By the end of this chapter, you will be able to:
Mastering these skills will help you confidently handle real-world fluid flow challenges in your trade, ensuring efficient and effective system designs.
Compiled by: Martin Karanja
Viscous flow of fluids is a critical aspect of automotive engineering, influencing how lubricants behave within engine components and how fuel or coolant circulates through various channels. Understanding viscous flow enables automotive engineers to optimize the performance and longevity of vehicles by minimizing energy losses and preventing component wear. In Kenya, where automotive workshops and manufacturing plants seek efficient fluid management, mastery of viscous flow principles ensures better maintenance of engines and hydraulic systems, contributing to cost savings and reliability.
Viscous flow is the movement of a fluid in which internal friction (viscosity) controls the velocity profile and the energy losses. It is the flow found in lubricating oil in engine bearings, in hydraulic fluid in brake and power steering lines, and in fuel and coolant in narrow passages. A mechanical or automotive technician uses the equations of viscous flow to estimate flow rates, friction forces, and power losses, and so to choose the right oil, clearance, and pipe size.
Dynamic viscosity (μ) is measured in pascal seconds (Pa·s). It is the ratio of shear stress to the velocity gradient: τ = μ × (du ÷ dy).
The flow of a viscous fluid in a circular pipe is the basis for understanding how oil, fuel, or coolant moves through lines and galleries. At low speed the flow is laminar, in smooth layers, and at higher speed it becomes turbulent. Example: a hydraulic repair workshop checks whether a replacement hose of smaller bore will still deliver enough oil to a steering rack.
The velocity distribution in laminar flow is parabolic, with the maximum velocity at the centre and zero velocity at the wall because of the no-slip condition. The maximum velocity is twice the mean velocity. The shear stress is zero at the centre and greatest at the wall.
The volumetric flow rate Q for laminar flow in a circular pipe is:
Q = π × ΔP × r⁴ ÷ (8 × μ × L)
where ΔP is the pressure difference, r is the pipe radius, μ is the dynamic viscosity, and L is the pipe length. Because r is raised to the fourth power, halving the radius reduces the flow to one sixteenth, which is why a partly blocked fuel line or oil gallery restricts flow so severely.
The Reynolds number decides the flow regime:
Re = ρ × V × D ÷ μ
where ρ is the density, V is the mean velocity, and D is the pipe diameter. Flow in a pipe is laminar below about Re = 2300 and may become turbulent above it, which raises the energy losses.
Oil of viscosity 0.05 Pa·s and density 850 kg/m³ flows through a pipe of radius 2 mm and length 0.5 m, with a pressure difference of 20 kPa.
Formula: Q = π × ΔP × r⁴ ÷ (8 × μ × L)
Substitution: Q = π × 20,000 × (0.002)⁴ ÷ (8 × 0.05 × 0.5)
Working: π × 20,000 × 1.6 × 10⁻¹¹ = 1.005 × 10⁻⁶, and 8 × 0.05 × 0.5 = 0.2, so Q = 5.03 × 10⁻⁶
Result: the flow rate is 5.03 × 10⁻⁶ m³/s, which is 5.03 cm³/s.
Check of the flow regime: V = Q ÷ (π × r²) = 5.03 × 10⁻⁶ ÷ (π × 0.002²) = 0.40 m/s, and Re = 850 × 0.40 × 0.004 ÷ 0.05 = 27. This is far below 2300, so the flow is laminar and the equation applies.
An annulus is the space between two concentric cylinders, for example a shaft in a bearing or a piston in its cylinder. Viscous flow through the annulus carries lubricant, controls leakage, and removes heat. Example: an engine rebuilding workshop checks the clearance between a plunger and its barrel, because a larger gap leaks more oil.
The passage has an inner radius ri and an outer radius ro. The velocity is zero at both walls when they are stationary, and it is greatest between them, nearer the inner wall. The shear stress acts on both surfaces.
