Question

Consider a 5-cm-diameter shaft rotating at \(5600 \mathrm{rpm}\) in a \(25-\mathrm{cm}\)-long bearing with a clearance of \(0.5 \mathrm{~mm}\). Determine the power required to rotate the shaft if the fluid in the gap is (a) air, (b) water, and (c) oil at \(40^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}$.

Short answer

Answer: The power required to rotate the shaft for each fluid is: 1. Air: 0.102 W 2. Water: 5.39 W 3. Oil: 350.77 W

Step-by-step solution

Write down the given information

Write down the information given in the problem statement: the diameter of the shaft (5 cm), the length of the bearing (25 cm), the clearance (0.5 mm), the rotation speed (5600 rpm), and the fluid temperature (\(40^{\circ} \mathrm{C}\)) and pressure (1 atm).

Convert units

Ensure all units are consistent. For example, convert cm to m and rpm to rad/s as shown below: 1. Diameter of the shaft: \(D = 5\,\text{cm} = 0.05\,\text{m}\) 2. Length of the bearing: \(L = 25\,\text{cm} = 0.25\,\text{m}\) 3. Clearance: \(c = 0.5\,\text{mm} = 0.0005\,\text{m}\) 4. Rotation speed: \(\omega = 5600\,\text{rpm} \times \frac{2\pi}{60} = 587.4\,\text{rad/s}\)

Calculate the shear stress

Calculate the shear stress acting on the shaft due to each fluid using the formula: \(\tau = \mu \frac{u}{c}\), where \(\tau\) is the shear stress, \(\mu\) is the dynamic viscosity of the fluid, \(u\) is the linear speed at the shaft surface, and \(c\) is the clearance. First, calculate the linear speed at the shaft surface: \(u = r\omega\), where \(r\) is the shaft radius (\(r = D/2 = 0.025\,\text{m}\)): $$u = r\omega = (0.025\,\text{m})(587.4\,\text{rad/s}) = 14.685\,\text{m/s}$$ Next, find or look up the dynamic viscosity \(\mu\) for each fluid at the given temperature and pressure. In this step, we will use the following values (units in \(\text{Pa}\cdot\text{s}\)): 1. Air: \(\mu_{\text{air}} = 1.9 \times 10^{-5}\) 2. Water: \(\mu_{\text{water}} = 0.001\) 3. Oil: \(\mu_{\text{oil}} = 0.065\) Now, calculate the shear stress for each fluid using \(\tau = \mu \frac{u}{c}\): 1. Air: \(\tau_{\text{air}} = \frac{1.9 \times 10^{-5}\,\text{Pa}\cdot\text{s} \times 14.685\,\text{m/s}}{0.0005\,\text{m}} = 0.553\,\text{Pa}\) 2. Water: \(\tau_{\text{water}} = \frac{0.001\,\text{Pa}\cdot\text{s} \times 14.685\,\text{m/s}}{0.0005\,\text{m}} = 29.37\,\text{Pa}\) 3. Oil: \(\tau_{\text{oil}} = \frac{0.065\,\text{Pa}\cdot\text{s} \times 14.685\,\text{m/s}}{0.0005\,\text{m}} = 1904.54\,\text{Pa}\)

Calculate power required to rotate the shaft

To calculate the power required to rotate the shaft, use the formula: \(P = \tau A u\), where \(P\) is the power, \(\tau\) is the shear stress, \(A\) is the contact area between the fluid and the shaft, and \(u\) is the linear speed at the shaft surface. First, calculate the contact area \(A = DL\). The contact area is: $$A = DL = (0.05\,\text{m})(0.25\,\text{m}) = 0.0125\,\text{m}^2$$ Now calculate the power required for each fluid: 1. Air: \(P_{\text{air}} = (0.553\,\text{Pa})(0.0125\,\text{m}^2)(14.685\,\text{m/s}) = 0.102\,\text{W}\) 2. Water: \(P_{\text{water}} = (29.37\,\text{Pa})(0.0125\,\text{m}^2)(14.685\,\text{m/s}) = 5.39\,\text{W}\) 3. Oil: \(P_{\text{oil}} = (1904.54\,\text{Pa})(0.0125\,\text{m}^2)(14.685\,\text{m/s}) = 350.77\,\text{W}\)

Present the results

The power required to rotate the shaft for each fluid is: 1. Air: \(P_{\text{air}} = 0.102\,\text{W}\) 2. Water: \(P_{\text{water}} = 5.39\,\text{W}\) 3. Oil: \(P_{\text{oil}} = 350.77\,\text{W}\)