Answer:
The overflow rate is 4.24×10^-4 m/s.
The detention time is 7069.5 s
Explanation:
Overflow rate is given as volumetric flow rate ÷ area
volumetric flow rate = 0.3 m^3/s
area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2
Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s
Detention time = volume ÷ volumetric flow rate
volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3
Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s
Answer:
A.) 0.08 kJ/kg.K
B.) 207.8 KJ/Kg
C.) 0.808
Explanation:
From the question, the use of fluids mechanic table will be required. In order to get the compressor processes, the kinetic energy and the potential energy will be negligible while applying the ideal gas model.
Since the steam is a closed system, the carbon dioxide will be compressed adiabatically.
Please find the attached files for the solution and the remaining explanation.
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, 
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area,
where d is the diameter
By substitution

To convert v to m/s from m/s, we simply divide it by 60 hence

Answer:
Minimum electrical power required = 3.784 Watts
Minimum battery size needed = 3.03 Amp-hr
Explanation:
Temperature of the beverages, 
Outside temperature, 
rate of insulation, 
To get the minimum electrical power required, use the relation below:

V = 5 V
Power = IV

If the cooler is supposed to work for 4 hours, t = 4 hours

Minimum battery size needed = 3.03 Amp-hr
Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2