For laminar flow driven by a pressure difference ΔP over a length L:
Q = π × ΔP ÷ (8 × μ × L) × [ro⁴ − ri⁴ − (ro² − ri²)² ÷ ln (ro ÷ ri)]
When the gap is thin, so that h = ro − ri is small compared with the radius, the annulus behaves like two parallel plates of width 2π × r (r is the mean radius):
Q = 2π × r × h³ × ΔP ÷ (12 × μ × L)
A plunger of mean radius 20 mm has a radial clearance of 0.2 mm and a length of 30 mm. The oil has a viscosity of 0.02 Pa·s, and the pressure difference is 100 kPa.
Formula: Q = 2π × r × h³ × ΔP ÷ (12 × μ × L)
Substitution: Q = 2π × 0.02 × (0.0002)³ × 100,000 ÷ (12 × 0.02 × 0.03)
Working: 0.1257 × 8 × 10⁻¹² × 100,000 = 1.005 × 10⁻⁷, and 12 × 0.02 × 0.03 = 0.0072, so Q = 1.40 × 10⁻⁵
Result: the leakage flow is 1.40 × 10⁻⁵ m³/s, which is 14.0 cm³/s.
Viscous flow between two parallel plates models thin oil films between flat surfaces, such as a slipper pad or a wet clutch disc. Two effects act together: the pressure gradient drives Poiseuille flow, and the movement of a plate drags the fluid in Couette flow. For a gap h, with the lower plate at y = 0 and the upper plate at y = h moving at a velocity U, the velocity is:
u = U × y ÷ h − (1 ÷ (2 × μ)) × (dp ÷ dx) × (h × y − y²)
The flow rate per metre width of the plates is q = U × h ÷ 2 − h³ ÷ (12 × μ) × (dp ÷ dx). Example: an automatic gearbox workshop considers the oil film between a clutch plate and its pressure plate as a thin gap between two flat surfaces.
When there is no pressure gradient, the velocity varies linearly from zero at the fixed plate to U at the moving plate. This is simple Couette flow. The shear stress is constant across the film: τ = μ × U ÷ h, and the force needed to move the plate is F = τ × A.
Worked example: a plate of area 0.02 m² slides at 2 m/s over an oil film of thickness 0.5 mm and viscosity 0.1 Pa·s.
Formula: τ = μ × U ÷ h, then F = τ × A
Substitution: τ = 0.1 × 2 ÷ 0.0005
Working: τ = 400 Pa, and F = 400 × 0.02 = 8
Result: the shear stress is 400 Pa and the force is 8 N.
When both plates are fixed, the flow is driven only by the pressure gradient. This is plane Poiseuille flow. The velocity profile is parabolic, with zero velocity at both plates:
u = (1 ÷ (2 × μ)) × (−dp ÷ dx) × y × (h − y)
The flow rate per metre width is q = h³ ÷ (12 × μ) × (−dp ÷ dx), the maximum velocity is at the centre, and the mean velocity is two-thirds of the maximum.
Worked example: a gap of 1 mm carries oil of viscosity 0.05 Pa·s, with a pressure drop of 50 kPa over 0.5 m.
Formula: q = h³ × (−dp ÷ dx) ÷ (12 × μ)
Substitution: −dp ÷ dx = 50,000 ÷ 0.5 = 100,000 Pa/m, so q = (0.001)³ × 100,000 ÷ (12 × 0.05)
Working: 1 × 10⁻⁴ ÷ 0.6 = 1.67 × 10⁻⁴
Result: the flow rate is 1.67 × 10⁻⁴ m³/s per metre width of the plates.
If the upper plate moves at +U and the lower plate at −U, and there is no pressure gradient, the velocity is linear from −U to +U across the gap, with zero velocity in the middle. The relative velocity of the plates is 2U, so the shear stress is τ = 2 × μ × U ÷ h. The net flow rate is zero, because the two halves of the film move in opposite directions.
Worked example: plates move at 1.5 m/s in opposite directions over a film of 0.4 mm and viscosity 0.08 Pa·s.
Formula: τ = μ × (U₁ + U₂) ÷ h
Substitution: τ = 0.08 × (1.5 + 1.5) ÷ 0.0004
Working: 0.08 × 3 = 0.24, and 0.24 ÷ 0.0004 = 600
Result: the shear stress is 600 Pa.
The force needed to keep the motion is the shear stress multiplied by the contact area. This is the basis for designing hydraulic clutches and thrust pads, where oil clearance and viscosity are set to give the required drag without wear.
