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ANTONII [103]
3 years ago
6

A water supply agency is planning to add two reservoirs to its system. Water will flow from Reservoir A to Reservoir B via a 10,

000-ft-long, 24-inch diameter steel pipe. The pipe will be installed beneath a road between the two reservoirs. The road crosses over a hill with a summit elevation of 375 ft above MSL. This summit is at a distance of 3000 ft from reservoir A.The elevation of the water surface in reservoir A will vary between 348.0 and 362.0 ft above MSL and the water surface in reservoir B will vary between 244.1 and 255.4 ft above MSL.Assume a temperature of 50°F. Neglect minor losses. Use a Hazen-Williams friction factor of 100 and an equivalent sand grain roughness of 0.060 in. These values are from Table 12.1.1 in the text.(a) Make a sketch that defines the geometry of the problem. Clearly label dimensions.(b) Determine which water level combination will lead to maximum discharge in the pipe and which water level combination will lead to minimum discharge in the pipe. This should be done by analyzing the relevant equations without doing any calculations. Show your work and explain your logic.

Engineering
1 answer:
NikAS [45]3 years ago
3 0

Attached is the solution to the above question.

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Georgia [21]
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5 0
3 years ago
Which of the following is most useful for doing research?
Ghella [55]

Answer:

Web Browser

Explanation:

Because you dont use a messaging app or presentation software to look up stuff its common knowledge

7 0
2 years ago
A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0
3 years ago
A hypereutectoid steel often presents hard and brittle cementite along the grain boundaries of pearlite. Which of the following
Anvisha [2.4K]

Answer:

(d) Spheroidizing

Explanation:

Spheroidizing

 This is the heat treatment process for steel which having carbon percentage more than 0.8 %.As we know that a hard and brittle material is having carbon percentage more than 0.8 %.That is why this process is suitable for the hard materials.

In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.

In this process grain become spheroidal and these grains are ductile.

6 0
3 years ago
A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,
juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

7 0
3 years ago
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