Drafting has been around a long time. We can safely assume that since we’ve had a tool in our hands, we’ve been describing plans and technical representations and doodling ideas. Let’s take a closer aspect at drafting and its advance from an under-the-radar part of the method to a very developed skill set.
<u>Explanation</u>
• 1970s – The beginning computer-aided design systems were included in the industry. Following the design engineers tried the learning curve of using CAD, their performance and productivity went through the roof. Over time, CAD software became affordable and more user-friendly, and its fame grew.
• 1990s – CAD software was expanded further to include 3-D characteristics, and quickly the technical designs of the past enhanced increasingly simulated and accessible to engineer.
• Present – The development of drafting has brought us to the present day, were using 3-D representations is the standard and the aim to generate full virtual prototypes.
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
Answer:
0.0659 A
Explanation:
Given that :
( saturation current )
at 25°c = 300 k ( room temperature )
n = 2 for silicon diode
Determine the saturation current at 100 degrees = 373 k
Diode equation at room temperature = I = Io 
next we have to determine the value of V at 373 k
q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1
Given that I is constant
Io = 
= 0.0659 A
Answer:
2.379m
Explanation:
The width = 23m
The depth = 3m
The radius is denoted as R
The wetted area is = A
The perimeter perimeter = P
Hydraulic radius
R = A/P
The area of a rectangular channel
= Width multiplied by Depth
A = 23x3
A = 69m²
Perimeter = (2x3)+23
P = 6+23
P= 29
Hydraulic radius R = 69/29
= 2.379m
This answers the question
Thank you!
