# Exercise 7

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```THERMOSIPHON
Statement
A thermo-siphon is a heating fluid loop working by natural convection. Model it as a single tube of 3 cm
in diameter and a vertical rectangular layout, i.e. two vertical pipes of 15 m each, joined by shorter pipes.
The vertical pipes are assumed thermally insulated; the lower pipe gets heat from an external water loop
of 1 kg/s, entering at 90 &ordm;C and exiting at 60 &deg;C; the upper pipe heats an air stream of 10 kg/s up to 20 &deg;C.
Pressure losses in the circuit are approximated by:
L1 2
 pt  
v
D2
where =0,025, L is hydraulic length, D hydraulic diameter,  density, and v the averaged speed in a
cross-section. As averaged thermal expansion coefficient for water take =5.10-4 K-1. Evaluate:
a)
b)
c)
d)
e)
Heat transferred to the air, and its input temperature.
Relation between friction-degraded energy, and the averaged speed and heating.
Averaged speed and mass flow rate generated.
Temperature increment in the loop, and comparison with the temperatures of the source and sink.
Entropy generation in each system and in the universe.
Se desea estudiar un circuito de calefacci&oacute;n por convecci&oacute;n natural de agua (termosif&oacute;n),
aproxim&aacute;ndolo a una &uacute;nica conducci&oacute;n de secci&oacute;n uniforme y 3 cm de di&aacute;metro en forma rectangular y
vertical, es decir, con dos tramos verticales de 15 m de altura unidos entre s&iacute; por dos tramos horizontales
de poca longitud. Se supone que los tramos verticales son adiab&aacute;ticos, que en el tramo de abajo recibe
calor de otra corriente de agua, &eacute;sta de 1 litro por segundo, que entra a 90 &deg;C y sale a 60 &deg;C, y que en el
tramo superior calienta una corriente de aire de 10 kg/s hasta 20 &deg;C. Se supone que la p&eacute;rdida de presi&oacute;n
total en el conducto puede calcularse con la f&oacute;rmula:
L1 2
 pt  
v
D2
siendo =0,025, L la longitud, D el di&aacute;metro,  la densidad y v la velocidad media en la secci&oacute;n. Para el
coeficiente de dilataci&oacute;n t&eacute;rmica del agua se toma un valor medio =5.10-4 K-1. Se pide:
a)
Calor cedido a la corriente de aire y temperatura de entrada de &eacute;ste.
a)
Relaci&oacute;n entre la energ&iacute;a mec&aacute;nica degradada por fricci&oacute;n y la velocidad y el calentamiento.
b)
c)
d)
Incremento de temperatura en el circuito y comparaci&oacute;n con las temperaturas de la fuente y el
sumidero t&eacute;rmicos.
Generaci&oacute;n de entrop&iacute;a en cada sistema y en el universo.
Solution.
This is a difficult problem because the working fluid departs from the usual models of
ideal liquids and ideal gases typical of piping problems. The problem might be easily solved by equating
the pressure loss to the pressure imbalance between the two vertical pipes, but it can be better understood
in terms of the generalised Bernoulli equation.
Thermosiphon
1
Fig. 1. Sketch of the main fluid loop and the input and output heat sources.
a)
b)
c)
d)
Heat transferred to the air, and its input temperature.
All energy input from the hot-water stream at the bottom is transported by the closed water loop to
the air stream above: Q  Qw  Qa  mwcw T90  T60   ma c pa T20  Ta1  , and substituting we get Q
=125 kW and Ta1=7.5 &ordm;C.
Relation between friction-degraded energy, and the averaged speed and heating.
dp
 em  emdf to the whole loop, w=0, em=0, and emdf&gt;0, one realises that the
Applying w  

source of motion is the non-constancy of the density in the pressure term. We may model the loop
as a hot left pipe (left because of the asymmetry in Fig. 1) of constant density, H, and a cold right
pipe of constant, but different, density, C. Neglecting the short pipes:
 pt
L1 2
p
p
 
v , although it is better to
0
 gL  emdf L , 0 
 gL  emdf L with emdf L 

D2

C
multiply each of the partial Bernoulli equations by its density and add them up, getting:
L 1 2
0  0  gL(     C )  2
 v , that with the linear expansion dilatable liquid model (4.30),
D2
L1 2
 v , that shows that the pressure head is the
  0 1   T  T0  yields: T  gL  2
D2
difference in weight of the two columns of liquid.
D2
cT
The other relation is between the heat input and the heating achieved: Q  mcT   v
4
=125 kW.
Averaged speed and mass flow rate generated.
From the two equation above we get v=0.8 m/s and T=54 &ordm;C.
Temperature increment in the loop, and comparison with the temperatures of the source and sink.
Is it possible to go up and down 54 &ordm;C when the heating water exists at 60 &ordm;C and the sink air at 20
&ordm;C? Fig. 2 shows that it is possible indeed, but countercurrent heat exchanger must be used (the
minimum thermal capacity is that of the loop: mc=2300 W/K against 4180 W/K for the hot source
water and 10000 W/K for the cold sink air).
Fig. 2. Combined temperature sketch for the two heat exchangers.
Thermosiphon
2
e)
Entropy generation in each system and in the universe.
Sloop=0
because
of
the