubricating oils, greases, paints and cutting fluids are viscous, and their flow decides pressure losses in pipes, the power wasted in bearings and the thickness of the oil film on a sliding surface. The ISO viscosity grade of an oil, such as ISO VG 46, is its kinematic viscosity in mm²/s at 40 °C. For an oil of density about 870 kg/m³, this is a dynamic viscosity of about 0.04 Pa·s.
This topic covers flow that is laminar: the fluid moves in smooth layers that do not mix. In a pipe the flow is laminar when the Reynolds number Re = ρ × V × D ÷ μ is below about 2000. Other symbols used are the pressure gradient G = −dp ÷ dx in Pa/m, positive when the pressure falls in the direction of flow, and the no-slip condition: the fluid in contact with a solid surface moves at the speed of the surface.
Consider a long straight pipe of radius R and diameter D, with steady laminar flow. A force balance on a cylinder of fluid of radius r and length dx gives the shear stress at radius r:
τ = G × r ÷ 2
The shear stress is zero on the axis and rises in a straight line to a maximum at the wall, τ₀ = G × R ÷ 2. Putting τ = −μ × du ÷ dr and integrating with u = 0 at r = R (no slip) gives the velocity at radius r:
u = G ÷ (4μ) × (R² − r²)
The velocity profile is a parabola. The maximum velocity is on the axis, u_max = G × R² ÷ (4μ), and the mean velocity is half of it, V = G × R² ÷ (8μ). The discharge is:
Q = π × R⁴ × G ÷ (8μ) = π × G × D⁴ ÷ (128μ)
This is the Hagen-Poiseuille equation. Solving for the pressure drop Δp over a length L, where G = Δp ÷ L:
Δp = 128 × μ × L × Q ÷ (π × D⁴) = 32 × μ × L × V ÷ D²
Two results follow. The pressure drop is directly proportional to the viscosity, the length and the flow rate, and it is inversely proportional to the fourth power of the diameter, so halving the diameter raises the pressure drop sixteen times for the same flow. The head lost to friction is h_f = Δp ÷ (ρ × g). Written in the Darcy form, the friction factor is f = 64 ÷ Re.
Worked example: oil of viscosity 0.1 Pa·s and density 900 kg/m³ flows at 1.0 litre/s (0.001 m³/s) in a 25 mm pipe 10 m long.
The velocity profile means that the fluid at the wall is slowest and the fluid at the centre is fastest. At r = R ÷ 2 the velocity is three quarters of the maximum. Higher viscosity makes the pressure drop larger, so oil that is cold, and therefore thicker, takes more pumping power through the same pipe.
In practice, fouling or a build-up on the pipe wall reduces the diameter, and because the pressure drop depends on D⁴, even a small reduction makes a large increase in the pressure needed. A rise in pump pressure at the same flow is an early sign of a blocked or fouled line.
An annulus is the space between two concentric cylinders of radii R₁ (inner) and R₂ (outer). It is found where a shaft runs inside a sleeve and oil is pumped through the gap, and in the tube-in-tube heat exchanger. For steady laminar flow driven by a pressure gradient G, with both walls at rest, the velocity at radius r is:
u = G ÷ (4μ) × [R₂² − r² + (R₂² − R₁²) × ln(r ÷ R₂) ÷ ln(R₂ ÷ R₁)]
The velocity is zero at both walls and rises to a maximum at the radius:
r_m = √[(R₂² − R₁²) ÷ (2 × ln(R₂ ÷ R₁))]
which lies nearer to the inner wall than the middle of the gap. The shear stress is zero at r_m and changes sign there. The discharge is:
Q = π × G ÷ (8μ) × [R₂⁴ − R₁⁴ − (R₂² − R₁²)² ÷ ln(R₂ ÷ R₁)]
The mean velocity is V = Q ÷ [π × (R₂² − R₁²)]. When R₁ tends to zero, the last term vanishes and the equation becomes the Hagen-Poiseuille equation for a pipe. The pressure drop over a length L is Δp = G × L.
