Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( J ) or Calories (Cal), where $$ 1 \mathrm{Cal}=4184 \mathrm{~J} $$. On the other hand, power is the rate at which energy is used and is measured in watts $$ (\mathrm{W} ; 1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) $$. Other useful units of power are kilowatts $$ (\mathrm{kW}) $$ and megawatts ( $$ 1 \mathrm{MW}=10^{6} \mathrm{~W} $$ ). If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour $$ (\mathrm{kWh}) $$, which is $$ 3.6 imes 10^{6} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$, where P is measured in MW and $$ \mathrm{t}=0 $$ corresponds to $$ 6: 00 $$ p.m. (see accompanying figure). Complete parts (a) through ( $$ (\mathrm{d}) $$ below. ( Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$. a. How much energy is consumed by this city in a typical 24 -hr period? Express the answer in MWh and in J . The energy consumed in a typical 24 -hour period in MWh is $$ \square $$ MWh. (Simplify your answer.)

Question

Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( J ) or Calories (Cal), where $$ 1 \mathrm{Cal}=4184 \mathrm{~J} $$. On the other hand, power is the rate at which energy is used and is measured in watts $$ (\mathrm{W} ; 1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) $$. Other useful units of power are kilowatts $$ (\mathrm{kW}) $$ and megawatts ( $$ 1 \mathrm{MW}=10^{6} \mathrm{~W} $$ ). If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour $$ (\mathrm{kWh}) $$, which is $$ 3.6 	imes 10^{6} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$, where P is measured in MW and $$ \mathrm{t}=0 $$ corresponds to $$ 6: 00 $$ p.m. (see accompanying figure). Complete parts (a) through ( $$ (\mathrm{d}) $$ below. ( Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$. a. How much energy is consumed by this city in a typical 24 -hr period? Express the answer in MWh and in J . The energy consumed in a typical 24 -hour period in MWh is $$ \square $$ MWh. (Simplify your answer.)

UpStudy AI · Accepted Answer

To find the total energy consumed by the city in a typical 24-hour period, we need to integrate the power function $$ P(t) = 345 - 230 \sin(\pi t / 12) $$ over the time period from 0 to 24 hours.

Given:
$$ P(t) = 345 - 230 \sin(\pi t / 12) $$

The total energy consumed in a 24-hour period can be calculated using the formula:
$$ \text{Total Energy} = \int_{0}^{24} P(t) dt $$

Substitute the given power function into the formula:
$$ \text{Total Energy} = \int_{0}^{24} (345 - 230 \sin(\pi t / 12)) dt $$

Now, we can calculate the total energy consumed by the city in a typical 24-hour period.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{24} \left(345-230\sin\left(\frac{\pi t}{12}\right)\right) dt$$
- step1: Evaluate the integral:
  $$\int \left(345-230\sin\left(\frac{\pi t}{12}\right)\right) dt$$
- step2: Use properties of integrals:
  $$\int 345 dt+\int -230\sin\left(\frac{\pi t}{12}\right) dt$$
- step3: Evaluate the integral:
  $$345t+\int -230\sin\left(\frac{\pi t}{12}\right) dt$$
- step4: Evaluate the integral:
  $$345t+\frac{2760}{\pi }\times \cos\left(\frac{\pi }{12}t\right)$$
- step5: Return the limits:
  $$\left(345t+\frac{2760}{\pi }\times \cos\left(\frac{\pi }{12}t\right)\right)\bigg |_{0}^{24}$$
- step6: Calculate the value:
  $$8280$$
The total energy consumed by the city in a typical 24-hour period is 8280 MWh.

To express the answer in Joules, we can use the conversion factor:
$$ 1 \text{ MWh} = 3.6 \times 10^6 \text{ J} $$

Therefore, the total energy consumed in Joules is:
$$ 8280 \text{ MWh} \times 3.6 \times 10^6 \text{ J/MWh} = 2985920000 \text{ J} $$

So, the total energy consumed by the city in a typical 24-hour period is 8280 MWh or 2985920000 J.
The city consumes 8280 MWh of energy in a typical 24-hour period, which is equal to 2,985,920,000 Joules.

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