How To Find Efficiency Of A Machine
The drivetrain (as well called driveline) is the sum of components which are delivering the engine power to the wheels. For case, on a rear-bicycle bulldoze (RWD) vehicle, the drivetrain consists of: clutch (or torque converter), gearbox (manual or automatic), propeller shaft, differential and drive shafts.
The efficiency of the drivetrain has a significant impact on the overall efficiency of the vehicle. The higher the efficiency of the drivetrain, the lower the fuel consumption of the vehicle (also lower CO2).
Paradigm: Drivetrain architecture (Audi A6 quattro) and primary components
Credit: Audi
In the commodity What is efficiency ? information technology is explained in detail how mechanical efficiency is calculated.
The electric current article is split up in two primary parts. In the starting time part, to empathise the concept of efficiency, we'll calculate the efficiency of a simple gear, office of the input/output power and torque, and in the second part we'll calculate the efficiency of each drivetrain component and the overall drivetrain efficiency.
Gear mesh efficiency
A simple gear mechanism has an input gear and an output gear meshed together. The input torque and angular speed are converted through the gear ratio in output torque and angular speed.
Paradigm: Simple gear input output
where:
Tin [Nm] – input torque
ωin [rad/s] – input speed
i [-] – gear ratio
Tout [Nm] – output torque
ωout [rad/s] – output speed
Nosotros can calculate the input power Pin [Due west] and output power Pout [W] as:
\[P_{in} = \omega_{in} \cdot T_{in} \tag{1}\]
\[P_{out} = \omega_{out} \cdot T_{out} \tag{two}\]
The efficiency is defined equally the ratio between the output power and input power:
\[\bbox[#FFFF9D]{\eta = \frac{P_{out}}{P_{in}}} \tag{3}\]
Whatsoever mechanical component/arrangement, which has moving parts, has friction. The friction is converting role of the energy into heat, which is dissipated in the surroundings environment, therefore lost. The overall friction tin can exist captured equally power loss of the components/arrangement. The output power is the divergence betwixt the input power and the power losses Ploss [Westward]:
\[P_{out} = P_{in} – P_{loss} \tag{4}\]
Replacing (4) in (three):
\[\eta = \frac{P_{out}}{P_{in}} = \frac{P_{in} – P_{loss}}{P_{in}}=\frac{P_{in}}{P_{in}} – \frac{P_{loss}}{P_{in}} = one – \frac{P_{loss}}{P_{in}}\]
gives the expression of the efficiency function of the input power and ability losses:
\[\bbox[#FFFF9D]{\eta = 1 – \frac{P_{loss}}{P_{in}}} \tag{5}\]
The mechanical efficiency of the unproblematic gear can as well be calculated function of input and output torque.
The output speed is equal with the input speed divided by the gear ratio:
\[\omega_{out} = \frac{\omega_{in}}{i} \tag{half-dozen}\]
Replacing (1) and (2) in (three) gives the expression of the efficiency office of input and output torque and speed:
\[\eta = \frac{P_{out}}{P_{in}} = \frac{\omega_{out} \cdot T_{out}}{\omega_{in} \cdot T_{in}} \tag{7}\]
Replacing (6) in (vii) gives:
\[\eta = \frac{\frac{\omega_{in}}{i} \cdot T_{out}}{\omega_{in} \cdot T_{in}}\]
from which we can write the final expression of the efficiency function of input/output torque and gear ratio:
\[\bbox[#FFFF9D]{\eta = \frac{T_{out}}{i \cdot T_{in}}} \tag{8}\]
Gearbox efficiency
The moving parts of a gearbox consists of gears (simple or planetary), synchronizes, shafts and bearings. The overall efficiency of a gearbox depends mainly on the gear mesh and bearings efficiency.
Epitome: Vi-speed manual gearbox components
Credit: ZF
Depending on the architecture, a gearbox has at least two shafts (input and output) and several simple gears. Each shaft is sustained in at to the lowest degree two ball bearings, one in each end. Therefore, when a gear is engaged, in that location are iv bearings and at to the lowest degree 1 gear mesh as sources of ability losses.
The overall efficiency of the gearbox can exist calculated every bit:
\[\eta_{gbx}= \eta_{brg}^{N_{brg}} \cdot \eta_{grm}^{N_{grm}} \tag{9}\]
where:
ηgbx [-] – gearbox efficiency
ηbrg [-] – bearing efficiency
ηgrm [-] – gear mesh efficiency
Nbrg [-] – number of bearings
Ngrm [-] – number of gear meshes
The efficiency of a brawl-begetting is around 0.99 and of a helical gear mesh around 0.98. With these numbers, we tin calculate the overall efficiency of the gearbox.
\[\eta_{gbx}= 0.99^{4} \cdot 0.98^{i} = 0.941\]
In reality the efficiency of the gearbox is not constant but information technology depends on the temperature and shaft speed. The minimum efficiency is usually obtained at low temperature (high oil viscosity) and loftier shaft speed. The maximum efficiency is obtained at high temperature (low oil viscosity) and depression shaft speed.
Propeller shaft efficiency
The propeller shaft is transmitting torque from the gearbox to the rear axle. Since the gearbox and rear axle have to movement relative to each other while transmitting torque, the propeller shaft needs at least 2 universal ("U") joints, ane at each end.
Image: Propeller shaft components (universal joints)
The efficiency of the propeller shaft depends on the number and efficiency of the U-joints and holding bearings. If the propeller shaft is made upwardly from two pieces, it needs at least ane center bearing and four U-joints.