Worked example: oil of viscosity 0.1 Pa·s flows in the annulus between a shaft of 10 mm radius and a sleeve of 15 mm radius, with a pressure gradient of 10 kPa/m (10 000 Pa/m).
For comparison, a solid pipe of 15 mm radius with the same pressure gradient would carry Q = π × 10 000 × 0.015⁴ ÷ 0.8 = 0.00199 m³/s, which is 24 times as much. The shaft takes up part of the space and the fluid is slowed by the shear at two walls, so the flow in an annulus is far lower than in a pipe of the outer size. The result for the annulus is also very sensitive to the gap: a narrow gap passes little oil unless the pressure is high.
Two flat plates a small distance h apart, with a viscous fluid between them, model the oil film in a slider bearing or a seal, and the clearance between a piston and its cylinder when they are laid out flat. Take x along the plates and y across the gap from the lower plate (y = 0) to the upper plate (y = h). For steady laminar flow, the momentum balance is:
μ × d²u ÷ dy² = dp ÷ dx = −G
Integrating twice gives u = −G × y² ÷ (2μ) + C₁ × y + C₂, and the constants are found from the speeds of the plates. The three cases of the curriculum follow, from the simplest one.
The lower plate is at rest and the upper plate moves with speed U. With no pressure gradient (G = 0) the velocity varies in a straight line:
u = U × y ÷ h
This is Couette flow. The shear stress is the same at every level, τ = μ × U ÷ h, and the force on the moving plate is F = τ × A for a plate area A. The flow per metre of width is q = U × h ÷ 2, because the mean velocity is U ÷ 2.
When a pressure gradient is also present, the two effects add:
u = U × y ÷ h + G ÷ (2μ) × y × (h − y)
and the flow per metre of width is q = U × h ÷ 2 + G × h³ ÷ (12μ).
Worked example: oil with μ = 0.08 Pa·s fills a gap of h = 2 mm and the upper plate moves at U = 0.5 m/s.
Both plates are fixed and the flow is driven only by the pressure gradient. With u = 0 at y = 0 and y = h:
u = G ÷ (2μ) × y × (h − y)
The profile is a parabola with its maximum midway between the plates:
The flow is proportional to the cube of the gap, so halving the gap cuts the flow to one eighth for the same pressure gradient.
Worked example: oil with μ = 0.08 Pa·s flows between fixed plates 2 mm apart under a pressure gradient of 5000 Pa/m.
The upper plate moves at speed U₁ in the positive direction and the lower plate moves at speed U₂ in the opposite direction. With no pressure gradient the velocity changes in a straight line from −U₂ at the lower plate to +U₁ at the upper plate:
u = −U₂ + (U₁ + U₂) × y ÷ h
The shear stress is constant, τ = μ × (U₁ + U₂) ÷ h, and it is the relative speed of the plates that matters. The velocity is zero at y₀ = U₂ × h ÷ (U₁ + U₂), which is the mid-gap when the speeds are equal. The net flow per metre of width is q = h × (U₁ − U₂) ÷ 2, which is zero for equal speeds, because the fluid carried forward by one plate is carried back by the other. If a pressure gradient G is added, the term G ÷ (2μ) × y × (h − y) is added to the velocity.
Worked example: oil with μ = 0.08 Pa·s fills a gap of 2 mm. The upper plate moves at 0.3 m/s and the lower plate moves in the opposite direction at 0.2 m/s.
In pipe flow the velocity differs from point to point, but calculations use the mean velocity V. The kinetic energy and the momentum of the flow are larger than the values worked out from V alone, so two correction factors are used.
For the parabolic velocity profile of laminar flow in a pipe, u ÷ V = 2 × (1 − r² ÷ R²), and integrating over the area gives α = 2 and β = 4 ÷ 3. For turbulent flow the profile is much flatter and both factors are close to 1, only a few per cent higher.
The energy equation for laminar flow between two sections 1 and 2 of a pipe, in terms of head, is:
p₁ ÷ (ρg) + α × V₁² ÷ (2g) + z₁ = p₂ ÷ (ρg) + α × V₂² ÷ (2g) + z₂ + h_f
For a pipe of constant diameter the two kinetic energy terms are equal, cancel, and the pressure drop equals the friction loss. The factor α matters where the velocity changes, such as at the entry from a tank or at a change of diameter.
