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Luda [366]
3 years ago
5

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

on is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Engineering
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

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stop and might even crash

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A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b
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B - Poor

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If you are trying to land the plane, which part(s) would you move? How would you move it/them?
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Answer: hope it helps

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7 0
3 years ago
2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7
adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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