Flat Plate Heat Exchanger. Find average convection HT coefficient

[cheme333]

Homework Statement

In this flat plate heat exchanger, rippled plates (0.3 m thick each) create channels that allow heat transfer between a hot and a cold fluid. Conduction occurs then from the heat temperature difference between the two fluids while convection occurs as the fluids move faster and turbulence occurs. The temperature distribution across one of the rippled plates at a certain instant of time is T(x) = a + bx + cx², where T is in Kelvin and x is in meters, a = 200K, b = -200K/m, and c = 30K/m². The material the plates are made of has a thermal conductivity of 1W/mK. Find the average convection heat transfer coefficient over time if the coldest surface of the middle plate is exposed to a fluid at 100K.

Homework Equations

I'm assuming all I will need is the heat transfer equations:

q_conv = h(T_s - T_$\infty$)

q_cond = -k (dT/dx)

The Attempt at a Solution

I'm not sure where to begin. I have assumed that the conduction flows from the hot fluid to the cold fluid and the convection occurs from the turbulence between the plates, but I do not know how to relate the two to solve for h.

Can someone offer any help?

Thanks.

 

[KataKoniK]

Thank you for your post. Based on the information provided, here is how you can solve for the average convection heat transfer coefficient:

1. First, we need to determine the temperature at the coldest surface of the middle plate. To do this, substitute x = 0 into the temperature distribution equation given (T(x) = a + bx + cx^2). This will give you the temperature at x = 0, which is the coldest surface of the middle plate.

2. Next, we can use the heat transfer equation for convection (qconv = h(Ts - T∞)) to calculate the heat transfer rate per unit area. Here, Ts is the temperature at the coldest surface of the middle plate (which we found in step 1) and T∞ is the temperature of the fluid (given as 100K).

3. The next step is to calculate the heat transfer rate per unit area due to conduction (qcond = -k(dT/dx)). Here, k is the thermal conductivity of the material and dT/dx is the temperature gradient in the x-direction (which can be calculated using the given temperature distribution equation).

4. Now, we can equate the two heat transfer rates (qconv = qcond) and solve for h, the average convection heat transfer coefficient.

I hope this helps. Let me know if you need further clarification or assistance. Good luck with your problem!