Answer:
The volume flow rate of air is 
Explanation:
A random duct is shown in the below attached figure
The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time
Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters
From the attached figure we can see that
The volume of the prism that the flow occupies in 1 second equals
Hence the volume flow rate is 
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:
(a) Relative Humidity = 48%,
Specific humidity = 0.0095
(b) Enthalpy = 65 KJ/Kg of dry sir
Specific volume = 0.86 m^3/Kg of dry air
(c/d) 12.78 degree C
(e) Specific volume = 0.86 m^3/Kg of dry air
Answer:
he must document or remember the order he took it apart so he put it back together
Explanation:
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
