Final answer:
To determine the power transmitted by a hollow shaft, calculate the mean torque within the permissible shear stress and then obtain the transmitted power at the specified rpm. Compare the solid and hollow shaft by ensuring they have equal polar moments of inertia for component strength comparisons.
Step-by-step explanation:
To calculate the power transmitted by a hollow steel shaft at 300 rpm, with a known external diameter (75 mm), internal diameter (50 mm), and permissible shear stress (70 N/mm²), we use the formula for power in terms of torque and angular velocity. First, however, we need to determine the torque acting on the shaft. Given the maximum torque is 1.3 times the mean torque, we need to find the mean torque that can be transmitted within the permissible shear stress limits.
Considering the shaft geometry, we can represent the maximum torque (T_max) as T_max = (τ * π * (D^4 - d^4)) / (16 * D), where τ is the shear stress, D is the external diameter, and d is the internal diameter. Subsequent calculations give the mean torque. Once we have the mean torque, the power (P) transmitted at an angular velocity (ω) is P = T_max * ω, where ω is in radians per second. To compare this hollow shaft to a solid shaft, we would consider geometric properties that yield similar strength, weight, and length, and then evaluate their ability to transmit power under similar conditions. Given that the weight and strength are the same, we must ensure that the solid shaft has an equal polar moment of inertia as the hollow one for a fair comparison. Lastly, we note that for such comparisons, hollow shafts generally provide better strength-to-weight ratios than solid shafts.