Page 503 - Advanced thermodynamics for engineers

20.9 PROBLEMS 495

where
h ¼ specific enthalpy of the fluid at temperature T;
u* ¼ the energy transported when there is no heat flow through thermal conduction;
v ¼ specific volume; and
T ¼ temperature.
P20.4 A thermocouple is connected across a battery, and a current flows through it. The cold
junction is connected to a reservoir at 0 C. When its hot junction is connected to a reservoir

at 100 C the heat flux due to the Peltier effect is 2.68 mW/A, and when the hot junction is at

200 C the effect is 4.11 mW/A. If the emf of the thermocouple due to the Seebeck effect is

2
given by ε ¼ at þ bt , calculate the values of the constants a and b. If the thermocouple is
used to measure the temperature effect based on the Seebeck effect, i.e. there is no current
flow, calculate the voltages at 100 C and 200 C.

2
[5.679 10 3 mV/K; 7.526 10 3 mV/(K) ; 0.6432 mV; 1.437 mV]
P20.5 A pure monatomic perfect gas with c p ¼ 5<=2 flows from one reservoir to another through a
porous plug. The heat of transport of the gas through the plug is <T=2. If the system is
adiabatic, and the thermal conductivities of the gas and the plug are negligible, evaluate the
temperature of the plug if the upstream temperature is 60 C.

[73 C]

P20.6 A thermal conductor with constant thermal and electrical conductivities, k and l respectively,
connects two reservoirs at different temperatures and also carries an electrical current of
density, J I . Show that the temperature distribution for one-dimensional flows is given by
2
d T J I s dT J I 2
þ ¼ 0
dx 2 k dx l
where s is the Thomson coefficient of the wire.
P20.7 A thermal conductor of constant cross-sectional area connects two reservoirs which are both
maintained at the same temperature, T 0 . An electric current is passed through the conductor,
and heats it due to Joulean heating and the Thomson effect. Show that if the thermal and
electrical conductivities, k and l, and the Thomson coefficient, s, are constant the
temperature in the conductor is given by
x
J I k J I k J I sL x
L 1 :
T T 0 ¼ e k ð Þ
J I sL
lsL L lsL e k 1
Show that the maximum temperature is acheved at a distance
8 9
J I sL
x k
<k e k 1 =
¼ ln :
L J I sL : J I sL ;
Evaluate where the maximum temperature will occur if J I sL ¼ 1, and explain why it is not in
k
the centre of the bar. Show that the maximum temperature achieved by Joulean heating alone is
in the centre of the conductor.
[x/L ¼ 0.541].