Size and cost of the machine will be reduced by selecting

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Option C: A higher value of Output Co-efficient & Synchronous Speed of the machine

Explanation & Concepts:
The size and cost of an electrical machine are inversely proportional to both its synchronous speed ($N_s$) and the output coefficient ($C_0$). A higher output coefficient allows for better utilization of active materials, while a higher speed reduces the required torque for a given power, leading to smaller dimensions.

Working Principle: The output equation of an electrical machine is given by $Q = C_0 D^2 L N_s$, where $Q$ is the output power, $D$ is the stator bore diameter, $L$ is the core length, and $N_s$ is the synchronous speed. Since $Q = C_0 (Volume) N_s$, rearranging for the volume of the machine ($Volume \propto \frac{Q}{C_0 N_s}$) shows that increasing $C_0$ and $N_s$ directly reduces the physical volume, which in turn lowers the amount of copper and steel required, thereby reducing total cost.

Key Formulas: $Q = C_0 D^2 L N_s$ — Output equation representing power capability; $Volume \approx \frac{Q}{C_0 N_s}$ — Relation showing volume inverse to speed and output coefficient

Key Points: Output coefficient ($C_0$) represents the specific magnetic and electric loading of the machine.; Higher $N_s$ reduces the torque requirements, allowing for a smaller rotor diameter.; Higher $C_0$ indicates a higher utilization of magnetic circuit and conductor space.; Reducing physical volume directly minimizes copper, iron, and insulation costs.

Advantages: Reduction in weight and floor space requirement; Lower overall manufacturing and material cost

Disadvantages: High synchronous speed may lead to mechanical stress and noise; High output coefficient may lead to heating issues due to high current density

Applications: Design optimization of high-speed industrial induction motors; Compact drive systems in space-constrained environments

Additional Info: The output coefficient $C_0$ is defined as $C_0 = 1.11 \pi^2 B_{av} ac 	imes 10^{-3}$, where $B_{av}$ is specific magnetic loading and $ac$ is specific electric loading.. Option A is incorrect because lower speed increases the required torque, necessitating a larger motor frame size.

Answer: C is correct — Increasing both the output coefficient and the synchronous speed reduces the volume and material requirements of the machine, leading to smaller size and lower cost.

Exam Tip: Always remember that in machine design, Power = Torque × Speed; thus, higher speed allows the same power to be delivered with less torque, which effectively reduces the size of the machine structure.

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Option A: Lower value of Synchronous Speed of the machine

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Option B: The lower value of Output Co-efficient of the machine

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Option D: A lower value of Output Co-efficient & Synchronous Speed of the machine

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