A contained in a horizontal cylinder fitted with a frictionless leak p...
Problem Statement
A contained fluid in a horizontal cylinder fitted with a frictionless leak-proof piston, is continuously agitated by means of a stirrer passing through the cylinder cover. The cylinder diameter is 0.40 m. During the stirring process lasting 10 minutes, the piston slowly moves out a distance of 0.485 m against the atmosphere. The net work done by the fluid during the process is 2. The speed of the electric motor is rpm. Determine the torque in the shaft and the power output of the motor.
Solution
Step 1: Calculation of Volume and Pressure
Given, the diameter of the cylinder is 0.40 m, which implies the radius is 0.20 m.
The volume of the cylinder is given by V = πr2h , where h is the height of the cylinder. Here, the height of the cylinder is not given but it can be calculated using the distance moved by the piston.
The distance moved by the piston is 0.485 m.
Therefore, the height of the cylinder is h = 0.485 m.
Hence, the volume of the cylinder is V = π(0.20)2(0.485) = 0.030 m3.
The net work done by the fluid is 2. This implies that the pressure difference between the fluid and the atmosphere is given by ΔP = 2/V = 2/0.030 = 66.67 kPa.
Step 2: Calculation of Torque and Power Output
The torque in the shaft can be calculated using the equation T = F × r, where F is the force exerted by the fluid on the piston and r is the radius of the cylinder. The force exerted by the fluid is given by F = ΔP × A, where A is the area of the piston. Here, the area of the piston is equal to the area of the cylinder, which is A = πr2 = π(0.20)2 = 0.1257 m2.
Hence, the force exerted by the fluid is F = ΔP × A = 66.67 × 0.1257 = 8.39 kN.
Therefore, the torque in the shaft is T = F × r = 8.39 × 0.20 = 1.68 kNm.
The power output of the motor is given by P = T × ω, where ω is the angular velocity of the motor in rad/s. The angular velocity of the motor in rpm can be converted to rad/s using the formula ω = 2πN/60, where N is the rpm.
Here, the speed of the electric motor is not given. Therefore, the power output of the motor cannot be calculated.
Conclusion
The torque in the shaft is 1.68 kNm. However, the power output of the motor could not be calculated due to the missing information.