My project is about finding the efficiency of solar trough (cylindrical parabola) concentrator that will boil water in a pipe located at the focus (check picture please). In the process of doing so (of course I made many assumptions, but I think they are irrelevant to my question now) I am trying to solve for heat loss from the pipe referred to as Q_loss (please always refer to my picture). I started by assuming Tco in equation3, and knowing T_ambient, T_sky I could find h_w and Q_loss. Using those result in eq2 I could find Tci(internal temperature of the glass envelope cover around the pipe). Now I want to substituting Tci in eq1 to find Q_loss which I will compare against the Q_loss I got from eq3 to see if my assumption about Tco was right, but the problem is that I don't have Tr(temperature on the pipe's surface which i assumed constant). How I would go about finding Tr please? I read many papers, but none of them pointed how to find it. They assume it is known or something! I have being thinking about this issue for days now. Please advise. Appreciate any guidance![enter image description here]1

  • $\begingroup$ Read Solar Engineering of thermal processes : Duffie and Beckman $\endgroup$
    – Solar Mike
    Commented Apr 9, 2018 at 21:23
  • $\begingroup$ Those equations you see in my picture I took them from his book and I could not find what is Tr from his book, or I miss-getting it. I was hoping if someone can enlighten me if it is there by explaining it to me or offer his own approach. Thank you $\endgroup$ Commented Apr 9, 2018 at 22:05
  • $\begingroup$ Tr is perhaps T radiation ? $\endgroup$
    – Solar Mike
    Commented Apr 10, 2018 at 4:01
  • $\begingroup$ Formatting your “wall of text” may make your question easier to understand. $\endgroup$
    – Solar Mike
    Commented Apr 10, 2018 at 4:18
  • $\begingroup$ If I summarize my question in 2 sentences: I want to calculate Q_loss(Heat loss from the internal pipe) using equations1,2,3, I will need Tr(temperature of the pipe's surface), how can I find Tr so that I can iterate to find Q_loss. Thank you $\endgroup$ Commented Apr 10, 2018 at 9:42

2 Answers 2



Energy Balance

Consider the picture below.

picture of tube system

This shows a $dz$ segment of the tube. Neglect radiation to/from the air and sky. The energy balance becomes

$$\dot{q}_r + \dot{q}_{fi} = \dot{q}_h + \dot{q}_{fo} $$

Expand this as

$$ A_{go} f_r \varepsilon \sigma T_r^4 = h_a A_{go} (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\Delta T_w(z) \\ 2 \pi r_{go} \Delta z f_r \varepsilon \sigma T_r^4 = h_a 2\pi r_{go} \Delta z (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\Delta T_w(z) $$

Allow this to go to a differential as

$$ 2\pi r_{go} f_r \varepsilon \sigma T_r^4 = 2 \pi r_{go} h_a (T_{go}(z) - T_a) + \dot{m}_w \tilde{C}_{pw}\frac{dT_w}{dz} $$

Tube Wall as Heat Exchanger

Write the heat flow from the water through the metal and glass as a heat exchanger expression with the overall heat transfer coefficient of the water + metal + glass (the air is already accounted in the energy balance).

$$\dot{q}_h = U A_{go} \left(T_{w}(z) - T_{go}(z)\right) = h_a A_{go}\left(T_{go}(z) - T_a\right)$$

This gives an expression for $T_{go}(z)$ as

$$ T_{go}(z) = R_U T_w(z) + R_h T_a \\ R_U = U / (U + h_a)\ \ \ R_h = h_a / (U + h_a)\ \ \ R_U + R_h = 1 $$

In this, $U$ is a combination of convection at the water/metal interface ($h_w$) + conductive resistances $k_j$ for metal tube and glass.

Tube Wall as Series Thermal Resisters

Write the heat flow from the water through the metal and glass as a series of thermal resisters for water + metal + glass.

$$ \left(T_w(z) - T_{go}(z)\right) = \dot{q}_h\left(R_w + R_t + R_g\right) \\ R_w = \frac{1}{2\pi r_{ti} \Delta z h_w} \hspace{0.5cm} R_j = \frac{\ln(r_{jo}/r_{ji})}{2\pi k_j \Delta z} $$

This leads to an expression similar to the above with $U$.

$$ \left(T_w(z) - T_{go}(z)\right) = \frac{r_{go} h_a}{\pi} \left(T_{go}(z) - T_a\right) \left(\frac{1}{r_t h_w} + \frac{\ln(r_{to}/r_{ti})}{k_t} + \frac{\ln(r_{go}/r_{gi})}{k_g}\right) $$

The relationship between $U$ and $\sum R_j$ can be derived using the two expressions for $\left(T_w(z) - T_{go}(z)\right)$.


