Maximum instantaneous power and current for a 3.0-hp pump at 240 V

[Soaring Crane]

What is the maximum instantaneous power dissipated by, and maximum current passing through, a 3.0-hp pump connected to a 240-V ac power source?

ANSWER: 6 hp, 13 A

Given:

P = 3 hp = 2238 W (1 hp = 246 W)
V = 240 V

I don't know which formula to use. At first, I attempted to use P = I*V, but, then, I got flustered over the provided formulas in my book. Peak current is mentioned, and there are different variations of P. Did I start off correctly? How am I supposed to execute this problem?

Thanks.

 

[xanthym]

What is the maximum instantaneous power dissipated by, and maximum current passing through, a 3.0-hp pump connected to a 240-V ac power source?

ANSWER: 6 hp, 13 A

Given:

P = 3 hp = 2238 W (1 hp = 246 W)
V = 240 V

Standard electrical specifications for AC Current and AC Voltage are expressed with "RMS" ("Root Mean Square") values of the AC sine wave quantities. Standard AC Power specifications are expressed with "Average Power" given by {(RMS Voltage)*(RMS Current)}. Following equations convert these standard specifications to alternate specifications concerning other properties of the AC sine wave:
{Peak AC Voltage} = (1.414)*{RMS AC Voltage}
{Peak AC Current} = (1.414)*{RMS AC Current}
{Peak AC Power} = {Peak AC Voltage}*{Peak AC Current} =
= (1.414)*{RMS AC Voltage}*(1.414)*{RMS AC Current} =
= 2*{Average AC Power}

For this problem, following values are given:
{RMS AC Voltage} = (240 V)
{Average AC Power} = (3 hp) = (2238 W)
from which can be derived:
{RMS AC Current} = {Average AC Power}/{RMS AC Voltage} = (2238)/(240) = (9.33 A)
{Peak AC Current} = (1.414)*{RMS AC Current} = (1.414)*(9.33 A) = (13.2 A)
{Peak AC Power} = (2)*{Average AC Power} = (2)*(3 hp) = (6 hp)


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[thepatientmental]

Firstly, it is important to note that power is typically measured in watts (W), not horsepower (hp). In this case, we can convert 3 hp to watts by multiplying it by the conversion factor of 746 W/hp, which gives us a power of 2238 W.

Now, to calculate the maximum instantaneous power dissipated by the pump, we can use the formula P = I*V, where P is power in watts, I is current in amperes (A), and V is voltage in volts (V). Substituting the given values, we get:

2238 W = I * 240 V

Solving for I, we get:

I = 2238 W / 240 V = 9.325 A

Therefore, the maximum instantaneous power dissipated by the pump is 2238 W and the maximum current passing through it is 9.325 A.

However, since the pump is rated at 3.0 hp, it is likely that this power refers to its rated or continuous power, not its maximum instantaneous power. So, we can also calculate the maximum power dissipated by the pump using the formula P = hp * 746 W/hp. Substituting the given values, we get:

P = 3 hp * 746 W/hp = 2238 W

This confirms that the maximum power dissipated by the pump is indeed 2238 W.

In terms of peak current, it is likely referring to the maximum current that the pump can handle without causing damage. This would also depend on the specific design and components of the pump. However, in this case, the maximum current passing through the pump is 9.325 A, which is well below the typical peak current ratings for household pumps.

In summary, the maximum instantaneous power dissipated by the 3.0-hp pump connected to a 240 V ac power source is 2238 W and the maximum current passing through it is 9.325 A.