Thermal Efficiency Apparatus
012-05443A
16
The Thermal Efficiency Apparatus is run with a load connected to measure P
H
(Figure
6) and then the load is disconnected and the power input into the hot reservoir is ad-
justed to maintain the temperatures (less power is needed when there is no load since
less heat is being drawn from the hot reservoir). See Figure 7. P
H(OPEN)
is the power
input to the hot reservoir when no load is present. Since, while there is no load, the hot
reservoir is maintained at an equilibrium temperature, the heat put into the hot reservoir
by the heating resistor must equal the heat radiated and conducted away from the hot
reservoir. So measuring the heat input when there is no load determines the heat loss
due to radiation and conduction. It is assumed this loss is the same when there is a load
and the heat engine is operating.
Having accounted for the obvious energy losses, the adjusted efficiency should match
the Carnot efficiency which assumes no energy loss. The adjusted efficiency is
e
adjusted
′
=
P
W
′
P
H
′
=
P
W
+ I
W
2
r
P
H
– P
H open
Calculate the internal resistance, r, using the equation
r =
V
P
– V
W
V
W
R
which is derived in the Indirect Measurement section. Record this resistance in Table
2.2. Then calculate the adjusted efficiency and record the result in Table 2.3.
Calculate the percent difference between the adjusted efficiency and the Carnot (maxi-
mum) efficiency
% Difference =
e
max
–
e
adjusted
e
max
×
100%
and record in Table 2.3.
Questions
➀
If the difference between the temperature of the hot side and the cold side was decreased,
would the maximum efficiency increase or decrease?
➁
The actual efficiency of this heat engine is very low and yet heat engines of this type are
used extensively in remote areas to run things. How can such an inefficient device be of
practical use?
➂
Calculate the rate of change in entropy for the system which includes the hot and cold
reservoirs. Since the reservoirs are at constant temperature, the rate of change in entropy is
∆
S
∆
t
=
∆
Q /
∆
t
T
=
P
T
for each reservoir. Is the total change in entropy positive or negative? Why?