Final Expression

Combine the energy balance and the heat flow through the tube walls (as a heat exchanger) to obtain the first order differential equation for the temperature of the water as a function of position along the tube.

$$ \dot{m}_w \tilde{C}_{pw}\frac{dT_w}{dz} + 2 \pi r_{go} h_a R_U T_w(z) - \left(R_U T_a + 2\pi r_{go} \sigma f_r \varepsilon T_r^4\right) = 0 $$

The expression has only one unknown $T_w(z)$. Only one boundary condition is required. The BC is $T_w(0) = T_{wi}$.


The first term is the enthalpy change of the water flowing through the $dz$ cross section.

The second term is the heat flow out of the water through the tube walls.

The last term is the heat flow to the air and the heat flow from the radiator.


Sketch answer, may be utterly wrong:

Do an energy balance for the pipe. At each point along the pipe, you have:

  • $Q_{in}$: Radiation heating the pipe element
  • $Q_{loss}$ as per your question, function of $T_r$ among other things
  • $Q_{trans}$ - power conducted to the medium, function of $T_r$, $T_{medium}$ and flow conditions

Assume short length elements so $T_{medium}$ is constant per length element. $Q_{loss}$ will of course not be constant along the whole length. $T_{medium}$ of course depends on conditions in the preceding element.

This also assumes that the pipe is thin, meaning the temperature across the wall of the pipe is constant.

I'll ingore convective heat transfer (space around pipe is evacuated, right?) and heat transfer along the pipe as negligible. With $Q_{in}$ known, $Q_{in} = Q_{loss} + Q_{trans}$ can maybe be solved numerically for each length element, using equation 1. As far as I can tell, $T_r$ should be the only free variable here. I have however little experience with numerical approaches to equations like these, so I'm not certain it's solvable.

If so, you'd solve for the first elemt in flow direction, find $T_{medium}$ for the second and so on.

  • $\begingroup$ Not so soon, try the solution first and tell me how it goes ;) $\endgroup$
    – mart
    Commented Apr 10, 2018 at 14:22
  • $\begingroup$ Sorry, that comment was a mistake when I hit enter key. $\endgroup$ Commented Apr 10, 2018 at 14:27
  • $\begingroup$ Thank you Mart. I don't think the equations are wrong. First, the glass pipe (2nd pipe) is not vacuumed and that is why I have K_eff in equation 1. I have water inside the pipe at 100C, the inlet water to the pipe is 20C, the other end of the pipe is closed. I assumed I have the same temperature on the receiver pipe because it is just a 2 m length pipe. Yes the incoming Q to the receiver pipe is known, but there are many other variables like Tr (inner side and outer side), Tci and Tco of the second pipe. I can probably iterate and find everything if I know Tr. I am not sure how to get that. $\endgroup$ Commented Apr 10, 2018 at 14:27
  • $\begingroup$ Then we need a term for convective heat loss, this should be a function of T_r and T_ci. Can you elaborate what T_co is? I don't understand what you mean with water at 100C but inlet is 20C? Are flow conditions known? What does pipe is closed on one end mean? You can edit your question, maybe you can provide a drawing indicating where which temeprature is measures? $\endgroup$
    – mart
    Commented Apr 10, 2018 at 14:44
  • $\begingroup$ Yes the convection term is needed. It is the 1st term of equation1. Tci is temperature of the inner glass pipe, Tco is the temperature of the outer glass pipe, Ta is ambiant temperature, Tr is the temperature of the inner pipe. the Inner pipe has water flowing in it. The inlet water temp is 20c. I boil that water up to about 100c to get saturated steam which i will be using to get distilled water. the pipe has one entrance, so basically water gets in, it starts evaporating, and the steam goes out from the inlet water entrance. The pipe is horizontal. $\endgroup$ Commented Apr 10, 2018 at 17:06

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