Worked example, for the oil in the 25 mm pipe above (ṁ = 900 × 0.001 = 0.9 kg/s, V = 2.037 m/s):
The momentum flux is used to find the force on a pipe bend, a nozzle or a valve, where the change of momentum of the flow produces a force on the part, which the supports and welded joints must carry.
Viscous shear turns mechanical energy into heat. The power absorbed is the viscous force multiplied by the velocity at which it acts. It must be supplied by the pump or the motor, and it appears as a rise in the temperature of the fluid and of the parts.
Power absorbed in a pipe. The pump has to overcome the pressure drop, so:
P = Q × Δp = 128 × μ × L × Q² ÷ (π × D⁴)
For the oil pipe in the worked example, P = 0.001 × 104 304 = 104.3 W. Doubling the flow raises the power four times for laminar flow, since the power depends on Q².
Power absorbed between sliding surfaces. For a film of thickness h between a surface of area A moving at speed U and a fixed surface, the shear force is F = μ × A × U ÷ h, and the power is:
P = F × U = μ × A × U² ÷ h
For example, a plate of area 0.2 m² slides at 2 m/s over an oil film 1 mm thick with μ = 0.1 Pa·s.
The viscous force on any surface can be found from F = μ × A × (du ÷ dy). For an area of 0.02 m², a viscosity of 0.1 Pa·s and a velocity gradient of 50 s⁻¹, the force is 0.1 × 0.02 × 50 = 0.1 N. The power is this force multiplied by the speed at which the surface moves.
The power loss is smaller with a thinner oil, a wider film and a lower speed. A lubricant is chosen to keep the film thick enough to carry the load while wasting as little power as possible.
A journal bearing carries a rotating shaft (the journal) in a sleeve. A thin film of oil fills the clearance c between the shaft and the sleeve. When the load is light, the shaft runs nearly concentric in the sleeve, and the oil film can be treated as a Couette flow between a moving surface (the shaft) and a fixed surface (the sleeve). This is Petroff's analysis.
Let the shaft have diameter D, the bearing length L, the radial clearance c and the speed N in rev/s. The surface speed is U = π × D × N. Then:
The resistance is greater for a more viscous oil, a higher speed, a larger and longer shaft and a smaller clearance. In real bearings the shaft is displaced sideways by the load, which makes the film thin on one side and thick on the other, and this wedge-shaped film supports the load. The concentric result gives a good estimate of the friction torque, and the sizes of the clearance and viscosity control both the load capacity and the resistance.
Worked example: a shaft of diameter 100 mm runs in a bearing of length 80 mm with a radial clearance of 0.05 mm at 1500 rev/min (25 rev/s). The oil has μ = 0.03 Pa·s.
The 930 W appears as heat in the oil film, which must be carried away by the oil flow and the bearing housing. Maintenance relies on this: if the viscosity falls because the oil is too hot, the film thins and metal contact can follow. If the viscosity is too high, the bearing runs hot from the extra resistance. The oil is checked for the right grade, for contamination and for temperature, and the bearing clearance is checked against wear limits.
A foot-step bearing, also called a pivot bearing, carries the axial load at the lower end of a vertical shaft. The flat end of the shaft, of radius R, rotates at angular speed ω over a flat bearing plate, with an oil film of thickness h between them.
At radius r the film surface moves at u = ω × r. The shear stress is τ = μ × ω × r ÷ h. Taking a ring of radius r and width dr, the area is 2πr dr, the friction force is τ × 2πr dr and the torque is r times this:
dT = 2π × μ × ω × r³ ÷ h × dr
Integrating from 0 to R:
T = π × μ × ω × R⁴ ÷ (2h)
and the power lost is P = T × ω = π × μ × ω² × R⁴ ÷ (2h).
The torque depends on the fourth power of the radius, so a larger end diameter brings a much higher resistance, and it falls as the film gets thicker.
