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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

Expert verified
Answer: The lowest possible effectiveness of a parallel-flow heat exchanger with an NTU of 2.5 is approximately 92%.

Step by step solution

01

Understanding the effectiveness-NTU relation for parallel-flow heat exchangers

The effectiveness of a heat exchanger is the ratio of the actual heat transfer to the maximum possible heat transfer. For parallel-flow heat exchangers, the effectiveness-NTU relationship is given by the following formula: ϵ=1eNTU(1C)(1CeNTU(1C)) where ϵ is the effectiveness, NTU is the Number of Transfer Units which is given as 2.5, and C is the capacity rate ratio defined as C=CminCmax where Cmin and Cmax are the minimum and maximum capacity rates, respectively.
02

Calculate the effectiveness for the lowest capacity rate ratio

Since the effectiveness depends on the capacity rate ratio C, we need to find the lowest possible value for C that will still result in a functional heat exchanger. The capacity rate ratio always lies between 0 and 1, with C=0 corresponding to no heat exchange and C=1 corresponding to equal capacity rates. For the lowest possible effectiveness, we will use the smallest valid value of C, which is just above 0. Let's call this value Cmin and use it in the effectiveness-NTU formula: ϵmin=1eNTU(1Cmin)(1CmineNTU(1Cmin)) When we plug in NTU=2.5 and take the limit as Cmin approaches 0, we will find the lowest possible effectiveness.
03

Finding the limit

As Cmin approaches 0, the term (1Cmin) approaches 1, and the term eNTU(1Cmin) approaches eNTU. Now we can rewrite the formula as: ϵmin=1e2.5(1Cmin)(1Cmine2.5(1Cmin)) Now, take the limit as Cmin approaches 0: limCmin01e2.5(1Cmin)(1Cmine2.5(1Cmin)) This limit results in the lowest possible effectiveness: ϵmin=1e2.5
04

Calculate the lowest possible effectiveness

Now we can compute the lowest possible effectiveness by calculating the numerical value of 1e2.5: ϵmin=1e2.50.918 This implies that the lowest possible effectiveness is around 91.8%.
05

Choose the correct option

Since the lowest possible effectiveness is approximately 91.8%, which is closest to 92%, we can conclude that the correct answer is: (e) 92%

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Exchanger
A heat exchanger is a system used to transfer heat between two or more fluids that are at different temperatures. These fluids can either be liquids or gases, and they flow through the heat exchanger either in direct contact or separated by a solid barrier to prevent mixing. Heat exchangers come in various designs and are used in a wide range of applications, including refrigerators, air conditioning units, power plants, and chemical processing.

One of the critical aspects in the design of heat exchangers is the rate of heat transfer, which is influenced by the surface area for heat exchange, the temperature difference between the fluids, the flow rate of the fluids, and the thermal properties of the fluids themselves. Efficient heat exchangers are designed to maximize this heat transfer while considering factors such as cost, size, and durability.
Parallel-Flow Heat Exchanger
In a parallel-flow or co-current heat exchanger, both the hot fluid and the cold fluid flow in the same direction. They enter the heat exchanger at the same end, move in parallel with each other, and exit at the same end. This design creates a temperature gradient that decreases along the length of the exchanger, as the hot fluid loses energy to the cold fluid.

Advantages and Disadvantages

One advantage of the parallel-flow design is that it tends to have a more uniform wall temperature, which can be beneficial for preventing thermal stress. However, its major disadvantage is that the temperature difference between the fluids, which drives the heat exchange, decreases along the length of the exchanger, usually resulting in lower effectiveness compared to counter-flow heat exchangers where fluids move in opposite directions. This means they often require larger surface areas to achieve similar heat transfer rates.
Number of Transfer Units (NTU)
The Number of Transfer Units, or NTU, is a dimensionless parameter that characterizes the size and effectiveness of a heat exchanger. It is defined as the ratio of the heat transfer surface area to the thermal capacity rate of the fluid with the minimum thermal capacity. The thermal capacity rate is the product of a fluid's mass flow rate and its specific heat capacity.