In viscous flow the velocity is not uniform across the section, so the true kinetic energy and momentum flow are different from the values calculated with the mean velocity. Correction factors are used to fix this. Example: a power steering design check that uses the mean velocity alone would underestimate the kinetic energy of laminar flow in a pipe by half.
The kinetic energy correction factor α is the ratio of the true kinetic energy flow to the kinetic energy flow calculated with the mean velocity V:
α = (1 ÷ A) × ∫ (u ÷ V)³ dA
The true kinetic energy per unit volume is α × ρ × V² ÷ 2. For laminar flow in a circular pipe α = 2, and for turbulent flow it is about 1.05 to 1.10.
The momentum correction factor β is the ratio of the true momentum flow to that calculated with the mean velocity:
β = (1 ÷ A) × ∫ (u ÷ V)² dA
The momentum flow is β × ρ × Q × V. For laminar flow in a circular pipe β = 4 ÷ 3, and for turbulent flow it is about 1.02.
Worked example: oil of density 850 kg/m³ has a mean velocity of 0.4 m/s in laminar flow in a pipe.
Formula: kinetic energy per unit volume = α × ρ × V² ÷ 2
Substitution: 2 × 850 × 0.4² ÷ 2
Working: 850 × 0.16 = 136
Result: the true kinetic energy is 136 J/m³, twice the 68 J/m³ given by the mean velocity alone.
Viscous forces turn mechanical energy into heat, so power must be supplied to keep the fluid moving. The power absorbed raises the temperature of the oil, reduces efficiency, and affects the choice of lubricant. Example: a workshop that tests a gear oil pump finds that thicker oil takes more drive power.
For a sheared film the power absorbed is:
P = τ × V × A
where τ is the shear stress, V is the velocity of the moving surface, and A is the area. For flow through a pipe the power needed to push the fluid is P = ΔP × Q.
Worked example for a film: the plate of the earlier example, with τ = 400 Pa, V = 2 m/s, and A = 0.02 m².
Formula: P = τ × V × A
Substitution: P = 400 × 2 × 0.02
Working: 800 × 0.02 = 16
Result: the power absorbed is 16 W, the same as F × V = 8 × 2.
Worked example for a pipe: with ΔP = 20 kPa and Q = 5.03 × 10⁻⁶ m³/s, P = 20,000 × 5.03 × 10⁻⁶ = 0.10 W.
Use an oil of the lowest viscosity that still keeps a full film at the working temperature, keep the clearances correct, and keep flow paths smooth and short. These measures reduce fuel use and extend component life.
A journal bearing supports a rotating shaft on a film of oil, and the viscous resistance of the film produces a friction torque that the shaft must overcome. The resistance sets the power loss and the heat generated in the bearing. Example: an engine workshop that rebuilds crankshafts selects the bearing clearance and oil grade so that the film is thick enough to prevent metal contact without excessive drag.
The clearance c between the shaft and the bearing forms a thin film around the shaft. When the shaft is lightly loaded and centred, the film is treated as two parallel plates moving relative to each other, and the shear stress is τ = μ × U ÷ c, with U = 2π × r × N for a shaft speed N in rev/s.
The friction force is F = τ × (2π × r × L), and the torque is T = F × r, which gives:
T = 4π² × μ × N × r³ × L ÷ c
where μ is the viscosity, N is the speed in revolutions per second, r is the shaft radius, L is the bearing length, and c is the radial clearance. The power loss is P = T × 2π × N.
Worked example: a shaft of radius 25 mm turns at 1500 rpm in a bearing of length 50 mm and clearance 0.1 mm, using oil of viscosity 0.03 Pa·s.
Formula: T = 4π² × μ × N × r³ × L ÷ c, with N = 1500 ÷ 60 = 25 rev/s
Substitution: T = 39.48 × 0.03 × 25 × (0.025)³ × 0.05 ÷ 0.0001
Working: 39.48 × 0.03 × 25 = 29.61, then 29.61 × 1.5625 × 10⁻⁵ = 4.627 × 10⁻⁴, then × 0.05 ÷ 0.0001 = 0.2313
Result: the viscous torque is 0.231 N·m, and the power loss is 0.2313 × 2π × 25 = 36.3 W.