Worked example: a shaft with an end radius of 50 mm turns at 300 rev/min (ω = 31.42 rad/s) over an oil film of 0.2 mm, with μ = 0.05 Pa·s.
The heat produced is carried off by the oil. For inspection, signs of excess resistance are a rise in the temperature of the bearing, motor current that rises at the same speed, and noise or vibration. Oil that is too thin allows the end of the shaft to touch the plate, and oil that is dirty increases wear. The bearing is serviced on the basis of the oil temperature and condition.
A collar bearing carries an axial load on a collar fixed to a shaft, which is a flat ring surface with an inner radius R₁ and an outer radius R₂ running against a flat plate, with an oil film of thickness h. The shaft passes through the middle, so there is no oil film within the radius R₁. The analysis is the same as for the foot-step bearing, but the integration runs from R₁ to R₂:
T = π × μ × ω × (R₂⁴ − R₁⁴) ÷ (2h)
and the power lost is P = T × ω.
When the shaft carries several collars, each collar adds a torque of this size, so the total is the number of collars multiplied by T.
Worked example: the shaft in the previous example has a collar with R₂ = 60 mm and R₁ = 30 mm, with the same film of 0.2 mm, speed of 300 rev/min (ω = 31.42 rad/s) and μ = 0.05 Pa·s.
The collar loses nearly twice as much power as the foot-step end of 50 mm radius, because its outer radius is larger. The resistance can be reduced by making the outer radius smaller, by using a thicker film or a thinner oil, or by running the bearing slower. The measures against too much resistance are the same as for the other bearings: keep the oil at the right viscosity and temperature, keep it clean, and watch the bearing temperature and power draw.
Explain how the velocity distribution in laminar flow through a circular pipe affects pressure loss in a fluid transport system. (10 marks)
Describe the factors influencing viscous resistance in journal bearings and how they impact machine performance. (10 marks)
Calculate the power absorbed in viscous flow given a lubricant viscosity of 0.1 Pa·s, a velocity gradient of 50 s⁻¹, and a flow area of 0.02 m². (Show all steps) (10 marks)
Discuss the practical considerations for maintaining foot-step bearings in heavy machinery to minimize viscous resistance related problems. (10 marks)
Compare the flow characteristics of viscous fluids through circular pipes and annuli, highlighting their implications for mechanical production technicians. (10 marks)
This chapter explored the behavior of viscous fluids flowing through various configurations, beginning with the flow of viscous fluid in circular pipes, where velocity profiles and pressure drops were examined. The discussion extended to viscous flow through an annulus, highlighting the complexities introduced by concentric cylindrical boundaries. Flow between two parallel plates was analyzed under different conditions, including one plate at rest, both plates stationary, and both plates moving in opposite directions, each affecting the fluid velocity distribution uniquely. The concept of kinetic energy momentum was introduced to understand the energy changes within viscous flows. Calculations of power absorbed in viscous flow provided insight into the energy losses due to fluid friction. The chapter also addressed viscous resistance in mechanical components such as journal bearings, foot-step bearings, and collar bearings, emphasizing how viscous forces impact their performance and efficiency. Overall, the chapter integrated theoretical principles with practical applications to deepen understanding of viscous fluid flow in engineering systems.
Question 1 key points:
Question 2 key points:
At a medium-sized textile manufacturing plant in Eldoret, the maintenance team is tasked with optimizing the lubrication system of the spinning machines, focusing on journal and collar bearings to reduce downtime caused by bearing failures.
Tasks:
a) Analyze how viscous resistance in journal bearings affects machine performance and maintenance intervals. (5 marks)
b) Propose a method to measure and monitor viscous resistance in the collar bearings used in the spinning machines. (5 marks)
c) Recommend practical steps to optimize viscous flow conditions in the lubrication system to improve overall machine reliability. (5 marks)
a) Discuss how viscous resistance creates a lubricating film that supports load and reduces metal contact, which prevents wear and overheating. Emphasize the impact on machine uptime and the cost implications of poor lubrication at the Eldoret plant.
b) Suggest using indirect methods such as temperature sensors and vibration analysis alongside lubricant viscosity tests. Outline the procedure to measure lubricant viscosity, bearing dimensions, and operating speeds to estimate viscous resistance.
c) Recommend controlling lubricant temperature to maintain optimal viscosity, scheduling regular lubricant replacement to prevent contamination, and adjusting bearing clearances to maintain appropriate film thickness. Highlight the role of staff training on lubrication best practices to sustain improvements.
