1. Power Flow Through A Manual Transmission

Sliding Mesh Transmission. Sliding mesh is the one of the earlier type of manual transmission technology, and the one which is easiest to understand. The most basic slidngmesh transmission mechanism is shown in the Fig.4. Here the input and output shafts are connected through a counter shaft. Power flow in automobile. Use a 3D animation to verify clutch states and shaft speeds in a dual-clutch transmission. Control animation.

MAIN FUNCTIONAL REQUIREMENT: Vary the torque and angular velocity of the power sent from the output shaft of the engine to the wheels.

DESIGN PARAMETER: Manual Transmission

GEOMETRY/STRUCTURE:

EXPLANATION OF HOW IT WORKS/ IS USED:

Power Flow Through A Manual Transmission

Engines provide varying amounts of power at different efficiencies, depending on the speed at which they are turning. A transmission lets the driver select the ratio between the engine and wheels, so the engine can be run at speeds that provide more power, or at speeds that may be less powerful but allow the engine to operate more efficiently.

Power enters the transmission through the input shaft. The input shaft is connected to the engine via the clutch, such that when the clutch is engaged, power goes straight from the engine to the input shaft of the transmission, and the crankshaft and input shaft rotate at the same speed.

Just inside the housing of the transmission, the input shaft is connected to the countershaft (also known as the layshaft), by gears on both shafts, such that whenever the input shaft turns, so does the countershaft, and always at a fixed speed ratio.

In addition to the gear that takes power from the input shaft, the countershaft has several gears on it, one for each of the car’s 'gears' including reverse. The diagram above depicts a five-speed manual transmission. All of these are connected to the countershaft, so they turn as a single piece.

A third shaft called the output shaft runs parallel to the countershaft, and has freely-rotating gears that are mounted on bearings and rotate independently of the output shaft. Each of these gears is paired with one of the countershaft gears, and is constantly in mesh with it. So actually the 'freely-rotating' gears are constrained to rotate at a constant ratio with respect to the countershaft. Each pair of countershaft and output shaft gears represents one 'gear' that the driver can select.

The free gears rotate at different speeds, depending on their size relative to that of the countershaft gear that powers them. This varying ratio between the countershaft gear and the output shaft gear is what ultimately determines the gear ratio. For the reverse gear there is a small 'reverse idler gear' between the countershaft gear and the output shaft gear. This causes the reverse gear to turn the opposite way from the other gears on the output shaft.

The output shaft is a splined shaft that is more or less (via the differential) connected to the wheels. On the output shaft, between every pair of 'free' gears is a collar. This is a round piece that is locked into the splines of the output shaft. By being attached to the shaft in this manner, the collar is forced to rotate with the shaft, but can slide up and down the shaft along the splines.

On the side surfaces of each collar are dog teeth that face toward the output shaft gears on either side. The 'free' gears on the output shaft have slots into which the dog teeth can lock, forcing the gear to rotate with the collar.

To briefly summarize the operation of the manual transmission: The clutch sends power to the input shaft. This turns the countershaft as well as the 'free' gears on the output shaft. When the driver selects a gear by moving the shift lever inside the car, a collar slides up to the appropriate 'free' gear, and the dog teeth lock the collar to that gear. Thus power is transferred from the countershaft to the output shaft through the selected gear, at the appropriate ratio.

DOMINANT PHYSICS:

Whenever gears are used to transfer energy betweeen shafts, the rotation may be changed by a certain ratio, which is calculated in this case by multiplying the ratios of all the gear pairs that the power goes through.

There is a ratio between the input shaft and the countershaft, and another ratio between the chosen output shaft gear and its partner countershaft gear. When these two ratios are multiplied, the ratio for that 'gear' of the transmission is obtained.

By changing the rotational speed of the engine’s power, the transmission is also changing the torque. Since torque and angular velocity vary inversely (assumingconstant power , Power = Torque * Rotational Speed), a low gear provides slow rotation with high torque, and a high gear provides faster rotation with less torque.

LIMITING PHYSICS:

There are certain limits that restrict the gearing of the manual transmission. The overall size of the gearbox is constrained, so that it can fit conveniently into the design of the vehicle. And since there are a finite number of gears to choose from (usually five), the optimal ratio cannot always be achieved.

Also the amount of torque input to the transmission cannot exceed a certain value, since excessive torque would cause failure of the gear teeth.

PLOTS/GRAPHS/TABLES:

None Submitted

WHERE TO FIND MANUAL TRANSMISSIONS:

These are found in all sorts of machinery, including race cars, buses, 18-wheelers, motorbikes, farm vehicles, and of course in passenger cars, where the hump between the seats is a constant reminder of the transmission’s presence.

