Chapter 27, Problem 28SP

An electric motor, which has 95 percent efficiency, uses 20 A at 110 V. What is the horsepower output of the motor? How many watts are lost in thermal energy? How many calories of thermal energy are developed per second? If the motor operates for 3.0 h, what energy, in MJ and in kW • h, is dissipated?

To determine

The horsepower output of the motor, which is $95\%$ efficient and uses $20\text{ A}$ at $110\text{ V}$, loss in thermal energy in $\text{Watts}$, the calories of thermal energy developed per second, and the energy in $\text{MJ and kW}\cdot \text{h}$ if the motor operates for $3\text{ h}$.

Answer to Problem 28SP

Solution:

$2.8\text{ hp}$, $0.11\text{ kW}$, $26\text{ cal/s}$, $6.6\text{ kW}\cdot \text{h}$, and $24\text{ MJ}$

Explanation of Solution

Given Data:

The efficiency of the motor is $95\%$.

The current supplied is $20\text{ A}$.

The voltage supplied is $110\text{ V}$.

Formula used:

The expression of electrical power is written as,

$P=VI$

Here, $P$ is the power, $I$ is the current and $V$ is the potential difference.

The expression of converting power from $\text{Watts}$ to $\text{cal/s}$ is

$P\left(\text{cal/s}\right)=P\left(\text{Watts}\right)\left(\frac{0.239\text{ cal/s}}{\text{1 Watts}}\right)$

Here, $P\left(\text{cal/s}\right)$ is power in $\text{cal/s}$ and $P\left(\text{Watts}\right)$ is power in $\text{Watts}$.

The expression of converting power from $\text{kW}$ to $\text{hp}$ is

$P\left(\text{hp}\right)=P\left(\text{kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)$

Here, $P\left(\text{hp}\right)$ is power in $\text{hp}$ and $P\left(\text{kW}\right)$ is power in $\text{kW}$.

The expression of energy is written as,

$W=Pt$

Here, $W$ is the energy and $t$ is the time.

The expression of converting power from $\text{kW}\cdot \text{h}$ to $\text{MJ}$ is

$W\left(\text{MJ}\right)=W\left(\text{kW}\cdot \text{h}\right)\left(\frac{3.6\text{ MJ}}{1\text{ kW}\cdot \text{h}}\right)$

Here, $W\left(\text{MJ}\right)$ is energy in $\text{MJ}$ and $W\left(\text{kW}\cdot \text{h}\right)$ is energy in $\text{kW}\cdot \text{h}$.

Explanation:

Recall the expression of power input.

$P_{in}=VI$

Here, $P_{in}$ is the power input.

Substitute $20\text{ A}$ for $I$ and $110\text{ V}$ for $V$

$\begin{array}{c}{P}_{in}=\left(110\text{ V}\right)\left(20\text{ A}\right)\\ =2200\text{ W}\left(\frac{0.001\text{ kW}}{1\text{ W}}\right)\\ =2.2\text{ kW}\end{array}$

Understand that the motor is $95\%$ efficient, that means that the complete input power will not be converted to output.

Recall the expression of power input.

$P_{out}=\eta P_{in}$

Here, $\eta$ is the efficiency and $P_{out}$ is the power output.

Substitute $2.2\text{ kW}$ for $P_{in}$ and $0.95$ for $\eta$

$\begin{array}{c}{P}_{out}=\left(0.95\right)\left(2.2\text{ kW}\right)\\ =2.09\text{ kW}\end{array}$

Recall the expression to convert the power from $\text{kW}$ to $\text{hp}$.

$P\left(\text{hp}\right)=P\left(\text{kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)$

Substitute $2.09\text{ kW}$ for $P\left(\text{kW}\right)$

$\begin{array}{c}P\left(\text{hp}\right)=\left(2.09\text{ kW}\right)\left(\frac{1.34\text{ hp}}{1\text{ kW}}\right)\\ =2.8\text{ hp}\end{array}$

The power output of the motor is $2.8\text{ hp}$.

Understand that the amount of loss in thermal energy is the loss of energy from the input to the output.

The expression of loss thermal energy in $\text{Watts}$ is

$P_{loss}=P_{in}-P_{out}$

Here, $P_{loss}$ is the loss thermal energy in $\text{Watts}$.

Substitute $2.2\text{ kW}$ for $P_{in}$ and $2.09\text{ kW}$ for $P_{out}$

$\begin{array}{c}{P}_{loss}=2.2\text{ kW}-2.09\text{ kW}\\ =0.11\text{ kW}\end{array}$

The loss thermal energy in $\text{Watts}$ is $0.11\text{ kW}$.

Understand that the due to energy balance, the amount of energy loss is equal to the energy developed.

Recall the expression of converting power from $\text{Watts}$ to $\text{cal/s}$.

$P\left(\text{cal/s}\right)=P\left(\text{Watts}\right)\left(\frac{0.239\text{ cal/s}}{\text{1 Watts}}\right)$

Substitute $0.11\text{ kW}$ for $P\left(\text{Watts}\right)$

$\begin{array}{c}P\left(\text{cal/s}\right)=\left(0.11\text{ kW}\left(\frac{1000\text{ W}}{1\text{ kW}}\right)\right)\left(\frac{0.239\text{ cal/s}}{1\text{ W}}\right)\\ =26.29\text{ cal/s}\\ \simeq \text{26 cal/s}\end{array}$

The calories developed per second is $26\text{ cal/s}$.

Recall the expression of energy.

$W=P_{in}t$

Substitute $2.2\text{ kW}$ for $P_{in}$ and $3\text{ h}$ for $t$

$\begin{array}{c}W=\left(2.2\text{ kW}\right)\left(3\text{ h}\right)\\ =6.6\text{ kW}\cdot \text{h}\end{array}$

Recall the expression of converting $\text{kW}\cdot \text{h}$ to $\text{MJ}$.

$W\left(\text{MJ}\right)=W\left(\text{kW}\cdot \text{h}\right)\left(\frac{3.6\text{ MJ}}{1\text{ kW}\cdot \text{h}}\right)$

Substitute $6.6\text{ kW}\cdot \text{h}$ for $W\left(\text{kW}\cdot \text{h}\right)$

$\begin{array}{c}W\left(\text{MJ}\right)=\left(6.6\text{ kW}\cdot \text{h}\right)\left(\frac{3.6\text{ MJ}}{1\text{ kW}\cdot \text{h}}\right)\\ =23.76\text{ MJ}\\ \simeq \text{24 MJ}\end{array}$

The energy dissipated if the motor works for $3\text{ h}$ in $\text{kW}\cdot \text{h}$ is $6.6\text{ kW}\cdot \text{h}$ and in $\text{MJ}$ is $24\text{ MJ}$.

Conclusion:

Therefore, the power output of the motor is $2.8\text{ hp}$, the loss thermal energy in $\text{Watts}$ is $0.11\text{ kW}$, the calories developed per second is $26\text{ cal/s}$, and the energy dissipated if the motor works for $3\text{ h}$ in $\text{kW}\cdot \text{h}$ is $6.6\text{ kW}\cdot \text{h}$ and in $\text{MJ}$ is $24\text{ MJ}$.