A foot-step bearing (a flat pivot) supports a vertical shaft on a flat circular end, with a thin oil film between the shaft end and the fixed surface. The film resists rotation and absorbs power. Example: a farm machinery workshop checks the oil film in the thrust pivot of a vertical shaft in a mixer.
For a shaft of end radius R turning at an angular speed ω over a film of thickness h, the shear stress at radius r is τ = μ × ω × r ÷ h. The friction torque on a ring of width dr is dT = τ × 2π × r × dr × r, and integrating from 0 to R gives:
T = π × μ × ω × R⁴ ÷ (2 × h) = π² × μ × N × R⁴ ÷ h
where N is the speed in rev/s. The power absorbed is P = T × ω.
Worked example: a shaft end of radius 50 mm turns at 300 rpm over a film of 0.2 mm, with oil of viscosity 0.1 Pa·s.
Formula: T = π² × μ × N × R⁴ ÷ h, with N = 300 ÷ 60 = 5 rev/s
Substitution: T = 9.870 × 0.1 × 5 × (0.05)⁴ ÷ 0.0002
Working: 9.870 × 0.1 × 5 = 4.935, then 4.935 × 6.25 × 10⁻⁶ = 3.084 × 10⁻⁵, then ÷ 0.0002 = 0.1542
Result: the viscous torque is 0.154 N·m, and the power loss is 0.1542 × 2π × 5 = 4.84 W.
A collar bearing supports an axial (thrust) load through a rotating collar on a shaft, which runs on a thin oil film against a fixed surface. Unlike the foot-step bearing, the contact is a ring between an inner radius r₁ and an outer radius r₂. Example: a gearbox workshop inspects the thrust collar of a shaft for scoring, because lack of oil film makes the collar overheat.
Integrating the shear stress over the ring from r₁ to r₂, for a film of thickness h:
T = π × μ × ω × (r₂⁴ − r₁⁴) ÷ (2 × h) = π² × μ × N × (r₂⁴ − r₁⁴) ÷ h
When r₁ is zero, this becomes the equation for the foot-step bearing.
Worked example: a collar with r₁ = 30 mm and r₂ = 50 mm turns at 5 rev/s over a film of 0.2 mm, with oil of viscosity 0.1 Pa·s.
Formula: T = π² × μ × N × (r₂⁴ − r₁⁴) ÷ h
Substitution: T = 9.870 × 0.1 × 5 × (0.05⁴ − 0.03⁴) ÷ 0.0002
Working: 0.05⁴ − 0.03⁴ = 6.25 × 10⁻⁶ − 0.81 × 10⁻⁶ = 5.44 × 10⁻⁶, then 4.935 × 5.44 × 10⁻⁶ = 2.685 × 10⁻⁵, then ÷ 0.0002 = 0.1342
Result: the viscous torque is 0.134 N·m.
This chapter explored the behavior of viscous fluid flow, beginning with the fundamental principles governing flow in circular pipes, where the velocity distribution and pressure drop are influenced by viscosity. It then extended these concepts to flow through an annulus, highlighting the effects of the concentric cylindrical boundaries on the fluid motion. The analysis of viscous flow between two parallel plates was detailed, considering cases where one plate moves and the other is stationary, both plates are at rest, and both plates move in opposite directions, each scenario affecting the shear and velocity profiles uniquely. The chapter also addressed the need for kinetic energy correction factors and the role of momentum in accurately describing viscous flows. The power absorbed due to viscous forces was examined, emphasizing its impact on system efficiency. Finally, the viscous resistance offered by mechanical elements such as journal bearings, foot-step bearings, and collar bearings was analyzed, demonstrating how viscosity influences frictional forces in rotating machinery. Together, these topics provide a comprehensive understanding of viscous flow phenomena relevant to various engineering applications.
Question 1 key points:
Question 2 key points:
At a Nairobi-based automotive parts manufacturer, engineers are tasked with improving the lubrication efficiency of journal bearings used in vehicle engines to reduce power losses and increase bearing life.