The overall efficiency of the propeller shaft tin be calculated every bit:
\[\eta_{prs}= \eta_{brg}^{N_{brg}} \cdot \eta_{uj}^{N_{uj}} \tag{10}\]
where:
ηprs [-] – propeller shaft efficiency
ηbrg [-] – bearing efficiency
ηuj [-] – universal articulation efficiency
Nbrg [-] – number of bearings
Nuj [-] – number of universal joints
For our example, nosotros are going to consider that the propeller shaft is one-piece, has ii universal joints and no center bearing. The efficiency of an universal joint is effectually 0.99. With these numbers, we can calculate the overall efficiency of the propeller shaft.
\[\eta_{prs}= 0.99^{0} \cdot 0.99^{2} = 0.98\]
In reality, the efficiency of the universal joint is non abiding but information technology depends mainly on the first (angle) betwixt the front and rear beam. The lower the starting time, the higher the efficiency.
Differential efficiency
The differential does the last gear reduction and the torque split between the right and left wheels. If the vehicle is driving on a directly line, only the final gear and bearings are adding power losses. There are 3 bearings (one on the input pinion, i on the left output shaft and 1 on the right output shaft) and 1 spiral bevel gear.
Image: Rear differential components
The overall efficiency of the differential tin exist calculated as:
\[\eta_{dif}= \eta_{brg}^{N_{brg}} \cdot \eta_{grm} \tag{11}\]
where:
ηdif [-] – differential efficiency
ηbrg [-] – begetting efficiency
ηgrm [-] – gear mesh efficiency
Nbrg [-] – number of bearings
The efficiency of a brawl-bearing is around 0.99 and of a spiral bevel gear mesh effectually 0.96. With these numbers, we can calculate the overall efficiency of the differential.
\[\eta_{dif}= 0.99^{3} \cdot 0.96 = 0.931\]
In reality the efficiency of the differential is not abiding only information technology depends on the temperature and shaft speed. The minimum efficiency is commonly obtained at low temperature (loftier oil viscosity) and loftier shaft speed. The maximum efficiency is obtained at high temperature (low oil viscosity) and low shaft speed.
Driveshaft efficiency
The driveshaft is transmitting the torque from the differential to the wheel. Each wheel has it's own driveshaft. At each end of the driveshaft there are abiding-velocity joints (CVJ), which are needed due to the relative movement between differential and wheel.
Prototype: Driveshaft components (constant-velocity joints)
The overall efficiency of the driveshaft tin can exist calculated every bit:
\[\eta_{drs}= \eta_{trp} \cdot \eta_{rzp} \tag{12}\]
where:
ηdrs [-] – driveshaft efficiency
ηtrp [-] – tripod joint efficiency
ηrzp [-] – rzeppa joint efficiency
The inner CVJ (on differential side) usually is a Tripod type joint, while the outer CVJ is a Rzeppa blazon joint. The efficiency of these joints is around 0.99. With these numbers, we can calculate the overall efficiency of the driveshaft.
\[\eta_{drs}= 0.99 \cdot 0.99 = 0.98\]
In reality, the efficiency of the constant-velocity joint is not constant but information technology depends mainly on the offset (angle) between the differential and bicycle. The lower the beginning, the higher the efficiency.
Overall efficiency of the drivetrain
Now that we accept the overall efficiency of each component, we can summate the overall efficiency of the drivetrain (driveline) equally:
\[\bbox[#FFFF9D]{\eta_{drv} = \eta_{gbx} \cdot \eta_{prs} \cdot \eta_{dif} \cdot \eta_{drs}} \tag{13}\]
Replacing the values obtained for each components, gives:
\[\eta_{drv} = 0.941 \cdot 0.98 \cdot 0.931 \cdot 0.98 = 0.841\]
From our parameters and methodology we got an overall efficiency of the drivetrain of 84.i %. This means that around 15.9 % of the engine power is lost through the drivetrain. The efficiency could be even lower for 4-wheel drive (4WD) vehicles which have a central differential.
Permit's see how much we get at the wheels Pout and what are the drivetrain ability losses Ploss, if the engine power at the clutch Pin is 150 kW and the drivetrain efficiency is 0.841.
From (3) we can summate the ability at the wheels (output ability):
\[P_{out} = \eta_{drv} \cdot P_{in} = 0.841 \cdot 150 = 126.xv \text{ kW}\]
From (4) we can calculated the power lost in the drivetrain:
\[P_{loss}=P_{in} – P_{out} = 150 – 126.15 = 23.85 \text{ kW}\]
The numbers evidence that the overall drivetrain efficiency has a significant impact on the dynamic performance of the vehicle since a significant office of the engine power is lost. As well, the lower the drivetrain efficiency, the higher the engine fuel consumption.
Front-wheel drive (FWD) vehicles usually have the highest drivetrain efficiency, mainly considering they don't contain a propeller shaft. At the opposite cease are the all-wheel bulldoze (AWD) and four-wheel drive (4WD) vehicles, with the everyman drivetrain efficiency (due to higher number of components).
Y'all can also check your results using the computer below.
Drivetrain Efficiency Figurer
| ηgbx [-] | ηprs [-] | ηdif [-] | ηdrs [-] | Pin [kW] |
| Drivetrain efficiency, ηdrv [-] = | ||||
| Output power, Pout [kW] = | ||||
| Loss of ability, Ploss [kW] = | ||||
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Source: https://x-engineer.org/drivetrain-losses-efficiency/
Posted by: freythum1941.blogspot.com
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