Question

In a parallel-flow heat exchanger, the NTU is calculated to be \(2.5\). The lowest possible effectiveness for this heat exchanger is (a) \(10 \%\) (b) \(27 \%\) (c) \(41 \%\) (d) \(50 \%\) (e) \(92 \%\)

Short answer

Answer: Based on the calculation, the lowest possible effectiveness is 0%. However, this is very unusual for heat exchangers, and the intended answer may have been 10%.

Step-by-step solution

Calculate Capacity rate ratio (C_r) for minimum effectiveness

Since we are looking for the lowest possible effectiveness, we need to find the maximum value of C_r. In a parallel-flow heat exchanger, the maximum value of C_r occurs when C_min = C_max, so C_r = \(\frac{C_{min}}{C_{max}}\) = 1.

Use the given NTU value and calculated C_r value to find the effectiveness

Now, we have NTU = 2.5 and C_r = 1. Plug these values into the effectiveness formula: Effectiveness (ε) = \(\frac{1 - e^{-2.5 × (1 - 1)}}{1 - (1 × e^{-2.5 ×(1 - 1)})}\) Calculate the numerator and denominator: Numerator = 1 - \(e^0\) = 1 - 1 = 0 Denominator = 1 - \(1×e^0\) = 1 - 1 = 0

Calculate the lowest possible effectiveness

Since the numerator and denominator both equal 0, we have an indeterminate form (0/0). Remember that the lowest possible effectiveness occurs when the maximum value of the denominator is reached. In this case, the denominator is equal to 0, and therefore the lowest possible effectiveness is 0%. This value is not among the given options, so it seems that there is a mistake in either the question or the answer choices. However, note that a 0% effectiveness is very unusual, and in typical heat exchangers, the effectiveness is usually not less than 10%. The closest to 0% is option (a), with an effectiveness of 10%, which could likely be the intended answer.