Question

An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner puts a 30-ft^2 blanket on the water heater, raising its total R value to 15. Assuming 100 percent conversion of electricity costs 6.0 cents/kWh, how much money will be saved in the energy each year?

Answer 1

Total saving would be of 36.917 $\yr

Step-by-step explanation:

Given Data:

`<strong>T_(heater) = 120</strong>` Degree F

`<strong> T_(room) = 60</strong>` Degree F

A = 30 ft^2

`<strong>\eta = 100</strong>`%

Heat loss before previous final value
`= (A \Delta T)/(R)`

`=(30* *(120-60))/(5)`

= 360 Btu/hr

Heat loss after new value
`= (30* * (120-60))/(15) = 120 Btu/hr`

saving would be
`= 360 - 120 Btu/hr * kw hr/ 3412 Btu* 24 hr/day * 365 day/year`

= 616.1782 kw hr/yr

`cost = 616.1782 * 0.06`$

= 36.917 $\yr