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

Expert verified
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

01

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: $$ A_o = 10\cdot A_i = 10\cdot(0.08\pi)\mathrm{~m^{2}} = 0.8\pi\mathrm{~m^{2}} $$
02

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: $$ \frac{1}{U\cdot(0.08\pi)} = \frac{1}{2000\cdot(0.08\pi)} + \frac{1}{150\cdot(0.8\pi)} $$
03

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) \(857\mathrm{~W/m^{2}K}\).

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Most popular questions from this chapter

A counter-flow heat exchanger is used to cool oil \(\left(c_{p}=\right.\) \(2.20 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\) from \(110^{\circ} \mathrm{C}\) to \(85^{\circ} \mathrm{C}\) at a rate of \(0.75 \mathrm{~kg} / \mathrm{s}\) by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)

There are two heat exchangers that can meet the heat transfer requirements of a facility. One is smaller and cheaper but requires a larger pump, while the other is larger and more expensive but has a smaller pressure drop and thus requires a smaller pump. Both heat exchangers have the same life expectancy and meet all other requirements. Explain which heat exchanger you would choose and under what conditions.

Explain how the maximum possible heat transfer rate \(\dot{Q}_{\max }\) in a heat exchanger can be determined when the mass flow rates, specific heats, and the inlet temperatures of the two fluids are specified. Does the value of \(\dot{Q}_{\max }\) depend on the type of the heat exchanger?

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

Consider a water-to-water double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold water enters at \(20^{\circ} \mathrm{C}\) and leaves at \(50^{\circ} \mathrm{C}\), while the hot water enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counterflow heat exchanger? Explain.

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