Question

A commercial customer calls to indicate that the air-conditioning unit in a small office building was running but suddenly shut off. The technician introduces himself to the customer and carefully listens to what he has to say. Before being invited inside, the technician puts protective covers over his shoes. The technician then goes to the space thermostat and finds it set at 74°F; the space temperature is 77°F. The fan is not running. The fan switch is moved to ON, but the fan motor does not start. It is decided that the control voltage power supply should be checked first. The power supply is in the roof condensing unit on this particular installation. The technician explains to the customer the reason for leaving the building and proceeds outside. Safety Precaution: The technician places a ladder against the building away from the power line entrance. Electrical test instruments are taken to the roof along with tools to remove the panels. The breakers are checked and seem to be in the correct ON position. The panel is removed where the low-voltage terminal block is mounted. A voltmeter check shows there is line voltage but no control voltage.

Answer 1

The subject of this question is Physics, and it involves calculating the effective resistance of an air conditioning unit and the cost of running the unit for a month.

The effective resistance can be calculated using the voltage and power consumed by the unit, while the cost can be calculated using the power, time, and electricity cost per kW.h.

The subject of this question is Physics, as it involves the operation and cost calculation of an air conditioning unit.

In order to determine the effective resistance of the air conditioning unit, we can use the formula:

Resistance (R) = Voltage (V) squared / Power (P)

Given that the unit operates on 408-V AC and consumes 50.0 kW of power, we can calculate:

1. Resistance (R) = (408 V)^2 / 50.0 kW = 3342.24 ohms

To calculate the cost of running the air conditioner during a hot summer month, we can use the formula:

Cost = Power (P) x Time (t) x Electricity Cost per kW.h

Given that the air conditioner is on 8.00 h per day for 30 days and the electricity cost is 9.00 cents/kW.h, we can calculate:

1. Cost = 50.0 kW x 8.00 h/day x 30 days x 9.00 cents/kW.h = $10,800