Energy cost of air displacement? 1

[ChrisGreaves]

New here. Aeronautics because I am curious about the cost of airflow around an object.
In my case a Toyota Rav4 driving down the highway; calm weather, dry etc. Just the Rav 4.
I first proposed this discussion on Eileen's Lounge, but a user suggested I check out at this site.

The Rav4 is five feet wide and five feet high in cross-section. Assume that there is no space between the road and the underside of the SUV. It moves down the road like a block of wood, 25 square feet in cross-section, pushing air out of the way.
If I assume that it pushes air five feet vertically, I calculate 4,500 foot pounds per second of work; that is lifting 60 square inches of air vertically by five feet, assuming 15 psi. At 60mph=88 fps= about 1,000 inches/second, this has left me contemplating that the car engine must be doing 4,500,000 foot-pounds of work each second.

But one horsepower is defined as 550 ft/lbs/second, which suggests that the car in my simple model is an 8,182 HP Toyota Rav4.

The Toyota RAV4 is 203 HP.
Please and Thank you, where is my error in thinking?

I can see that air might be pushed to the side rather than vertically, but then that air must do work to push other air to the side.. For me the bottom line is that where there was air occupying a 5'x5' cross section, there is now no air in the cross section - just steel, plastic, rubber etc.

I am trying to determine how much work is done just to make a car occupy air-space on a 3.5 hour trip along the highway.
Thanks for any advice on the calculation, also any tips about a more appropriate forum.
Cheers, Chris

 

[EdStainless]

But your car isn't a block.
All shapes will have a drag coefficient, cv.
For simple geometric shapes (sphere or cylinder) you can look these up.
Most motor vehicles have cv's in the range of 0.3-0.4
The air drag is a function of the square of the speed.
So the drag at 85 is double that at 60

 

[GregLocock]

The air isn't pushed permanently out of the way, it flows around, under, and over the car and tumbles in behind it. So the air stays at roughly the same height overall. Otherwise there would be a vacuum behind it.

 

[WKTaylor]

A Wind tunnel [or analytical equivalent] is needed to separate-out aero drag from rolling drag, cooling-drag, etc.

Start with a few SAE books and documents...

SAE R-114 Fundamentals of Vehicle Dynamics
SAE R-366 Road Vehicle Dynamic
SAE R-392 Theory and Applications of Aerodynamics for Ground Vehicles
SAE R-430 Aerodynamics of Road Vehicles

SAE HS-1566 - Aerodynamic Flow Visualization Techniques and Procedures

SAE J1321 Fuel Consumption Test Procedure
SAE J1594 Vehicle Aerodynamics Terminology
SAE J2071 Aerodynamic Testing of Road Vehicles - Open Throat Wind Tunnel Adjustment, Information Report
SAE J2084 Aerodynamic Testing of Road Vehicles -Testing Methods and Procedures, Information Report

 

[3DDave]

It would take that (assuming calcs were correct) if a total vacuum was established behind the car as if the car was a piston in a tube. Instead the HP required is to have a breeze that blows air around the car as an obstacle.

Consider that 14.5 psi is pushing the rear of the vehicle forward and slightly more is pushing the vehicle back.

 

[MintJulep]

The force needed to move your car through the air is given by the drag equation.

Drag equation - Wikipedia

en.m.wikipedia.org

Work is force x distance.

 

[rb1957]

your assumptions are too simplistic and unreasonable, as is the answer they produce. At least you have the good sense to realise this (that a 200hp car can't generate 8000hp). Many references above will explain the problems with your assumptions. By guesswork I'd be surprised if it took more than 50hp to maintain a speed of 60mph, that most of your engine power is used to accelerate the car or to drive uphill. A commonsense view of this would be where is the accelerator pedal (to maintain 60mph) ? how does the engine sound ? Neither of this will be "balls to the wall", so the engine is producing much less than 200 hp

 

[XR250]

\By guesswork I'd be surprised if it took more than 50hp to maintain a speed of 60mph, that most of your engine power is used to accelerate the car or to drive uphill. A commonsense view of this would be where is the accelerator pedal (to maintain 60mph) ?

