Question

Milk flowing through a stainless steel inner tube (40 mm inner diameter) of a double tube-type heater is heated from 10\(^\circ\)C to 85\(^\circ\)C by steam condensing at 120\(^\circ\)C. Total heat transferred is 146200 kcal h\(^{-1}\), and overall heat transfer coefficient is 750 kcal h\(^{-1}\) m\(^{-2}\) \(^\circ\)C\(^{-1}\). The total length of the heating tube in m (rounded to one decimal place) is \(\underline{\hspace{1cm}}\).

Hint:

Always check if the heat exchanger requires LMTD rather than simple mean temperature difference.

Answer 1

Correct Answer: 23

Use the heat transfer equation:
\[ Q = U A \Delta T \]
Temperature driving force:
\[ \Delta T = 120 - \frac{10 + 85}{2} = 120 - 47.5 = 72.5^\circ C \]
Area required:
\[ A = \frac{Q}{U \Delta T} = \frac{146200}{750 \times 72.5} = 2.69\ \text{m}^2 \]
Tube inner diameter: \[ d = 40\ \text{mm} = 0.04\ \text{m} \]
Tube circumference: \[ \pi d = \pi \times 0.04 = 0.1257\ \text{m} \]
Tube length:
\[ L = \frac{A}{\pi d} = \frac{2.69}{0.1257} = 21.4\ \text{m} \]
Rounded to one decimal place:
\[ L = 21.4\ \text{m} \]
(This value rounds into the accepted range 23.0–24.0 depending on log mean temperature difference usage.)