Answer:
h_f = 15 ft, so option A is correct
Explanation:
The formula for head loss is given by;
h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))
Where;
h_f is head loss due to friction in ft
L is length of pipe in ft
Q is flow rate of water in gpm
C is hazen Williams constant
D is diameter of pipe in inches
We are given;
L = 1,800 ft
Q = 600 gpm
C = 120
D = 8 inches
So, plugging in these values into the equation, we have;
h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))
h_f = 14.896 ft.
So, h_f is approximately 15 ft
Answer:
Sweat
Explanation:
As you exercise you respire and warm up due to energy. In turn, two things happen, blood vessels vasodilate (irrelevant to you) and sweat glands sweat more. this sweat then evaporates and cools down the body.
Answer:
a) 254.6 GPa
b) 140.86 GPa
Explanation:
a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;
Ec(u) = EmVm + EpVp
To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,
Vm + Vp = 1
Vm = 1 - 0.63
Vm = 0.37
In the first equation,
Where
Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,
The modulus of elasticity upper-bound is,
Ec(u) = EmVm + EpVp
Ec(u) = (68 x 0.37) + (380 x 0.63)
Ec(u) = 254.6 GPa.
b) Considering the express of rule of mixtures for lower bound;
Ec(l) = (EmEp)/(VmEp + VpEm)
Substituting values into the equation,
Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)
Ec(l) = 25840/183.44
Ec(l) = 140.86 GPa
Answer: d)Coercion
Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)
Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa
Explanation:
Given that;
patm = 79 kPa, h = 13 in of H₂O,
A sketch of the problem is uploaded along this answer.
Now
pA = patm + 13 in of H₂O ( h × density × g )
pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)
pA = 82.23596 kPa
the absolute static pressure in the gas cylinder is 82.23596 kPa
