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Answer:
Weight of cement = 10968 lb
Weight of sand = 18105.9 lb
Weight of gravel = 28203.55 lb
Weight of water = 5484 lb
Explanation:
Given:
Entrained air = 7.5%
Length, L = 40 ft
Width,w = 12 ft
thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft
Specific gravity of sand = 2.60
Specific gravity of gravel = 2.70
The volume will be:
40 * 12 * 0.5 = 240 ft³
We need to find the dry volume of concrete.
Dry volume = wet volume * 1.54 (concrete)
Dry volume will be = 240 * 1.54 = 360ft³
Due to the 7% entarained air content, the required volume will be:
V = 360 * (1 - 0.07)
V = 334.8 ft³
At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:
Total of ratio = 1+2+3 = 6
Their respective volume will be =
Volume of cement =
Volume of sand =
Volume of gravel =
To find the pounds needed the driveway, we have:
Weight = volume *specific gravity * density of water
Specific gravity of cement = 3.15
Weight of cement =
55.8 * 3.15 * 62.4 = 10968 pounds
Weight of sand =
111.6 * 2.60 * 62.4 = 18105.9 lb
Weight of gravel =
167.4 * 2.7 * 62.4 = 28203.55 lb
Given water to cement ratio of 0.50
Weight of water = 0.5 of weight of cement
= 1/2 * 10968 = 5484 lb
Answer:
Head loss is 1.64
Explanation:
Given data:
Length (L) = 200 m
Discharge (Q) = 0.16 m3/s
According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m
We know,
where, f = Darcy friction factor
V = flow velocity
g = acceleration due to gravity
We know, flow rate Q = A x V
solving for V
obtained Darcy friction factor
calculate Reynold number (Re) ,
where, = density of water
= Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2
so reynold number is
= 6.4 x 10^5
For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm
Relative roughness (e/D) = 0.0015 / 318 = 0.00005
from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126
Therefore head loss is
HL = 1.64 m