Answer:
Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds
ANSWER : 104.904 ft^3/sec
Explanation:
General concept : Average flow rate is the volume of fluid per unit time through an area
Hence the average flow rate of the air conditioning unit of this room
Volume of the room / time taken for the air to cycle the room = v / t
assuming the time taken = 25 seconds
volume of room = width * length * height
= 14.1 * 15.5 * 12 = 2622.6 ft^3
Average flow rate = V/ t
= 2622.6 / 25 = 104.904 ft^3/sec
Answer: B
Explanation: unless newer models added wingding to code inside fused computer...wingdings on a window ...not a motor
Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h

b) The coefficient of performance is:

c) The coefficient of performance of a reversible heat pump is:

Th=20+273=293 K
Tl=10+273=283K
Replacing:

Answer:
the pressure gradient in the x direction = -15.48Pa/m
Explanation:
- The concept of partial differentiation was used in the determination of the expression for u and v.
- each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.