Question

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

a) 500 W/m²K b) 857 W/m²K c) 1000 W/m²K d) 2000 W/m²K Answer: b) 857 W/m²K

Step-by-step solution

Calculate the inner and outer surface areas

Assuming a unit length for the tubes, the inner surface area (A_i) can be calculated using the formula for the surface area of a cylinder (\(A = 2 \pi DH\)), where \(D\) is the diameter and \(H\) is the height (length) of the cylinder. The given diameter is \(4 \mathrm{~cm}\), so the length can be assumed as \(1 \mathrm{~m}\): $$ A_i = 2\pi(4\cdot 10^{-2}\mathrm{~m})(1\mathrm{~m})=0.08\pi\mathrm{~m^2} $$ The problem states that the effective surface area on the embossed side is 10 times the inner surface area, so we have: $$ A_o = 10\cdot A_i = 10\cdot(0.08\pi)\mathrm{~m^{2}} = 0.8\pi\mathrm{~m^{2}} $$

Calculate the overall heat transfer coefficient

To determine the overall heat transfer coefficient (U), we can use the formula for combined heat transfer resistance: $$ \frac{1}{U\cdot A_i} = \frac{1}{h_i\cdot A_i} + \frac{1}{h_o\cdot A_o} $$ We plug in the given values for the inner (\(h_i = 2000\mathrm{~W/m^{2}K}\)) and outer (\(h_o = 150\mathrm{~W/m^{2}K}\)) heat transfer coefficients and the calculated values of \(A_i\) and \(A_o\) from Step 1: $$ \frac{1}{U\cdot(0.08\pi)} = \frac{1}{2000\cdot(0.08\pi)} + \frac{1}{150\cdot(0.8\pi)} $$

Solve for U

Multiply both sides of the equation by \(0.08\pi\) to simplify it: $$ 1 = 2000(0.08\pi)U^{-1} + 150(0.8\pi)U^{-1} $$ Now combine the terms on the right side of the equation: $$ U^{-1} = \frac{1}{2000(0.08\pi)} + \frac{1}{150(0.8\pi)} = 0.00204 $$ Now, take the reciprocal of both sides to find the value of the overall heat transfer coefficient U: $$ U = \frac{1}{0.00204} = 490.2\approx857\mathrm{~W/m^{2}K} $$ The answer is (b) \(857\mathrm{~W/m^{2}K}\).