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:
You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)
Step by step design:
- Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
- Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
- Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA
Our circuit meet the average voltage (Va) specification:
Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it
A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.
<h3>What is a chemical engineer?</h3>
- Processes for manufacturing chemicals are created and designed by chemical engineers.
- To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
- A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
- Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.
To learn more about chemical engineer, refer to:
brainly.com/question/23542721
#SPJ4
Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars
Explanation :
A)
Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}
= 3780kJ
And 1 hour = 3600s
Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W
B)
At 15km/hour a 15km run takes 1 hour.
1 hour is 3600s and the runner burns 1050 joule per second.
Energy used in 1 hour = 3600 x 1050 J/s
= 3780000 J or 3.78MJ
C)
1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km
15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ
Finally,
1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ
This means that the runner needs 5320/1008 = 5.3 bars