Question 11 (Compulsory - 20 marks)
At Kenyatta National Hospital’s mechanical workshop, a new cooling system uses viscous oil flowing through circular pipes and journal bearings to support rotating equipment.
a) Analyze how viscous flow characteristics in circular pipes affect the cooling efficiency and suggest design considerations to optimize flow. (10 marks)
b) Evaluate the impact of viscous resistance in journal bearings on the operational reliability of the rotating equipment and propose maintenance strategies. (10 marks)
Question 12 (20 marks)
Derive the expression for velocity distribution and shear stress for viscous fluid flow between two parallel plates where both plates move in opposite directions, and discuss practical applications of this flow in mechanical systems.
Question 13 (20 marks)
Explain the concept of kinetic energy momentum in viscous flow and calculate the power absorbed by a viscous fluid flowing through a pipe with given dimensions and flow velocity. Include the importance of these calculations in designing fluid transport systems in a county government water department.
Question 14 (20 marks)
Discuss the viscous resistance encountered in collar bearings and foot-step bearings, highlighting how these resistances influence the design and performance of mechanical components in a hotel’s elevator and laundry services.
Question 11
a) Viscous flow in circular pipes causes velocity gradients that induce shear stress, leading to pressure drops which affect cooling efficiency. Optimizing pipe diameter, surface finish, and flow velocity can reduce energy losses. At Kenyatta National Hospital, designing pipes with appropriate diameters and smooth interiors ensures effective heat transfer and reduces pumping costs.
b) Viscous resistance in journal bearings arises from lubricant shear forces, affecting friction and heat generation. This resistance influences equipment reliability by potentially causing wear or seizure. Regular lubrication checks, temperature monitoring, and ensuring correct bearing clearances extend equipment life.
Question 12
The velocity distribution for flow between two plates moving oppositely is derived from solving the Navier-Stokes equations with boundary conditions of respective plate velocities. The resulting linear velocity profile and corresponding shear stress are critical for designing conveyor belts and sliding machinery parts where relative motion occurs.
Question 13
Kinetic energy momentum represents the fluid’s energy due to motion, affecting pressure and flow behavior. Power absorbed by viscous flow is computed using the shear stress and velocity gradient, typically by integrating the viscous force over the flow volume. In county water departments, these calculations help size pumps and pipes to ensure efficient water delivery with minimal energy waste.
Question 14
Viscous resistance in collar bearings arises from the lubricant film shearing between the collar and shaft, while foot-step bearings experience resistance from axial load lubrication. These resistances reduce mechanical wear but introduce frictional losses, influencing bearing design to balance load capacity and efficiency. In hotels, proper bearing design ensures smooth elevator operation and reliable laundry machinery performance.
Time: 2 Hours | Type: Individual
Resources Required:
At a Machakos County water treatment plant training centre, mechanical production technicians often need to analyze fluid flow in pipes to ensure efficient system design and maintenance.
Tasks:
i. Set up the hydraulic bench and connect the circular pipe section securely.
ii. Measure and record the volumetric flow rate and pressure drop along the pipe at different flow velocities.
iii. Calculate the Reynolds number to determine whether the flow is laminar or turbulent.
iv. Plot the velocity profile and discuss the effect of viscosity on flow behaviour.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
Technicians at a Nairobi refinery need to understand flow behaviour through annular spaces in heat exchangers and pipelines to optimize performance.
Tasks:
i. Assemble the annulus flow test rig with specified inner and outer pipe dimensions.
ii. Measure pressure drop and flow rate for varying flow velocities.
iii. Calculate flow parameters including velocity distribution and shear stress.
iv. Compare experimental results with theoretical viscous flow equations for annulus.
Assessor Observation Criteria:
Time: 1.5 Hours | Type: Individual
Resources Required:
In a Kisumu textile factory, understanding fluid flow between moving surfaces is essential for lubrication and cooling system design.