Tasks:
a) Analyze how viscous resistance affects power loss in journal bearings and suggest design modifications to minimize it. (8 marks)
b) Explain the impact of varying lubricant viscosity on the flow characteristics between bearing surfaces. (6 marks)
c) Propose a method to experimentally determine the kinetic energy correction factor for lubricant flow in the bearing assembly. (6 marks)
a) Discuss that viscous resistance creates shear stress opposing shaft rotation, leading to power loss; recommend design changes such as optimizing bearing clearance, surface finish, and lubricant selection to reduce viscous drag.
b) Explain that higher viscosity increases shear forces and energy loss, while too low viscosity may reduce the lubricant film thickness, risking metal contact; balance is critical for optimal flow and protection.
c) Propose measuring velocity profiles using flow visualization or sensors within a test rig, then integrating velocity data to calculate kinetic energy correction factor accounting for non-uniform flow.
Question 11 (Compulsory - 20 marks)
At Equity Bank vehicle maintenance workshop, engineers are designing a lubrication system for a diesel engine. The system involves oil flowing through circular pipes and journal bearings.
a) Calculate the velocity distribution of the oil in a pipe of 15 mm diameter assuming laminar viscous flow, given the volumetric flow rate and oil viscosity. (10 marks)
b) Discuss the viscous resistance encountered in the journal bearings of the engine’s crankshaft and its implications for bearing design and engine efficiency. (10 marks)
Question 12 (20 marks)
Explain the flow of viscous fluid between two parallel plates when both plates move in opposite directions. Derive the velocity profile and discuss how this principle is applied in automotive brake systems.
Question 13 (20 marks)
Describe the concept of kinetic energy correction factor in viscous flow and demonstrate how it affects momentum calculations in fluid flow through automotive cooling system pipes. Include relevant formulae and a worked example.
Question 14 (20 marks)
Compare the viscous resistance characteristics of foot-step bearings and collar bearings in automotive applications. Explain how these resistances influence the performance and maintenance of vehicle transmission and steering systems.
Question 11
a) Using Hagen-Poiseuille equation and velocity profile for laminar flow, velocity varies parabolically with maximum at pipe centerline. Calculations use given flow rate \(Q\), diameter \(d=15\) mm, and dynamic viscosity \(\mu\).
b) Viscous resistance in journal bearings arises from the lubricant film creating a drag force opposing shaft rotation. This resistance affects bearing load capacity, heat dissipation, and overall engine efficiency. Proper clearance and lubricant viscosity selection minimize losses and wear.
Question 12
When two parallel plates move in opposite directions at velocities \(U\) and \(-U\), the velocity profile is linear, crossing zero at mid-plane. This shear flow models brake fluid behavior between moving components, influencing braking force transmission and heat generation. Derivation involves solving Navier-Stokes equations under steady-state, incompressible, laminar flow conditions.
Question 13
The kinetic energy correction factor \(\alpha\) accounts for non-uniform velocity profiles when calculating momentum flux: \(\alpha = \frac{\int_A u^3 dA}{(\int_A u dA)^3/A^2}\). For laminar flow in pipes, \(\alpha = 2\). This factor ensures accurate force and pressure drop calculations critical for radiator pipe design. A numerical example demonstrates the effect on momentum calculation.
Question 14
Foot-step bearings exhibit viscous resistance mainly due to lubricant shear in axial load transmission, affecting gearbox torque losses and heat. Collar bearings’ viscous resistance arises from circumferential lubricant film shear, impacting steering system friction and responsiveness. Both influence maintenance schedules: excessive resistance leads to increased wear and energy consumption, requiring timely lubrication and inspection.
Time: 2 Hours | Type: Individual
Resources Required:
At the Technical Training Institute in Kisumu, automotive engineering students often analyze fluid flow characteristics in cooling systems. This practical simulates viscous flow through a circular pipe as found in engine coolant circuits.
Tasks:
i. Set up the circular pipe flow apparatus ensuring all connections are tight and leak-free.
ii. Measure and record the flow rate of the fluid using the flow meter and stopwatch with measuring cylinder.
iii. Measure the pressure difference between two points along the pipe using the manometer.
iv. Calculate the pressure drop per unit length and discuss how viscosity affects the flow resistance.
Assessor Observation Criteria:
Time: 3 Hours | Type: Pairs
Resources Required:
At the Jomo Kenyatta University of Agriculture and Technology workshop, understanding flow between concentric pipes is crucial for students designing automotive lubrication systems where oil flows through bearing clearances.
