Review Problems

1-121 Cold water is to be heated in a 1200-W teapot. The time needed to heat the water is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the teapot and the water are constant. 3 Heat loss from the teapot is negligible.
Properties The average specific heats are given to be 0.6 kJ/kg.(C for the teapot and 4.18 kJ/kg.(C for water.
Analysis We take the teapot and the water in it as our system that is a closed system (fixed mass). The energy balance in this case can be expressed as [pic]
Then the amount of energy needed to raise the temperature of water and the teapot from 18(C to 96(C is [pic]
The 1500 W electric heating unit will supply energy at a rate of 1.2 …ver más…

The cost of this heat loss to the home owner is [pic]
Discussion The heat loss from the heating ducts in the attic is costing the homeowner 1.2 cents per hour. Assuming the heater operates 2,000 hours during a heating season, the annual cost of this heat loss adds up to $24. Most of this money can be saved by insulating the heating ducts in the unheated areas.

1-123
"GIVEN"
L=4 "[m]"
D=0.2 "[m]"
P_air_in=100 "[kPa]"
T_air_in=65 "[C]"
"Vel=3 [m/s], parameter to be varied"
T_air_out=60 "[C]" eta_furnace=0.82 Cost_gas=0.58 "[$/therm]"

"PROPERTIES"
R=0.287 "[kJ/kg-K], gas constant of air"
C_p=CP(air, T=25) "at room temperature"

"ANALYSIS" rho=P_air_in/(R*(T_air_in+273)) A_c=pi*D^2/4 m_dot=rho*A_c*Vel Q_dot_loss=m_dot*C_p*(T_air_in-T_air_out)*Convert(kJ/s, kJ/h)
Cost_HeatLoss=Q_dot_loss/eta_furnace*Cost_gas*Convert(kJ, therm)*Convert($, cents)

|Vel [m/s] |CostHeatLoss [Cents/h] |
|1 |0.3934 |
|2 |0.7868 |
|3 |1.18 |
|4 |1.574 |
|5 |1.967