Chapter 22.2, Problem 31PP

a.

To determine

Power used by heater.

Answer to Problem 31PP

  $P=1.8\text{kW}$

Explanation of Solution

Given:

Current drawn by heater, $I=15\text{A}$

Source voltage, $V=120\text{V}$

Operation time is 5 h/day

Formula used:

Power consumed is given as,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

Calculation:

On substituting the given values in above relation

  $\begin{array}{l}P=120\times 15\\ =1.8\text{kW}\end{array}$

Conclusion:

Therefore, the power used by heater is $P=1.8\text{kW}$ .

b.

To determine

Energy consumed in given time.

Answer to Problem 31PP

  $E=270\text{kWh}$

Explanation of Solution

Given:

Current drawn by heater, $I=15\text{A}$

Source voltage, $V=120\text{V}$

Operation time is 5 h/day

Time is $t=30\text{days}$

Formula used:

The energy power relation is given as,

  $E=Pt$

Where, $E$ is energy, $P$ is power and $t$ is time

Calculation:

As, $t=30\text{days}$ and time of operations is 5 h/day

So total operating time in hour is $\left(30\times 5\right)\text{hours}=150\text{hours}$

And from previous subpart $P=1.8\text{kW}$

So, on substituting the values in above relation

  $\begin{array}{l}E=\left(1.8\times 150\right)\text{kWh}\\ =270\text{kWh}\end{array}$

Conclusion:

Therefore, the energy consumed is $E=270\text{kWh}$ .

c.

To determine

Cost of operation for given time and rate.

Answer to Problem 31PP

Operation cost is $\$32.4$

Explanation of Solution

Given:

Current drawn by heater, $I=15\text{A}$

Source voltage, $V=120\text{V}$

Operation time is 5 h/day

Time is $t=30\text{days}$

Rate of electricity is $ 0.12/kWh

Formula used:

The cost is given as,

  $\text{Cost = kWh}\times \text{rate}$

Calculation:

As from previous subpart, the energy consumed in 30 days is 270 kWh.

So, the cost of electricity is

  $\text{Cost = 270}\times \$0.12=\$32.4$

Conclusion:

Therefore, the cost of electricity is $\$32.4$ .