Suppose you must remove an average of 3.4 X 10 ^8 J of thermal energy per day to keep your house cool during the summer. If you upgrade from an old air conditioner with a COP of 2 to a new one with a COP of 6, by how many joules is the required mechanical work reduced each day? If the cost of supplying your air condition with mechanical work is $0.12 per 3.6 X 10^6J, how much money will you save during a 92 day cooling season?
To find out the amount of mechanical work reduced each day when upgrading from an old air conditioner to a new one, we first need to calculate the amount of energy that needs to be supplied to the air conditioner daily.
Given that the average thermal energy to be removed per day is 3.4 × 10^8 J, we can use the Coefficient of Performance (COP) to determine the amount of mechanical work required.
The COP of an air conditioner is defined as the ratio of thermal energy removed from the space to the mechanical work input. In this case, the COP of the old air conditioner is 2, and for the new one, it is 6.
For the old air conditioner:
COP = Thermal energy removed (Q) / Work input (W)
2 = 3.4 × 10^8 J / W
Solving for W, the work input, we have:
W = (3.4 × 10^8 J) / 2
W = 1.7 × 10^8 J
So, the old air conditioner requires 1.7 × 10^8 J of mechanical work each day.
Similarly, for the new air conditioner:
COP = 6 = 3.4 × 10^8 J / W
Solving for W, the work input, we have:
W = (3.4 × 10^8 J) / 6
W = 5.67 × 10^7 J
So, the new air conditioner requires 5.67 × 10^7 J of mechanical work each day.
The reduction in required mechanical work each day is:
Reduction = Work input (old) - Work input (new)
Reduction = 1.7 × 10^8 J - 5.67 × 10^7 J
Reduction = 1.13 × 10^8 J
Therefore, the required mechanical work is reduced by 1.13 × 10^8 J each day.
To determine the cost savings during a 92-day cooling season, we need to calculate the total reduced mechanical work.
Total reduced mechanical work = Reduction × Number of days
Total reduced mechanical work = 1.13 × 10^8 J/day × 92 days
Total reduced mechanical work = 1.04 × 10^10 J
Next, we calculate the cost savings based on the cost to supply the air conditioner with mechanical work.
Cost savings = Total reduced mechanical work / Mechanical work per cost
Cost savings = 1.04 × 10^10 J / (3.6 × 10^6 J/$0.12)
Cost savings = 2.89 × 10^3 dollars
Therefore, you will save $2,890 during the 92-day cooling season.
To calculate the reduction in required mechanical work each day, we need to first find the change in thermal energy removed per day.
The change in thermal energy removed per day is given as:
ΔQ = (COP_new – COP_old) * Q
where
COP_new = 6 (new air conditioner COP)
COP_old = 2 (old air conditioner COP)
Q = 3.4 * 10^8 J (average thermal energy to be removed per day)
Substituting the values, we get:
ΔQ = (6 - 2) * 3.4 * 10^8 J
= 4 * 3.4 * 10^8 J
= 1.36 * 10^9 J
The reduced mechanical work required each day is equal to the change in thermal energy removed per day, so it's 1.36 * 10^9 J.
To calculate the money saved during the 92-day cooling season, we need to find the total mechanical work saved and then convert it to cost.
The total mechanical work saved is given by:
W = ΔQ * t
where
ΔQ = 1.36 * 10^9 J (reduced mechanical work required per day)
t = 92 days (cooling season duration)
Substituting the values, we get:
W = 1.36 * 10^9 J * 92 days
= 1.2512 * 10^11 J
To convert the mechanical work to cost, we divide it by the conversion factor for cost:
Cost = W / (3.6 * 10^6 J per cost unit)
Substituting the values, we get:
Cost = (1.2512 * 10^11 J) / (3.6 * 10^6 J per cost unit)
= 3.47 * 10^4 cost units
To find the cost in dollars, we multiply by the cost per unit:
Cost in dollars = (3.47 * 10^4 cost units) * ($0.12 per 3.6 * 10^6 J)
= 0.12 * 3.47 * 10^4
= 4.16 * 10^3 dollars
Therefore, you would save approximately $4,160 during the 92-day cooling season by upgrading to the new air conditioner.