Chapter 18, Problem 47P

(a)

To determine

The maximum instantaneous dissipated power by a pump.

Answer to Problem 47P

The maximum instantaneous dissipated power by the pump is $4476\text{W}$ .

Explanation of Solution

Given:

Power of pump, $P=3\text{hp}$

Voltage supply, $V_{rms}=240\text{V}$

Formula used:

Write the expression to calculate the value of rms voltage.

  $V_{rms}=\frac{V_o}{\sqrt{2}}$  ... (1)

Where,

  $V_o$ is the peak voltage.

Write the expression to calculate the value of rms current.

  $I_{rms}=\frac{I_o}{\sqrt{2}}$  ... (2)

Where,

  $I_o$ is the peak current.

Write the expression to calculate the peak power.

  $P_o=V_o{I}_o$  ... (3)

Where,

  $P_o$ is the maximum instantaneous dissipated power by the pump.

Calculation:

Consider the conversion factor to convert the horsepower to watt.

  $\begin{array}{c}3\text{hp}=3\times 746\text{W}\\ 3\text{hp}=2238\text{W}\end{array}$

Calculate the peak voltage from equation (1).

  $\begin{array}{c}240=\frac{V_o}{\sqrt{2}}\\ V_o=240\times \sqrt{2}\\ =339.41\text{V}\end{array}$

Since, the peak current is the maximum current passing through the pump. Use power formula to calculate the value of rms current.

  $P=V_{rms}{I}_{rms}$

Substitute the values in the above expression.

  $\begin{array}{c}2238=240\cdot I_{rms}\\ I_{rms}=\frac{2238}{240}\\ =9.325\text{A}\end{array}$

Substitute the value of the rms current in equation (2) to calculate the peak current.

  $\begin{array}{c}9.325=\frac{I_o}{\sqrt{2}}\\ I_o=9.325\times \sqrt{2}\\ =13.19\text{A}\end{array}$

Substitute the values in equation (3) to calculate the maximum instantaneous dissipated power by the pump.

  $\begin{array}{c}{P}_o=339.41\times 13.19\\ =4475.9\text{W}\\ \approx 4476\text{W}\end{array}$

Conclusion:

Therefore, the maximum instantaneous dissipated power by the pump is $4476\text{W}$ .

(b)

To determine

The maximum current passing through the pump.

Answer to Problem 47P

The maximum current passing through the pump is $13.19\text{A}$ .

Explanation of Solution

Given:

Power of pump, $P=3\text{hp}$

Voltage supply, $V_{rms}=240\text{V}$

Formula used:

Write the expression to calculate the value of rms current.

  $I_{rms}=\frac{I_o}{\sqrt{2}}$  ... (4)

Where,

  $I_o$ is the peak current.

Calculation:

From part (a), the value of the peak current is $13.19\text{A}$ . Therefore, the maximum current passing through the pump is $13.19\text{A}$ .

Therefore, the maximum current passing through the pump is $13.19\text{A}$ .