This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Calculate the torque (T) transmitted by the shaft. The power (P) transmitted by a shaft is given by P = (2 N T)/(60), where N is the rotational speed in rpm. Given P = 2.2 MW = 2.2 × 10^6 W and N = 60 rpm. T = (P × 60)/(2 N) T = 2.2 × 10^6 W × 602 × 60 rpm T = (2.2 × 10^6)/(2 ) N·m T ≈ 350140.85 N·m Step 2: Calculate the diameter of the solid shaft (D_s). For a solid shaft, the maximum shear stress (_max) is given by _max = (16 T)/( D_s^3). Given _max = 85 MPa = 85 × 10^6 N/m^2. D_s^3 = (16 T)/( _max) D_s^3 = 16 × 350140.85 N·m × 85 × 10^6 N/m^2 D_s^3 = (5602253.6)/( × 85 × 10^6) m^3 D_s^3 ≈ 0.02098 m^3 D_s = (0.02098)^1/3 m D_s ≈ 0.2758 m = 275.8 mm The required diameter of the solid shaft is 275.8 mm. Step 3: Calculate the external diameter (D_e) of the hollow shaft. For a hollow shaft, the maximum shear stress (_max) is given by _max = (16 T)/( D_e^3 (1 - k^4)), where k = (D_i)/(D_e) is the ratio of internal to external diameter. Given 3D_e = 4D_i, which means k = (D_i)/(D_e) = (3)/(4) = 0.75. The hollow shaft transmits the same torque T ≈ 350140.85 N·m and has the same allowable shear stress _max = 85 × 10^6 N/m^2. D_e^3 = (16 T)/( _max) (1 - k^4) First, calculate 1 - k^4: 1 - k^4 = 1 - (0.75)^4 = 1 - 0.31640625 = 0.68359375 Now, substitute the values: D_e^3 = 16 × 350140.85 N·m × 85 × 10^6 N/m^2 × 0.68359375 D_e^3 = (5602253.6)/( × 85 × 10^6 × 0.68359375) m^3 D_e^3 ≈ 0.03068 m^3 D_e = (0.03068)^1/3 m D_e ≈ 0.3130 m Step 4: Calculate the internal diameter (D_i) of the hollow shaft. D_i = k × D_e = 0.75 × 0.3130 m D_i ≈ 0.23475 m Step 5: Calculate the cross-sectional areas of the solid and hollow shafts. The weight per meter length is proportional to the cross-sectional area for the same material. Area of solid shaft (A_s): A_s = ()/(4) D_s^2 = ()/(4) (0.2758 m)^2 A_s ≈ 0.05973 m^2 Area of hollow shaft (A_h): A_h = ()/(4) (D_e^2 - D_i^2) = ()/(4) ((0.3130 m)^2 - (0.23475 m)^2) A_h = ()/(4) (0.097969 - 0.0551075625) m^2 A_h = ()/(4) (0.0428614375) m^2 A_h ≈ 0.03367 m^2 Step 6: Calculate the percentage saving in weight per meter length. Percentage Saving = (A_s - A_h)/(A_s) × 100\% Percentage Saving = 0.05973 m^2 - 0.03367 m^20.05973 m^2 × 100\% Percentage Saving = (0.02606)/(0.05973) × 100\% Percentage Saving ≈ 43.63\% The percentage saving in weight per meter length is 43.63\%. What's next?