I am guessing it is much less than that.

]
The air drag is a function of the square of the speed.
So the drag at 85 is double that at 60

This makes sense but why is my fuel mileage not significantly worse at 85 mph versus 60? In my truck it goes from maybe 19 to 14 mpg.

 

[MintJulep]

In my truck it goes from maybe 19 to 14 mpg

(19 - 14) / 19 = .263

So, a 26% increase in fuel use. That seems significantly worse.

 

[MintJulep]

Horsepower & Aerodynamic Drag Calculator

www.stealthmotorsport.co.uk

 

[WKTaylor]

* MJ Your post, #10... RE: Cybersecurity risk. Please be careful about posting non-secure websites... http: VS https: .... my company is beating-us-up for following any non-secure link... especially with secondary internal data entry of any kind within that link. I almost had a 'Rut-Ro'... moment in my interest.

** ChrisGreaves -Student-. Long ago I learned to do some basic self-study research on a topic I needed help with... so I would ask 'smartly-framed' questions of the experts... gaining the expert's interest and enthusiasm and trust for talking with 'lowly-me', IE: my asking from a basis of 'basic knowledge' meant I was serious... not 'flirtatious'.

It seems that you may have done very little technical self-study on this subject for 'familiarization'... before asking this relatively complex question.

I've found a load of basic science/design related to 'automotive aerodynamics', by simple Google search. Most college engineering libraries should have one or more of these titles on the shelf... in addition to the Info I posted [see #4] earlier...

Vehicle Aerodynamics - Testing, Modification & Development: For road, racing and alternative transport
A Century of Car Aerodynamics: – the science and art of cars and airflow
Competition Car Aerodynamics
Race Car Aerodynamics: Designing for Speed

 

[ChrisGreaves]

The air drag is a function of the square of the speed.

I apologize for my delay in getting back to this topic.

Ed, my basic physics classes in high school said that while the kinetic energy of a particle is given by 1/2 m v^2, the kinetic energy of a fluid is 1/2 m v^3.
That is, for air, water etc, we need to use the cube of the velocity (since the mass per unit time is proportional to the velocity).
This fits in with my understanding of why riot police use fire hoses, and why vehicles crossing a creek can be swept away.
Thanks, Chris

 

[ChrisGreaves]

The air isn't pushed permanently out of the way, it flows around, under, and over the car and tumbles in behind it. So the air stays at roughly the same height overall. Otherwise there would be a vacuum behind it.

Hi Greg. I understand this. But I was considering the energy required to displace (say) a linear foot of air, displaced as the car moved forward one foot along the roadway.
That air must be displaced, sideways, or upwards, or below the car.

We know, do we not, that air resistance consumed energy, specifically the energy required to move a specific mass of air?
Thanks, Chris

 

[ChrisGreaves]

It would take that (assuming calcs were correct) if a total vacuum was established behind the car as if the car was a piston in a tube. Instead the HP required is to have a breeze that blows air around the car as an obstacle.

Consider that 14.5 psi is pushing the rear of the vehicle forward and slightly more is pushing the vehicle back.

Hi Dave.
I wasn't assuming that a total vacuum was established at all.
I know from experience that air is pushed out of the way and flows around the car.
The car isn't a piston in a (rigid?) tube; the car is a particle immersed in a fluid medium (air)

If I watch a goldfish immersed in water, there is no obvious vacuum behind that goldfish; I am not sure why we need to consider that a car has a vacuum (let alone a total vacuum) behind it as it moves along a roadway.
Cheers, Chris

 

[ChrisGreaves]

The force needed to move your car through the air is given by the drag equation.

Drag equation - Wikipedia

en.m.wikipedia.org

Work is force x distance.

Hello mintjulep.
I will try using the drag co-efficient, which is probably a perfect fluid engineering solution to the cost of moving a car through air.
I will have to think about this.
I was trying to develop the image of a solid rectangular block ("car") pushing air upwards.

