Exercise 5 - Portal del DMT

WORK NEEDED AND WORK LOST
Statement
With a small 1 kW heater switched on, a pot of water happens to go from 80 ºC to 85 ºC in one minute.
At this moment, the heater is switched off and the water is observed to cool down to 80 ºC in another
minute. Ambient was at 20 ºC. Find:
a) Energy balance for the water, indicating the hypothesis assumed.
b) Mass of water in the pot.
c) Minimum electrical work that would have been required for the first process.
d) Work potentially lost in the second process.
Con un pequeño calentador de 1 kW sumergido en un cazo con agua ésta pasa de 80 ºC a 85 ºC en un
minuto, momento en el que se desconecta el calentador y el agua vuelve a 80 ºC en un minuto, estando el
ambiente a 20 ºC. Se pide:
a)
Balance energético del agua, indicando las hipótesis asumidas.
b)
Masa de agua.
c)
Energía eléctrica mínima que hubiese sido necesario aportar para el calentamiento antedicho
(límite termodinámico en ese ambiente).
d)
Irreversibilidad en ambos procesos (i.e. trabajo perdido, potencialmente).
Solution.
a) Energy balance for the water, indicating the hypothesis assumed.
The energy balance for a control mass, i.e. assuming there is no evaporation, is:
ΔE  W  Q
Assuming a perfect-liquid model (incompressible and with constant thermal capacity), E=mcT.
No work flows through the water frontier.
Heat flow through the water frontier may be split in two parts: through the heater and through the walls.
Naming 1, 2 and 3 the three states described in the statement, the energy balance for each process is:
Heating, from 1 to 2:
mc(T2  T1 )  Qin  Qout ,12 , with Qin  Wele t and Qout ,12   KA(T  Tamb )dt
Cooling, from 2 to 3:
mc(T3  T2 )   Qout ,23 , with Qout ,23   KA(T  Tamb )dt
where an expression for the heat transfer through the walls in terms of the temperature jump is given just
to make more clear that heat losses can be assumed equal during both the heating period and the cooling
period, because the heat rate is the same (same temperature jump) and acts during the same amount of
time.
b) Mass of water in the pot.
With the assumptions above, we have a system of two equations with two unknowns:
Work needed and work lost
1
mc(T2  T1 )  Welet  Qout
mc(T3  T2 ) 
 Qout

 m  4200(85  80)  1000  60  Qout 
m  1.44 kg

 m  4200(80  85)  Qout
Qout  30 kJ  Qout  500 W


i.e., the hot-water-pot looses heat to the ambient at a rate of 500 W, and the 1000 W of the heater solit in
500 W to compensate the losses and 500 W to increase the water temperature.
c) Minimum electrical work that would have been required for the first process.
The minimum required work is the exergy increase:
Wumin  ΔE  p0ΔV  T0ΔS = mc(T2  T1 )  0  T0 mc ln T2 / T1  
 1.44  4200  85  80  (20  273) 1.44  4200ln 358/ 353  5.3 kJ
i.e., it might be enough to draw 5.3 kJ of electricity (an average of 88 W, instead of the 1000 W drawn by
the heater), and, with a suitable machine that does not consume itself, help to draw another (30-5.3)=24.7
kJ from the ambient, to pass water from 80 ºC to 85 ºC without losses. Notice, by the way, the great error
that might occur by mixing up the temperature scales when substituting in the expression above (in doubt,
it is better to strictly keep to absolute temperatures and never use the Celsius scale).
d) Work potentially lost in the second process.
If, instead of just letting the water to loose energy by heat transfer to the environment, this energy were to
be fully used to obtain a maximum of work, the limit for the latter would be (the exergy decrease):
Wumin  ΔE  p0ΔV  T0ΔS = mc(T3  T2 )  0  T0mc ln T3 / T2  
 1.44  4200  80  85  (20  273) 1.44  4200ln 353/ 358   5.3 kJ
meaning that 5.3 kJ of work might have been produced (the minus sign) by wisely cooling 1.44 kg of
water from 85 ºC to 80 ºC (by using an appropriate heat engine working in between that source and the
environment).
Back
Work needed and work lost
2