REFERENCES/MORE INFORMATION:

Birch, Thomas. 'Manual Drive Trains and Axles, 2^nd ed.' Columbus: Prentice Hall, 1999.

Brain, Marshall. 'How Manual Transmissions Work.' How Stuff Works. BYG Publishing, Inc. 1998.

Power Flow through a Transmission Line:

Power Flow through a Transmission Line – So far the transmission line performance equation was presented in the form of voltage and current relationships between sending-and receiving-ends. Since loads are more often expressed in terms of real (watts/kW) and reactive (VARs/ kVAR) power, it is convenient to deal with transmission line equations in the form of sending- and receiving-end complex power and voltages. While the problem of flow of power in a general network will be treated in the next chapter, the principles involved are illustrated here through a single transmis­sion line (2-node/2-bus system) as shown in Fig. 5.17.

Let us take the receiving-end voltage as a reference phasor (V_R= V_R ∠0^°) and let the sending-end voltage lead it by an angle δ(V_s= V_s ∠δ). The angle δ is known as the torque angle whose significance has been explained in Chapter 4 and will further be taken up in Chapter 12 while dealing with the problem of stability.

The complex power leaving the receiving-end and entering the sending-end of the transmission line can be expressed as (on per phase basis)

Receiving- and sending-end currents can, however, be expressed in terms of receiving- and sending-end voltages [see Eq. (5.1)1 as

Let A, B, D, the transmission line constants, be written as

Therefore, we can write

In the above equations S_R and S_s are per phase complex voltamperes, while V_R and V_s are expressed in per phase volts. If V_R and V_s are expressed in kV line, then the three-phase receiving-end complex power is given by

This indeed is the same as Eq. (5.58). The same result holds for S_s. Thus we see that Eqs. (5.58) and (5.59) give the three-phase MVA if V_s and V_R are expressed in kV line.

If Eq. (5.58) is expressed in real and imaginary parts, we can write the real and reactive powers at the receiving-end as

Similarly, the real and reactive powers at sending-end are

It is easy to see from Eq. (5.61) that the received power P_R will be maximum at

Thus the load must draw this much leading MVAR in order to receive the maximum real power.

Consider now the special case of a short line with a series impedance Z. Now

Substituting these in Eqs. (5.61) to (5.64), we get the simplified results for the short line as

for the receiving-end and for the sending-end

The above short line equation will also apply for a long line when the line is replaced by its equivalent- π (or nominal- π) and the shunt admittances are lumped with the receiving-end load and sending-end generation. In fact, this technique is always used in the load flow problem to be treated in the next chapter.

Halloween games for toddlers. From Eq.(5.66), the maximum receiving-end power is received, when δ=θ

Normally the resistance of a transmission line is small compared to its reactance (since it is necessary to maintain a high efficiency of transmission), so that θ= tan^-1 X/R 90′; where Z = R + jX. The receiving-end Eqs. (5.66) and (5.67) can then be approximated as

Equation (5.72) can be further simplified by assuming cos δ≈1, since δ is normally small^*. Thus

Let V_S - V_R = ΔV , the magnitude of voltage drop across the transmission line.

Several important conclusions that easily follow from Eqs. (5.71) to (5.74) are enumerated below:

• For R≈0 (which is a valid approximation for a transmission line) the real power transferred to the receiving-end is proportional to sin δ(≈δ for small values of δ ), while the reactive power is proportional to the magnitude of the voltage drop across the line.

• The real power received is maximum for δ = 90° and has a value V_s V_R /X. Of course, δ is restricted to values well below 90° from considerations of stability to be discussed in Chapter 12.
• Maximum real power transferred for. a given line (fixed X) can be increased by raising its voltage level. It is from this consideration that voltage levels are being progressively pushed up to transmit larger chunks of power over longer distances warranted by large size generating
• For very long lines voltage level cannot be raised beyond the limits placed by present-day high voltage technology. To increase power transmitted in such cases, the only choice is to reduce the line reactance. This is accomplished by adding series capacitors in the line. This idea will be pursued further in Chapter 12. Series capacitors would of course increase the severity of line over voltages under switching conditions.

• As said in 1 above, the VARs (lagging reactive power) delivered by a line is proportional to the line voltage drop and is independent of δ. Therefore, in a transmission system if the VARs demand of the load is large, the voltage profile at that point tends to sag rather sharply. To maintain a desired voltage profile, the VARs demand of the load must be met locally by employing positive VAR generators (condensers). This will be discussed at length in Sec. 5.10.

A somewhat more accurate yet approximate result expressing line voltage drop in terms of active and reactive powers can be written directly from Eq. (5.5), i.e.

This result reduces to that of Eq. (5.74) if R = 0.