Proving Heat exchanger equations

In summary, the potential problem with the operating equation for S when R = 1 can be solved by substituting y = 1 - R and using the approximation exp(cy) ≈ 1 + cy. As R approaches 1, the exponential term disappears and the operating equation reduces to a simplified function of (UA/MaCa). However, when R = 1, the equation becomes indeterminate and further analysis may be necessary to solve it.
  • #1
metalscot
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For a double-pipe heat exchanger operating with countercurrent flow, the operating equation is:

S = (1 - exp X)/ (R - exp X) X = ((U*A)/(Ma*Ca)) (1-R)

where R and S are defined by:

R = (Ma*Ca)/( Mb*Cb) = (Tb2 - Tb1)/(Ta1 - Ta2)

S = (Ta1 - Ta2)/(Ta1- Tb1)

(i) Show how there is a potential problem using the above equation if the value of R is equal to 1.

The potential problem with the equation for S can be solved using an appropriate substitution. For small values, the exponential can be written as: exp(cy)≈ 1+ cy

(ii) Use the substitution

y=1-R

to show that as the value of R approaches 1, the exponential disappears and the operating equation reduces to a simplified function of (UA/MC).


Attemped solution:

(i) For R to = 1

R = (Ma*Ca)/( Mb*Cb) = (Tb2 - Tb1)/(Ta1 - Ta2) = 1

This would mean that (Tb2 - Tb1) and (Ta1 - Ta2) were the same, say Tb2=30 Tb1=10 Ta1=40 and Ta2=20

For the value of S however because Ta1 is used twice, S would equal 2/3

But when R= 1 in the operating equation S must also equal 1.

(ii) exp(cy)≈ 1+ cy

y=1-R, as R approaches 1 y approaches 0.
when y becomes 0 cy becomes 0 meaning that exp(cy)≈1

when exp (cy) is 1, X=0 in the operating equation when X = 0, (UA/MaCa) or (1-R) must equal zero. From the previous equation we know that it is 1-R that equals 0 leaving us with the simplified (UA/MaCa)

Can someone please tell me if my methods and answers are correct?
Thank you in advance :)
 
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  • #2


Your methods and answers are mostly correct. However, there are a few minor mistakes and clarifications that can be made.

(i) When R = 1, it does not necessarily mean that (Tb2 - Tb1) and (Ta1 - Ta2) are the same. It just means that the ratio of (Ma*Ca)/(Mb*Cb) is equal to 1. This could happen if (Tb2 - Tb1) and (Ta1 - Ta2) are both equal to 10, or if they are both equal to 100, for example. So, the specific values of Tb2, Tb1, Ta1, and Ta2 are not necessarily important in this case. The important thing to note is that when R = 1, the ratio between the temperatures (Tb2 - Tb1) and (Ta1 - Ta2) is equal to 1, which means they are the same.

(ii) The substitution y = 1 - R is correct, but the explanation is a bit unclear. Instead of saying "when y becomes 0, cy becomes 0," you can say "when R approaches 1, y approaches 0, which means that the exponential term becomes 1, and X becomes 0." This is because when R = 1, y = 0, so the exponential term becomes exp(0) = 1, and X becomes 0.

Also, it should be noted that when R = 1, the operating equation becomes S = 0/0, which is an indeterminate form. This means that the equation cannot be solved as is, and further analysis or substitutions may be necessary to find a solution.

Overall, your understanding and approach to the problem seem to be correct. Just be careful with the notation and make sure to clearly explain your steps and reasoning.
 

1. How do you determine the heat transfer coefficient for a heat exchanger?

The heat transfer coefficient for a heat exchanger can be determined through experimental testing or through analytical calculations. Experimental testing involves measuring the inlet and outlet temperatures, flow rates, and other parameters of the heat exchanger, and then using these values to calculate the heat transfer coefficient. Analytical calculations involve using equations and models to estimate the heat transfer coefficient based on the design and geometry of the heat exchanger.

2. What is the difference between a parallel flow and a counterflow heat exchanger?

In a parallel flow heat exchanger, the hot and cold fluids flow in the same direction, while in a counterflow heat exchanger, they flow in opposite directions. This affects the overall efficiency of the heat exchanger, as counterflow arrangement allows for a larger temperature difference between the two fluids, leading to a higher heat transfer rate.

3. How do you calculate the heat transfer rate for a heat exchanger?

The heat transfer rate for a heat exchanger can be calculated using the equation Q = UAΔTm, where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer surface area, and ΔTm is the log mean temperature difference between the two fluids.

4. What are the factors that affect the performance of a heat exchanger?

The performance of a heat exchanger is affected by several factors, including the design and geometry of the heat exchanger, the properties of the fluids being used, the flow rates and temperatures of the fluids, and the type of heat transfer mechanisms involved (e.g. conduction, convection, radiation).

5. How can you improve the efficiency of a heat exchanger?

There are several ways to improve the efficiency of a heat exchanger, such as increasing the heat transfer surface area, increasing the flow rates of the fluids, using more efficient heat transfer fluids, and optimizing the design and placement of the heat exchanger. Regular maintenance and cleaning of the heat exchanger can also help to improve its efficiency.

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