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

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

$T_{heater} = 120$ Degree F

$T_{room} = 60$ Degree F

A = 30 ft^2

$\eta = 100$%

Heat loss before previous final value $= \frac{A \Delta T}{R}$

                                                              $=\frac{30\times *(120-60)}{5}$

                                                              = 360 Btu/hr

Heat loss after new value$= \frac{30\times \times (120-60)}{15} = 120 Btu/hr$

saving would be $= 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year$

                           = 616.1782 kw hr/yr

$cost = 616.1782 \times 0.06$$

         = 36.917 $\yr