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The power absorbed by the motor is: Pm = Ea Ia = 01n io = 01 500 90 = 45 kW The power delivered by the voltage supply is: PS = VS io = 0567 120 90 = 612 kW 2 Analysis of regenerative operation To analyze regenerative operation of the chopper, we refer to Figure 1137 and apply KVL to the motor side, noting that now the current is owing in the reverse direction: vo = Ra Ia + Ea = Ra io + Ea = 90 02 + 01 380 = 20 V We now turn to equation 1122, and observe that in this equation the motor acts as the source, and the supply voltage as the load, thus: VS = vo 1 tT1 1 or 120 = 1 20 1 300t1

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In this section, we analyze each of the four types of estimate: how often projects are underestimated and overestimated, what overruns or underruns are typical, what types of projects are likely to be overestimated or underestimated, and so on The samples analyzed here are smaller than the full set of 850 projects Some projects contained information on the estimation techniques used, but did not give the estimates themselves Of the remaining projects that provide at least one estimate, few give all four estimates

We can then compute t1 = 28 ms from the above equation The duty cycle for the step-up chopper is now: 5 t1 = = 0833 T 6 The power absorbed by the armature is: =

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The power absorbed by the motor will now be negative, since current is owing in the reverse direction; the motor is in fact generating power: Pm = Ea Ia = 01n io = 01 380 ( 90) = 342 kW The power delivered by the voltage supply is: PS = VS io = 0567 120 ( 90) = 18 kW This power is negative because the supply is actually absorbing power, not delivering it

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In C#, you use the switch-case statement to do much the same thing as VB .NET s Select-Case statement. The switch-case statement works on a number or a string, using one over the other determines switch-case s governing data type. Once the governing type is set, the same data type must be used throughout that particular switch-case statement.

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1 Note that the sum of the motor and armature power losses is equal to the power delivered by the source; this is to be expected, since we have assumed ideal (lossless) switches In a practical chopper, the chopping circuit would actually absorb power; heat dissipation is therefore an important issue in the design of choppers This chopper operates in quadrant I (see Figure 1140) 2 Note that in the regenerative case the equivalent duty cycle is greater than 1 Note also that now the power absorbed by the motor is a negative quantity; that is, the motor delivers power to the rest of the circuit However, the power absorbed by the armature resistance is still a positive quantity because the armature resistance dissipates power regardless of the direction of the current ow through it Here VS might, for example, represent a battery pack in an electric vehicle, which would be recharged at the rate of 18 kW The source of energy capable of producing this power is the inertial energy stored in the vehicle: when the vehicle decelerates, this mechanical energy causes the electric motor to act as a generator (see 17), producing the 90-ampere current in the reverse direction This chopper operates in quadrant IV (see Figure 1140)

Inverters (DC-AC Converters) As will be explained in 17, variable-speed drives for AC motors require a multiphase variable-frequency, variable-voltage supply Such drives are called DC-AC converters, or inverters Inverter circuits can be quite complex, so the objective of this section is to present a brief introduction to the subject, with the aim of illustrating the basic principles A voltage source inverter (VSI) converts the output of a xed DC supply (eg, a battery) to a variable-frequency AC supply Figure 1141 depicts a half-bridge VSI; once again, the switches can be either bipolar or MOS transistors, or thyristors The operation of this circuit is as follows When switch S1 is turned on, the output voltage is in the positive half-cycle, and vo = VS /2 To generate the negative half-cycle, switch S2 is turned on, and

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