Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
For pipe 2
Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV
head loss (h)
Now putting the all values
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
1) 1.4(D + F)
2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)
3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)
4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)
5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S
6) 0.9D + 1.6W + 1.6H
7) 0.9D + 1.0E + 1.6H
Explanation:
Load and Resistance Factor Design
there are 7 basic load combination of LRFD that is
1) 1.4(D + F)
2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)
3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)
4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)
5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S
6) 0.9D + 1.6W + 1.6H
7) 0.9D + 1.0E + 1.6H
and
here load factor for L given ( * ) mean it is permitted = 0.5 for occupancies when live load is less than or equal to 100 psf
here
D is dead load and L is live load
E is earth quake load and S is snow load
W is wind load and R is rain load
Lr is roof live load
What is the question? It looks like a statement...
Answer:
11.2mm or 0.45in
Explanation:
The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.
Please go through the attached file for a step by step solution to this question.
