Your HP-10C: A Problem Solver 9
q is the heat loss from the water
heater (Btu per hour).
h is the heat-transfer coefficient.
A is the total surface area of the
cylinder.
T is the temperature difference
between the cylinder and the
surrounding air.
Example: Assume you have a 52-gallon
cylindrical water heater and you wish to
determine how much energy is being lost because of poor insulation. In initial
measurements, you found an average temperature difference between the heater
surface and surrounding air of 15° F. The surface area of the tank is 30 square
feet and the heat transfer coefficient is approximately 0.47. To calculate the heat
loss of the water heater, simply press the following keys in order.
Keystrokes Display
15v
30
15.0000
30.
Input temperature
difference (T) and area
of water heater {A).
*
450.0000
Calculates A × T.
.47
0.47
Heat-transfer coefficient
(h).
*
211.5000
Heat loss in Btu per hour
(h × AT).
Programmed Solutions
The heat loss for the water heater in the preceding example was calculated for a
15° temperature difference. But suppose you want to calculate the heat loss for
several temperature differences? You could perform each heat loss calculation
manually. However, an easier and faster method is to write a program that will
calculate the heat loss for any temperature difference.
Writing the Program. The program is the same series of keystrokes you
executed to solve the problem manually.
Loading the Program. To load the instructions of the program into the HP-10C
press the following keys in order. The calculator records (remembers) the
instructions as you key them in. (The display gives you information you will
find useful later, but which you can ignore for now.)
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