Recent content by jackcarroll434

The heat transfer coefficients and thermal conductivity values were all taken from the internet. The heat trasnfer coefficeint for air assumes only free convection is taking place, while the heat transfer coefficient for water assumes slow laminar flow of water in a tube. I intend to use these...

Thanks for the help Chester!

My T* came to 287.91 K - slightly different

I would ignore the value and derivation I used in post #1. Since post #2 we have modelled the total heat supplied as the sum of heat in and heat out. The heat out term (##q_{out}##) deals with the convective losses to the environment (1 term), whereas the heat in (##q_{in}##) term deals with...

Interesting - since your first post (post #2) I've been working on the basis that $$ U_{inner} = \frac{1}{ \frac{1}{h_i} + \frac{r_1 ln( \frac{r_2}{r_1})}{k_{pvc}L} + \frac{r_1 ln( \frac{r_3}{r_2})}{k_{alu}L} }$$ Therefore I do not think the outside resistance has been coutned twice, as it's...

Putting ## dx ## on the RHS and the ##T## terms on the LHS, integrating between the start and end boundary conditions of ## T=20 ^{\circ}C ## at ## x=0##, and ## T = 37 ^{\circ}C## at ## x=0.2m##, I get a value of 72.36 W for ## q ##. How does this compare to your solution?

Ah I see. It makes sense now that you have introduced the new term $$U = \frac{1}{\frac{1}{U_{inner}}+\frac{r_1}{r_3h_0}}$$ So this would make the differential equation: $$\dot{m}C\frac{dT}{dx} = 2\pi r_1U[T^*+T_\infty - T]$$ I now follow

If I substite ##q_{out}=q-q_{in}## then: $$ q_{in}=2 \pi r_1U_{inner}( \frac{q-q_{in}}{2 \pi r_3 h_o} + T_\infty - T)$$ $$q_{in}= \frac{r_1U_{inner}q}{r_3h_0} - \frac{r_1U_{inner}q_{in}}{r_3h_0} + 2 \pi r_1U_{inner}T_\infty - 2 \pi r_1U_{inner}T$$ Unless the term ## \frac{r_1U_{inner}}{r_3h_0}##...

As per your suggestion, I found $$ T_1 = \frac{q_{in}}{2 \pi r_1 h_i} + T$$ $$ T_2 = \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} + \frac{q_{in}}{2 \pi r_1 h_i} + T $$ $$ T_3 = \frac{q_{in} \ln \frac{r_3}{r_2}}{2 \pi k_{al}} + \frac{q_{in} \ln \frac{r_2}{r_1}}{2 \pi k_{pvc}} +...

Hi Chester, I understand your approach and follow what you have done, however I have not yet quite worked through the derivation of Q(in) in terms of T*, T(atm) and T. I will have another go later this evening.

Homework Statement: Calculate the power of heating source required in a heat exchanger Relevant Equations: Heat transfer for LMTD heat exchanger. I have a real-world problem whereby: Water is flowing at a constant flow rate of 10 mL/min through a PVC tube, inner radius of 1.25mm and outer...