It would be ten because your ten feet away
Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.
Answer:
Option A
Chemical engineering
Explanation:
Chemical engineering mainly encompass the study of behavior of different particles such as petroleum, water, drugs and other products. When Anne is involved in a study with engineers who study flow of particles, the flow, viscosity and other properties are among the behavior that chemical engineers are involved in.
Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( 
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
