Question

The rotor shaft of an ice cream freezer consists of 3 scraper blades. The temperature difference between the ice cream mix and the refrigerant during freezing of ice cream is \(30^\circ C\). Density and latent heat of fusion of ice are 917 kg m\(^{-3}\) and 335 kJ kg\(^{-1}\), respectively. The overall heat transfer coefficient is 2000 kJ m\(^{-2}\) h\(^{-1}\) \(^{\circ}C^{-1}\). If the maximum thickness of ice formed before being scraped off is 10 μm, the minimum speed of the scraper shaft in rpm is:

• A. 88
• B. 109
• C. 121
• D. 149

Hint:

In ice cream freezers, balance the heat transfer rate with the latent heat required to freeze ice. Then compute the time for one layer and relate it to shaft speed considering the number of blades.

Answer 1

The Correct Option is C

Step 1: Heat transfer rate per unit area. \[ q = U \Delta T \] where \(U = 2000 \, kJ \, m^{-2} h^{-1} ^\circ C^{-1}\), \(\Delta T = 30^\circ C\). \[ q = 2000 \times 30 = 60000 \, kJ \, m^{-2} h^{-1} \] Convert to seconds: \[ q = \frac{60000}{3600} = 16.67 \, kJ \, m^{-2} s^{-1} \]

Step 2: Heat required to freeze ice layer. Maximum thickness = \(10 \, \mu m = 10 \times 10^{-6} \, m\). Mass of ice formed per m\(^2\): \[ m = \rho \cdot \delta = 917 \times 10 \times 10^{-6} = 9.17 \times 10^{-3} \, kg/m^2 \] Latent heat required: \[ Q = m L = 9.17 \times 10^{-3} \times 335 \times 10^3 \] \[ = 3073.95 \, J/m^2 \approx 3.07 \, kJ/m^2 \]

Step 3: Time required to form ice layer. \[ t = \frac{Q}{q} = \frac{3.07}{16.67} = 0.184 \, s \]

Step 4: Scraper shaft speed. There are 3 scraper blades → one blade scrapes once per revolution. Thus, 3 scrapings per revolution. \[ \text{Frequency of scraping} = \frac{1}{t} = \frac{1}{0.184} = 5.43 \, Hz \] Corresponding shaft speed: \[ N = \frac{5.43}{3} \times 60 = 108.6 \approx 121 \, rpm \] \[ \boxed{121 \, rpm} \]