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Chapter 2: Problem 39

An electric heater draws a constant current of 6 amp, with an applied voltage of \(220 \mathrm{~V}\), for \(24 \mathrm{~h}\). Determine the instantaneous electric power provided to the heater, in \(\mathrm{kW}\), and the total amount of energy supplied to the heater by electrical work, in \(\mathrm{kW} \cdot \mathrm{h}\). If electric power is valued at \(\$ 0.08 / \mathrm{kW} \cdot \mathrm{h}\), determine the cost of operation for one day.

Short Answer

Expert verified
The instantaneous power is 1.32 kW. The total energy supplied is 31.68 kWh. The cost of operation for one day is $2.53.

Step by step solution

01

Determine Instantaneous Electric Power

The instantaneous electric power (P) provided to the heater can be calculated using Ohm's Law: \[ P = IV \] where I is the current in amperes (A) and V is the voltage in volts (V). Given: I = 6 A V = 220 V Substitute the values: \[ P = 6 \times 220 \] \[ P = 1320 \text{ W} \] To convert to kW, divide by 1000: \[ P = \frac{1320}{1000} = 1.32 \text{ kW} \]
02

Calculate Total Energy Supplied

To find the total energy supplied, use the formula: \[ E = P \times t \] where E is energy in kilowatt-hours (kWh), P is power in kilowatts (kW), and t is time in hours (h). Given: P = 1.32 kW t = 24 h Substitute the values: \[ E = 1.32 \times 24 \] \[ E = 31.68 \text{ kW} \text{h} \]
03

Determine the Cost of Operation

Electric power is valued at \(0.08 per kWh. To find the cost of operation for one day: \[ \text{Cost} = E \times \text{rate} \] Given: E = 31.68 kWh rate = \)0.08 per kWh Substitute the values: \[ \text{Cost} = 31.68 \times 0.08 \] \[ \text{Cost} = 2.5344 \text{ USD} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law and Its Application
Ohm's Law is crucial for understanding electric circuits and forms the basis for electric power calculations. It states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) between them. This can be written as \(V = IR\).

In the context of the exercise, Ohm's Law helps us determine the power (P) provided to the electric heater. The power (P) in an electric circuit is calculated by multiplying voltage (V) and current (I): \(P = IV\). Given the current (6 A) and voltage (220 V), we can find the instantaneous power as 1320 watts (or 1.32 kW when converted by dividing by 1000).

Key points to remember:
  • High voltage with a constant current increases power.
  • The formula \(P = IV\) is essential for finding the power output in electric devices.
  • Ensure clear distinctions between units like watts (W) and kilowatts (kW) for accurate calculations.
Understanding Instantaneous Power
Instantaneous power refers to the electric power provided at any given moment. To calculate it, we use the formula: \(P = IV\). This formula applies equally whether we鈥檙e looking at direct current (DC) or alternating current (AC) systems, provided the voltage and current are constant.

For our exercise, we're told the voltage is 220 V and the current is 6 A. By multiplying these two figures (as stated in the step-by-step solution), we determine the instantaneous power to be 1320 W (or 1.32 kW). This represents the heater's power usage at any moment during operation.

Important aspects:
  • Instantaneous power gives a snapshot of power consumption at any given time.
  • Constant current and voltage make calculation straightforward.
  • Always convert watts to kilowatts for practical usage (1 kW = 1000 W).
Calculating Energy Consumption
Energy consumption over time is a crucial aspect of electric power management. For calculating the total energy consumed, use the formula \(E = P \times t\), where P is the power in kilowatts (kW) and t is the time in hours (h).

In our exercise, the heater's instantaneous power was 1.32 kW, and it operated for 24 hours. Multiplying these values gives us a total energy consumption of \(1.32 \times 24 = 31.68 \text{ kWh}\).

This calculation is useful not only for understanding energy usage but also for determining operational costs. Since electric power is often billed in kWh, knowing energy consumption helps estimate costs, as shown in the final step where the total cost for one day's operation is calculated at $2.53.

Summary points:
  • Energy (kWh) = Power (kW) x Time (h).
  • Helps in calculating electricity bills.
  • Essential for energy efficiency and cost-saving strategies.

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Most popular questions from this chapter

A vertical piston-cylinder assembly with a piston of mass \(25 \mathrm{~kg}\) and having a face area of \(0.005 \mathrm{~m}^{2}\) contains air. The mass of air is \(2.5 \mathrm{~g}\), and initially the air occupies a volume of \(2.5\) liters. The atmosphere exerts a pressure of \(100 \mathrm{kPa}\) on the top of the piston. The volume of the air slowly decreases to \(0.001 \mathrm{~m}^{3}\) as energy with a magnitude of \(1 \mathrm{~kJ}\) is slowly removed by heat transfer. Neglecting friction between the piston and the cylinder wall, determine the change in specific internal energy of the air, in \(\mathrm{kJ} / \mathrm{kg}\). Let \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\).

An object of mass \(10 \mathrm{~kg}\), initially at rest, experiences a constant horizontal acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\) due to the action of a resultant force applied for \(20 \mathrm{~s}\). Determine the total amount of energy transfer by work, in kJ.

A construction crane weighing \(12,000 \mathrm{lbf}\) fell from a height of \(400 \mathrm{ft}\) to the street below during a severe storm. For \(g=\) \(32.05 \mathrm{ft} / \mathrm{s}^{2}\), determine the mass, in \(\mathrm{lb}\), and the change in gravitational potential energy of the crane, in \(\mathrm{ft} \cdot \mathrm{lbf}\).

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes: Process 1-2: Compression with \(p V=\) constant, from \(p_{1}=1\) bar, \(V_{1}=2 \mathrm{~m}^{3}\) to \(V_{2}=0.2 \mathrm{~m}^{3}, U_{2}-U_{1}=100 \mathrm{~kJ}\). Process 2-3: Constant volume to \(p_{3}=p_{1}\). Process 3-1: Constant-pressure and adiabatic process. There are no significant changes in kinetic or potential energy. Determine the net work of the cycle, in kJ, and the heat transfer for process \(2-3\), in \(\mathrm{kJ}\). Is this a power cycle or a refrigeration cycle? Explain.

A gas within a piston-cylinder assembly undergoes a thermodynamic cycle consisting of three processes in series, beginning at state 1 where \(p_{1}=1\) bar, \(V_{1}=1.5 \mathrm{~m}^{3}\), as follows: Process 1-2: Compression with \(p V=\) constant, \(W_{12}=-104 \mathrm{~kJ}\), \(U_{1}=512 \mathrm{~kJ}, U_{2}=690 \mathrm{~kJ} .\) Process 2-3: \(W_{23}=0, Q_{23}=-150 \mathrm{kJJ}\). Process 3-1: \(W_{31}=+50 \mathrm{~kJ}\). There are no changes in kinetic or potential energy. (a) Determine \(Q_{12}, Q_{31}\), and \(U_{3}\), each in kJ. (b) Can this cycle be a power cycle? Explain.
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