Chapter 20, Problem 14P

To determine

Find the motor which is recommended to purchase based on the given information.

Answer to Problem 14P

Bases on the given information, the Motor – Y is recommended to purchase.

Explanation of Solution

Given data:

The power of the Motor – X $\left(P_1\right)$ and Motor – Y $\left(P_2\right)$ is $1.5\text{kW}$,

The pump is expected to run $4200\text{hours}$ every year,

Average electricity cost of the motor is $11\text{cents per kWh}$,

The number of years for the life of Motor – X $\left(n_1\right)$ and Motor – Y $\left(n_2\right)$ is $5\text{years}$,

Initial cost for Motor – X and Motor – Y is $\$300\text{and \$}\text{400}$,

Operating point efficiency of the Motor – X $\left({\epsilon }_1\right)$ and Motor – Y $\left({\epsilon }_2\right)$ is $0.75$ and $0.85$,

Maintenance cost of Motor – X and Motor – Y is $\$1\text{2 per year}\text{and \$}\text{10 per year}$.

Calculation:

Assume the interest rate as $8\%$, and the system doesn’t have salvage value.

Case (i): For Motor - X

Initially calculate the operating cost for the Motor- X using the power, efficiency at the operating point, electricity cost and pump life time.

Convert the electricity cost from cent per kWh to dollar per kWh using the following relationship.

$1\text{cents}=\text{0}\text{.01}\text{dollars}$

Therefore,

$\begin{array}{c}11\text{cents per kWh}=11\left(0.01\$\right)\text{per kWh}\\ =\$0.11\text{per kWh}\end{array}$

The average electricity cost of the motor is $\$0.11\text{per kWh}$.

Formula to calculate the operating cost of the Motor – X is,

$\text{Operating cost}=\left(\frac{P_1}{{\epsilon }_1}\right)\left(\text{Electricity}\text{cost}\right)\left(\text{Pump life time}\right)$ (1)

Here,

$P_1$ is the power of the Motor – X,

${\epsilon }_1$ s the operating efficiency of the Motor – X.

Substitute $1.5\text{kW}$ for $P_1$, $0.75$ for ${\epsilon }_1$, $\$0.11\text{per kWh}$ for $\text{Electricity}\text{cost}$, and $4200\text{hours}$ every year for $\text{Pump life time}$ in equation (1) to find $\text{Operating cost}$.

$\begin{array}{c}\text{Operating cost}=\left(\frac{1.5\text{kW}}{0.75}\right)\left(\frac{\$0.11}{\text{kWh}}\right)\left(\frac{4200\text{hours}}{\text{year}}\right)\\ =\$924\text{per year}\end{array}$

Therefore, the present worth of the Motor – X can be calculated using the following expression:

$PW=-\left(\text{Initial cost}\right)-\left(\text{Operating cost}+\text{Maintenence cost}\right)\left(\frac{P}{A},i,n_1\right)$ (2)

Here,

$P$ is the present cost,

$A$ is the uniform series payment,

$i$ is the interest rate,

$n_1$ is the number of years of life of Motor – X.

Substitute $\$300$ for $\text{Initial cost}$, $\$924$ for $\text{Operating cost}$, $\$1\text{2}$ for $\text{Maintenence cost}$, $8\%$ for $i$, and $5$ for $n_1$ in equation (2) to find $PW$.

$\begin{array}{c}PW=-\left(\$300\right)-\left(\$924+\$1\text{2}\right)\left(\frac{P}{A},8\%,5\right)\\ =-\left(\$300\right)-\left(\$936\right)\left(\frac{P}{A},8\%,5\right)\end{array}$ (3)

Refer to the Table 20.9 in the textbook, the value for $\left(\frac{P}{A},i,n\right)$ with the interest rate of $8\%$, and number of years of $5$, is $3.9928$.

Substitute $3.9928$ for $\left(\frac{P}{A},8\%,5\right)$ in equation (3) to find $PW$.

$\begin{array}{c}PW=-\left(\$300\right)-\left(\$936\right)\left(3.9928\right)\\ =-\left(\$300\right)-\left(\$3737.26\right)\\ =-\$4037.26\end{array}$ (4)

Hence, the present worth of the Motor – X is $-\$4037.26$.

Case (ii): For Motor - Y

Initially calculate the operating cost for the Motor- Y using the power, efficiency at the operating point, electricity cost and pump life time.

Convert the electricity cost from cent to dollar using the following relationship.

$1\text{cents}=\text{0}\text{.01}\text{dollars}$

Therefore,

$\begin{array}{c}11\text{cents per kWh}=11\left(0.01\$\right)\text{per kWh}\\ =\$0.11\text{per kWh}\end{array}$

The average electricity cost of the motor is $\$0.11\text{per kWh}$.

Formula to calculate the operating cost of the Motor – Y is,

$\text{Operating cost}=\left(\frac{P_2}{{\epsilon }_2}\right)\left(\text{Electricity}\text{cost}\right)\left(\text{Pump life time}\right)$ (5)

Here,

$P_2$ is the power of the Motor – Y,

${\epsilon }_2$ s the operating efficiency of the Motor – Y.

Substitute $1.5\text{kW}$ for $P_2$, $0.85$ for ${\epsilon }_2$, $\$0.11\text{per kWh}$ for $\text{Electricity}\text{cost}$, and $4200\text{hours}$ every year for $\text{Pump life time}$ in equation (5) to find $\text{Operating cost}$.

$\begin{array}{c}\text{Operating cost}=\left(\frac{1.5\text{kW}}{0.85}\right)\left(\frac{\$0.11}{\text{kWh}}\right)\left(\frac{4200\text{hours}}{\text{year}}\right)\\ =\$815.3\text{per year}\end{array}$

Therefore, the present worth of the Motor – Y can be calculated using the following expression:

$PW=-\left(\text{Initial cost}\right)-\left(\text{Operating cost}+\text{Maintenence cost}\right)\left(\frac{P}{A},i,n_2\right)$ (6)

Here,

$n_2$ is the number of years of life of Motor – Y.

Substitute $\$400$ for $\text{Initial cost}$, $\$815.3$ for $\text{Operating cost}$, $\$10$ for $\text{Maintenence cost}$, $8\%$ for $i$, and $5$ for $n_2$ in equation (6) to find $PW$.

$\begin{array}{c}PW=-\left(\$400\right)-\left(\$815.3+\$1\text{0}\right)\left(\frac{P}{A},8\%,5\right)\\ =-\left(\$400\right)-\left(\$825.3\right)\left(\frac{P}{A},8\%,5\right)\end{array}$ (7)

Refer to the Table 20.9 in the textbook, the value for $\left(\frac{P}{A},i,n\right)$ with the interest rate of $8\%$, and number of years of $5$, is $3.9928$.

Substitute $3.9928$ for $\left(\frac{P}{A},8\%,5\right)$ in equation (7) to find $PW$.

$\begin{array}{c}PW=-\left(\$400\right)-\left(\$825.3\right)\left(3.9928\right)\\ =-\left(\$400\right)-\left(\$3295.26\right)\\ =-\$3695.26\end{array}$ (8)

Hence, the present worth of the Motor – Y is $-\$3695.26$.

Compare the equations (4) and (8), from the analysis it is clear that the Motor – Y provides the least present value than the Motor – X.

Therefore, the Motor – Y should be recommended to purchase.

Conclusion:

Thus, the Motor – Y should be recommended to purchase.