Question

if a barrel plating run uses 200.0 a for exactly 6 hours for an electroplating application at 0.30 v, how many kilowatt-hours have been used in the run? if the voltage is 0.90 v, what is the power usage (in kwh)?

Answer 1

For an electroplating application, 0.36 kWh of energy was used at 0.30 V and 1.08 kWh at 0.90 V during a 6-hour barrel plating run.

Step-by-step explanation:

Calculating Kilowatt-hours for Electroplating

To solve the mathematical problem of calculating energy usage in kilowatt-hours (kWh) for an electroplating application, we employ the formula E = Pt, where E is energy in kWh, P is power in kilowatts (kW), and t is time in hours.

For the given parameters in the question:

Current (I) = 200.0 amperes (A)

Voltage (V) = 0.30 volts (V) for the first scenario, and 0.90 V for the second

Time (t) = 6 hours

First, we need to convert amps to kilowatts using the equation P = VI, where V is voltage and I is current. Therefore, at 0.30 V:

P = 0.30 V × 200.0 A = 60.0 W

Since 1 kW = 1000 W, 60.0 W is 0.060 kW.

Next, we calculate the energy used:

E = Pt = 0.060 kW × 6 h = 0.36 kWh.

Now, with a voltage of 0.90 V:

P = 0.90 V × 200.0 A = 180.0 W

Again converting to kilowatts: 180.0 W is 0.180 kW.

Calculating the energy:

E = Pt = 0.180 kW × 6 h = 1.08 kWh.

To solve the mathematical problem completely, we performed unit conversions and applied the relationship between power, time, and energy to find that 0.36 kWh of energy was used at 0.30 V and 1.08 kWh at 0.90 V during a 6-hour barrel plating run.