And I was not able to parse what you wrote there. Can you please restate that or elaborate?doke said:i got confused because if the connections btw the parts
Do you know how power, rotational speed and torque are related for a rotational system? If the gear train is required to output 9 kW can you then from that requirement alone say how much power should (at least) be in the input?doke said:im confused with only Part A of the Question
If you by "P=TW" means power equal torque times angular speed, all using compatible units like power in W, torque in Nm and angular speed in s-1), then you are correct.doke said:P=TW but in the question he is saying to obtain power again while i have net power so i got confused here. also i have no idea which formula can obtain the rpm also question provides Rpm
The problem text seems to suggest you can define you assumptions. Since there are no mentioning of friction or losses in the text, the simplest is often to assume no friction, but that is up to you.doke said:does efficiency has a role to obtain the required power?
Given that engine-to-pump gear ratio is 10:1 then yes. But reading the text I come up with a different gear ratio, just so you know.doke said:if i use gear ratio 10:1 that means 200rpm*10 which is the ratio so the motor must run at 2000rpm ? am i correct here ?
Is there a special reason you want to address efficiency? The problem text does not seem to imply you need to address anything regarding efficiency.doke said:regarding to the power :i think i must use concept of efficiency which is P motor=Ppump/efficiency . is that correct
In order to work properly, the shaft of your pump needs to rotate at 200 rpm while driven by a minimum torque of 430 N-m.doke said:... drives a water pump should run at 200Rpm and 9kw net power...
thanks for reaching dear, so in which way i can get the required motor powerLnewqban said:In order to work properly, the shaft of your pump needs to rotate at 200 rpm while driven by a minimum torque of 430 N-m.
Moving bacwards toward the motor, you will keep the rpm rates that you choose to have the input rpm that the manufacturer of the motor specifies for certain torque.
Simultaneously, you will be needing increasing transferred power at each step, due to lost torque in friction losses; therefore, your motor will need to deliver more than the 9 KW needed by the pump.
Please, check above link.doke said:thanks for reaching dear, so in which way i can get the required motor power
See link in post #14 above.doke said:as a result so T Nm = 9000Watt * 9.549 / 4000 rpm = 21.48Nm .then ω=60*4000rpm×2π≈418.88rad/s so
P=21.48×418.88
P≈8997.8W
Am i correct now ?
The relationship between RPM (revolutions per minute) and power in an electric motor is defined by the formula: Power (Watts) = Torque (Nm) × Angular Velocity (rad/s). As RPM increases, the angular velocity increases, which can increase the power output, assuming torque remains constant. However, power output is also limited by the motor's design and efficiency.
You can calculate the RPM of a motor using the formula: RPM = (Power (Watts) × 60) / (Torque (Nm) × 2π). This formula rearranges the relationship between power, torque, and angular velocity to isolate RPM. It allows you to find the RPM if you know the power output and the torque produced by the motor.
A gearbox alters the RPM and torque output of a motor. When using a gearbox, the output RPM is reduced while the torque is increased based on the gear ratio. For example, a gear ratio of 2:1 will halve the RPM but double the torque. The power output remains constant (minus losses) because power is conserved: Power in = Power out, adjusted for efficiency losses in the gearbox.
A pulley system can change the RPM and torque delivered to a load based on the diameter of the pulleys used. If a smaller pulley is connected to the motor and a larger pulley drives the load, the RPM will decrease, and torque will increase, similar to a gearbox. The power transferred through the system will remain consistent, minus losses due to friction and other inefficiencies.
Efficiency losses in these systems can arise from several factors, including friction in the motor bearings, heat generation in the motor windings, inefficiencies in the gearbox (typically around 5-15% loss), and friction in the pulley system. These losses reduce the effective power output and can impact the overall performance, so it’s important to consider these factors in system design and calculations.