Solving Transfer of Torque with \mu = 0.38 Ns/m^2

[KillerZ]

Homework Statement

The clutch system shown in Fig. P10-22 is used to transmit torque through a 3-mm-thick oil film with $$\mu = 0.38 N \bullet s/m^{2}$$ between two identical 30-cm-diameter disks. When the driving shaft rotates at a speed of 1450 rpm, the driven shaft is observed to rotate at 1398 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque.

Homework Equations

$$T = \mu\left(\frac{4\pi^{2}R^{3}\dot{n}L}{l}\right)$$

The Attempt at a Solution

$$\mu = 0.38 N \bullet s/m^{2}$$
$$L = 30cm = 0.3m$$
$$l = 3mm = 0.003m$$

This is where I am stuck I am not sure what R is. I think I have to find T for the two rpms and find the difference between them for the transferred torque.

 

[CFDFEAGURU]

The units should tell you what R is. You have to get N-m out of this equation, so based on what you know (the given values) you can figure out what unit R will have and that will tell you what it's value should be.

Thanks
Matt

 

[Mene]

I would like to clarify a few things before approaching the solution. Firstly, it is important to note that the given value of \mu is the dynamic viscosity of the oil, not the coefficient of friction. Secondly, the equation provided in the problem statement is for the torque transmitted through a thin film of fluid, not for the transfer of torque between two disks. Therefore, the equation may not be directly applicable in this scenario.

To solve this problem, we need to consider the following assumptions:
1. The oil film between the two disks is thin enough that we can assume a linear velocity profile.
2. The disks are identical and have the same radius, denoted by R.
3. The torque is transmitted from the driving shaft to the driven shaft without any losses.

With these assumptions in mind, we can approach the solution as follows:

1. Firstly, we need to find the angular velocity, \omega, of the driving and driven shafts. We can do this by converting the given values of rpm to rad/s.
\omega_{d} = 1450\text{ rpm} \times \frac{2\pi}{60} = 152.36\text{ rad/s}
\omega_{dr} = 1398\text{ rpm} \times \frac{2\pi}{60} = 146.33\text{ rad/s}

2. Next, we can use the linear velocity profile to find the velocity, v, of the oil film at the mid-plane of the disks.
v = \frac{R(\omega_{d} - \omega_{dr})}{2} = \frac{0.3m(152.36 - 146.33)}{2} = 0.906\text{ m/s}

3. Now, we can use the equation for the torque transmitted through a thin film of fluid to find the transmitted torque, T.
T = \mu\left(\frac{4\pi^{2}R^{3}v}{l}\right) = 0.38\text{ Ns/m}^{2} \times \left(\frac{4\pi^{2}(0.3)^{3}(0.906)}{0.003}\right) = 196.14\text{ Nm}

Therefore, the transmitted torque through the oil film is 196.14 Nm. As a scientist