Tasks:
i. Assemble the annulus flow setup and fill with water mixed with dye tracer.
ii. Operate the pump at different flow rates and observe the velocity profile by dye movement.
iii. Sketch the velocity distribution across the annulus gap at each flow rate.
iv. Calculate shear stress at the inner and outer cylinder walls using measured velocity gradients.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At the Eldoret Technical Training Institute, students study viscous flow between plates to understand lubrication films in automotive clutch plates and brakes.
Tasks:
i. Fill the parallel plate setup with viscous fluid ensuring no air bubbles.
ii. Set the top plate to move at a specified speed and measure the force required to maintain motion.
iii. Record the velocity and force data for different plate speeds.
iv. Calculate the fluid viscosity using the measured shear forces and plate velocities.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
Automotive students at Meru Technical College examine fluid flow in narrow channels to simulate oil flow in engine galleries where both surfaces remain stationary.
Tasks:
i. Set up the horizontal channel and establish steady water flow at different flow rates.
ii. Introduce dye at the inlet and observe the flow pattern between the plates.
iii. Measure the velocity of the dye front at several points along the channel.
iv. Explain the velocity distribution and how viscosity influences flow in this setup.
Assessor Observation Criteria:
Time: 3 Hours | Type: Pairs
Resources Required:
At the Kenya Utalii College workshop, this practical models viscous shear in automotive transmission systems where components move relative to each other.
Tasks:
i. Fill the apparatus with viscous fluid and ensure proper sealing to prevent leaks.
ii. Operate the plates moving in opposite directions at preset speeds.
iii. Measure and record the forces acting on each plate during motion.
iv. Calculate the shear stress and discuss the implications for lubrication design.
Assessor Observation Criteria:
Time: 1.5 Hours | Type: Individual
Resources Required:
At the Kisumu County Referral Hospital engineering workshop, technicians use kinetic energy correction factors to improve fluid flow accuracy in cooling and hydraulic systems.
Tasks:
i. Measure fluid velocity distribution in a pipe section using flow meter and pressure readings.
ii. Calculate actual kinetic energy and compare it with the simplified average velocity kinetic energy.
iii. Determine the kinetic energy correction factor for the viscous flow.
iv. Discuss the significance of this factor in automotive fluid systems.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At the National Oil Corporation training facility in Nairobi, understanding power losses due to viscous flow helps optimize fuel pump and lubrication system efficiency.
Tasks:
i. Set up the flow apparatus with the viscous fluid and start the pump.
ii. Measure the flow rate and torque on the pump shaft at steady state.
iii. Calculate the power absorbed by the fluid due to viscous resistance.
iv. Analyze how viscous losses affect overall system performance.
Assessor Observation Criteria:
Time: 3 Hours | Type: Individual
Resources Required:
At the Technical University of Kenya, students examine viscous resistance in journal bearings to improve automotive engine bearing design and maintenance.
Tasks:
i. Measure bearing dimensions and set clearance on the test rig.
ii. Fill the bearing with lubricant and run the shaft at varying speeds.
iii. Record torque required to rotate the shaft at each speed.
iv. Calculate viscous resistance and relate it to bearing performance.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
Automotive technicians at the Kenya Railways workshop analyze viscous resistance in footstep bearings to ensure reliable performance in locomotive engines.
Tasks:
i. Assemble the footstep bearing with specified lubricant film thickness.
ii. Rotate the bearing shaft at incremental speeds and record torque values.
iii. Calculate the viscous resistance from the torque-speed relationship.
iv. Recommend lubrication adjustments to minimize resistance.
Assessor Observation Criteria:
Time: 2 Hours | Type: Individual
Resources Required:
At the Kenya Power and Lighting Company training centre, collar bearing viscous resistance tests help maintain efficient operation of electrical generator rotating parts.
Tasks:
i. Assemble the collar bearing on the test rig with lubricant applied.
ii. Run the bearing shaft at different rotational speeds and measure torque.
iii. Determine viscous resistance values from torque data.
iv. Analyze how lubricant viscosity affects bearing resistance.
Assessor Observation Criteria:
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