The NTU can be calculated using the expression:
NTU=UACmin
where U is the overall heat transfer coefficient, A is the heat transfer area, and Cmin is the minimum thermal capacity rate of the two fluids. The effectiveness of a heat exchanger is directly linked to the NTU value; a higher NTU means a larger surface area relative to the fluid's thermal capacity, which often leads to a higher effectiveness.
Capacity Rate Ratio
The capacity rate ratio (C) is a critical parameter in determining the performance of a heat exchanger. It is defined as the ratio of the minimum capacity rate of the two fluids to the maximum capacity rate:
C=CminCmax
The value of C affects how the temperature of the two fluids changes throughout the heat exchanger. In the case of a parallel-flow heat exchanger, as we see in the exercise, the capacity rate ratio plays a significant role in calculating the effectiveness of heat transfer, especially when it comes to determining the lowest possible effectiveness.

Implications of C Value

When C is low, which occurs when there's a large discrepancy between the fluid capacity rates, the fluid with the lower capacity rate undergoes a more significant temperature change. As C approaches 1, the fluids have similar capacity rates, which can lead to closer exit temperatures for both fluids. The value of C can vary between 0 and 1, with C=0 meaning no heat is transferred, and C=1 indicating maximum heat transfer efficiency within the constraints of the heat exchanger design.

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Most popular questions from this chapter

In a parallel-flow, liquid-to-liquid heat exchanger, the inlet and outlet temperatures of the hot fluid are 150C and 90C while that of the cold fluid are 30C and 70C, respectively. For the same overall heat transfer coefficient, the percentage decrease in the surface area of the heat exchanger if counter-flow arrangement is used is (a) 3.9% (b) 9.7% (c) 14.5% (d) 19.7% (e) 24.6%

Consider the flow of saturated steam at 270.1kPa that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter 1.25 cm at a rate of 0.25 kg/s through each tube. The water enters the tubes of heat exchanger at 20C and exits at 60C. Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is 1500 W/m2K, while the fouling resistance for the steam and water may be taken as 0.00015 and 0.0001 m2K/W, respectively. Using the NTU method, determine (a) effectiveness of the heat exchanger, (b) length of the tube, and (c) rate of steam condensation.

Cold water (cp=4180 J/kgK) enters the tubes of a heat exchanger with 2 -shell passes and 23 -tube passes at 14C at a rate of 3 kg/s, while hot oil (cp=2200 J/kgK) enters the shell at 200C at the same mass flow rate. The overall heat transfer coefficient based on the outer surface of the tube is 300 W/m2K and the heat transfer surface area on that side is 20 m2. Determine the rate of heat transfer using (a) the LMTD method and (b) the εNTU method.

The radiator in an automobile is a cross-flow heat exchanger (UAs=10 kW/K) that uses air (cp=1.00 kJ/kgK) to cool the engine-coolant fluid (cp=4.00 kJ/kgK). The engine fan draws 30C air through this radiator at a rate of 10 kg/s while the coolant pump circulates the engine coolant at a rate of 5 kg/s. The coolant enters this radiator at 80C. Under these conditions, the effectiveness of the radiator is 0.4. Determine (a) the outlet temperature of the air and (b) the rate of heat transfer between the two fluids.

A shell-and-tube heat exchanger with 2-shell passes and 8 -tube passes is used to heat ethyl alcohol (cp=2670 J/kgK) in the tubes from 25C to 70C at a rate of 2.1 kg/s. The heating is to be done by water (cp= 4190 J/kgK) that enters the shell at 95C and leaves at 60C. If the overall heat transfer coefficient is 800 W/m2K, determine the heat transfer surface area of the heat exchanger using (a) the LMTD method and (b) the εNTU method.

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