She should create a computer animated view of the design to walk the client through it so that client will understand and get the picture of the design.
This question is incomplete, the complete question is;
the water depths upstream and downstream of a hydraulic jump is 0.3 m and 1.2 m. Determine flow rate ( in m³ ) if the rectangular channel is 20 m wide.
Answer:
the flow rate is 32.549 m³/sec
Explanation:
Given that
y₁ = 0.3 m
y₂ = 1.2 m
β = 20 m
Now for Rectangular Channel, we know that;
2q²/g = y₁y₂( y₁ + y₂)
where g = 9.81 m/s²
and q = Q/β
so
2(Q/β)²/g = y₁y₂( y₁ + y₂)
we substitute our given values
2(Q/20)²/9.81 = 0.3 × 1.2( 0.3 + 1.2)
2(Q²/400)/9.81 = 0.36(1.5)
2(Q²/400) = 0.54 × 9.81
Q²/400 = 5.2974 / 2
Q²/400 = 2.6587
Q² = 1059.48
Q = √1059.48
Q = 32.549 m³/sec
Therefore the flow rate is 32.549 m³/sec
Answer:
10.984mm
Explanation:
by elastic modulus
stress=modulus of elasticity*strain
stress=loading/area area" cross-section"
11mm=0.011m
area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter
stress=55000/(9.503*10^-5)=578.745 MPa
convert MPa and GPa to pascal.
strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain
v=Poisson ratio
lateral strain=(-v)*axial strain= -0.31*0.00463
lateral strain= -1.4353*10^-3=change in diameter/ original diameter
change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m
negative indicates decrease in diameter.
decrease in dia.=0.01578mm
new diameter=11-0.01578= 10.984mm
Answer:
Magnitude of P=8.624 lb
Explanation:
The complete solution to the problem is given in the attachments.