Answer:
See attachment for chart
Explanation:
The IPO chart implements he following algorithm
The expressions in bracket are typical examples
<u>Input</u>
Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module
<u>Processing</u>
Assign variable to the input number (x)
Calculate the square (x = 5 * 5)
Display the result (25) ----> This will be passed to the output module
<u>Output</u>
Display 25
Answer:
hello your question is incomplete attached below is the complete question
answer: There is a hydraulic jump
Explanation:
First we have to calculate the depth of flow downstream of the gate
y1 = ----------- ( 1 )
Cc ( concentration coefficient ) = 0.61 ( assumed )
Yg ( depth of gate opening ) = 0.5
hence equation 1 becomes
y1 = 0.61 * 0.5 = 0.305 m
calculate the flow per unit width q
q = Q / b ----------- ( 2 )
Q = 10 m^3 /s
b = 2 m
hence equation 2 becomes
q = 10 / 2 = 5 m^2/s
next calculate the depth before hydraulic jump y2 by using the hydraulic equation
answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel
attached below is the remaining part of the solution
Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
<u>a) Determine the pressure at inlet to expansion process</u>
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp) ]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%