This function takes kW, % utilization, $/kWhr and determines power cost.
Usage
solvecost_power(power, utilization = 100, cost, time = "day")
Arguments
- power
Power consumed in kW
- utilization
Amount of time equipment is running in percent. Defaults to continuous.
- cost
Power cost in $/kWhr
- time
Desired output units, one of c("day", "month", "year"). Defaults to "day".
Value
A numeric value for power, $/time.
Examples
powercost <- solvecost_power(50, 100, .08)
cost_data <- data.frame(
power = seq(10, 50, 10),
utilization = 80
) %>%
dplyr::mutate(costs = solvecost_power(power = power, utilization = utilization, cost = .08))