In practice I know that air can flow under a car, and can be displaced sideways, but I was visualizing a car as a theoretical solid block that, either directly or indirectly needed to move air upwards.

"Indirectly" in the sense that while some air might be displaced sideways, that displaced ai would need to displace its neighbouring air, either upwards, or again to the side,
Cheers, Chris

 

[ChrisGreaves]

your assumptions are too simplistic and unreasonable, as is the answer they produce. At least you have the good sense to realise this (that a 200hp car can't generate 8000hp). Many references above will explain the problems with your assumptions. By guesswork I'd be surprised if it took more than 50hp to maintain a speed of 60mph, that most of your engine power is used to accelerate the car or to drive uphill. A commonsense view of this would be where is the accelerator pedal (to maintain 60mph) ? how does the engine sound ? Neither of this will be "balls to the wall", so the engine is producing much less than 200 hp

rb, thank you for this response.
I agree that my assumptions are simplistic.
My thinking started while driving a car on a highway, and were formed as a mental exercise to get what is sometimes called a first approximation to energy requirements. Back then I thought to get a "worst-case value", which is my 8,000 hp result, and then determine how best to determine solutions of more complexity than my solid rectangular block, by making the (mental) model more realistic.

This is the sort of problem I would expect algorithmic students to develop by successively more accurate steps in estimation.

One approach that I had considered was to use an estimate I have read that says 80% of the gasoline used by a car is exhausted as waste heat through the engine-block cooling system.
I could measure the energy content used in 3.5 hours of driving and calculate the energy used in that drive to determine a working value of horse-power for the trip.

That approach has limitations on account of the energy loss in a 300 Km very hilly trip across part of Newfoundland. https://www.google.ca/maps/dir/Bona...try=ttu&g_ep=EgoyMDI0MTIxMS4wIKXMDSoASAFQAw==

Cheers, Chris

 

[GregLocock]

Well OK. take a 4 sq m vehicle moving at 30 m/s. Every second it has to accelerate 30*4*1.2 kg of air up to 30 m/s, F=m.a, hence F=30*4*1.2*30=4320 N, so that's 130 kW (175 hp).

So obviously imagining a car as a piston without considering what happens afterwards is not at all accurate.

The standard aero equation for Cd=0.35 says F=1/2*Cd*A*v^2, 630N, 19 kW.

The drag can usefully be split into two parts, the 'lost ' KE of the air as it backfills behind the car, and skin friction. In practice the first part, the inertial drag or pressure recovery drag as we sometimes call it, dominates. You can reduce it by using a streamliner type shape, but then the skin friction increases, which is the idea behind a kamm tail. That reduces the skin friction while still allowing a fair bit of pressure recovery.

There's also various parasitic drags such as the wheels in the wheelwells, flow through the engine compartment and cabin, and the flow around the horrible mess under the floor.

 

[3DDave]

The OP initial assumption of 15 psi delta is only happening if there is a vacuum.

Hi Dave.
I wasn't assuming that a total vacuum was established at all.

You were making that assumption, otherwise there could not be:

If I assume that it pushes air five feet vertically, I calculate 4,500 foot pounds per second of work; that is lifting 60 square inches of air vertically by five feet, assuming 15 psi.

For me the bottom line is that where there was air occupying a 5'x5' cross section, there is now no air in the cross section

The lack of air is called a vacuum.

There are many resources discussing drag on automobiles. Search for them and read up.

 

[IRstuff]

This fits in with my understanding of why riot police use fire hoses, and why vehicles crossing a creek can be swept away.

No, police use fire hoses because it's a non-lethal means of stopping people and does not require reloading, while rubber bullets and tasers do. Moreover, being hit by 100 gallons per minute will knock down someone down, given that it equates to 830 lbs of water hitting someone per minute.

1/2 * mass * v² is kinetic energy, period, there is no v³
The fact that you willy-nilly mess up simple dimensional analysis makes everything you've been doing completely suspect and dubious.

 

[3DDave]

I think pressure times velocity would be power per unit area, though it's not usual to calculate that directly.