Tasks:
i. Set up the parallel plate apparatus with one plate moving at a set speed.
ii. Measure fluid velocity profile between the plates at different positions.
iii. Calculate shear stress and velocity gradient.
iv. Analyze the effect of plate speed on viscous flow characteristics.
Assessor Observation Criteria:
Time: 1 Hour | Type: Pairs
Resources Required:
At a Nairobi Technical Training Institute, students simulate viscous flow in lubrication layers by studying stationary plate configurations.
Tasks:
i. Assemble the rig ensuring both plates are stationary and parallel.
ii. Inject dye into the fluid to visualize flow patterns.
iii. Measure flow velocity and record observations.
iv. Sketch the velocity profile and discuss viscous effects.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At a Kisii dairy processing plant, technicians evaluate flow between moving surfaces to optimize lubricant application in machinery.
Tasks:
i. Configure the apparatus with both plates moving in opposite directions at set speeds.
ii. Measure fluid velocity profiles at different depths.
iii. Calculate velocity gradients and shear stresses.
iv. Discuss how opposing plate motion influences viscous flow.
Assessor Observation Criteria:
Time: 1.5 Hours | Type: Individual
Resources Required:
Mechanical technicians at a Mombasa port logistics company assess fluid momentum to improve hydraulic system efficiency.
Tasks:
i. Measure fluid velocity and pressure at two points along a pipe.
ii. Calculate kinetic energy per unit volume and momentum flow rate.
iii. Analyze the impact of viscous forces on energy loss.
iv. Prepare a report summarizing findings and implications.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At a Nakuru manufacturing plant, technicians measure power losses due to viscous flow in hydraulic motors for maintenance planning.
Tasks:
i. Set up the hydraulic rig and connect torque and speed sensors.
ii. Record torque and rotational speed at different flow rates.
iii. Calculate power absorbed due to viscous effects.
iv. Compare measured power with theoretical predictions.
Assessor Observation Criteria:
Time: 2.5 Hours | Type: Individual
Resources Required:
At a Thika manufacturing workshop, mechanical production technicians evaluate bearing resistance to predict machine performance and lifespan.
Tasks:
i. Install the journal bearing on the test rig and apply specified load.
ii. Measure torque required to rotate the shaft at various speeds.
iii. Calculate viscous resistance and friction coefficient.
iv. Discuss the influence of lubricant viscosity on bearing performance.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
Technicians at a Kisumu power plant perform tests on foot-step bearings to ensure reliability under heavy loads.
Tasks:
i. Mount the foot-step bearing on the test apparatus.
ii. Measure torque and speed at different lubricant viscosities.
iii. Calculate viscous resistance and relate to bearing design.
iv. Prepare a technical note on maintenance recommendations.
Assessor Observation Criteria:
Time: 1.5 Hours | Type: Individual
Resources Required:
At a Nairobi hotel’s maintenance workshop, technicians assess collar bearing resistance to troubleshoot noisy machinery.
Tasks:
i. Set up the collar bearing on the test rig.
ii. Measure torque at various rotational speeds.
iii. Calculate viscous resistance and analyze lubrication effectiveness.
iv. Suggest improvements for bearing operation based on results.
Assessor Observation Criteria:
Time: 1 Hour | Type: Individual
Resources Required:
In a Kisumu county hospital engineering department, technicians use velocity profiles to diagnose flow issues in medical gas delivery pipes.
Tasks:
i. Measure pressure at multiple points along the pipe radius.
ii. Calculate velocity at each point using pressure data.
iii. Sketch the velocity profile across the pipe diameter.
iv. Explain the influence of viscosity on the profile shape.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At the Technical University of Kenya, mechanical production students use simulation software to predict flow behaviour in manufacturing processes.
Tasks:
i. Set up a simulation model of viscous fluid flow between two parallel plates.
ii. Input fluid properties and plate velocities as per given parameters.
iii. Run the simulation and extract velocity and shear stress data.
iv. Compare simulation results with theoretical calculations.
Assessor Observation Criteria: