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Makovka662 [10]
3 years ago
5

A groundwater contains the following cations (expressed as the cation):

Engineering
2 answers:
Hunter-Best [27]3 years ago
6 0

Answer:

208 mg/L

Explanation:

Only Mg++ and Ca++ causes hardness not Na+

Given

Mg++ = 20 mg/L

Ca++ = 50 mg/L

Hardness\ of\ water = C^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ C^{2+}}\ +\ mg^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ mg^{2+}}

= 50\times\frac{50}{20}\ +\ 20\times\frac{50}{12}

= 208.33 mg/L \approx 208mg/L

Oliga [24]3 years ago
5 0

Answer:

50mg/L

Explanation:

CaCO3 -> Ca++ + CO2

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Answer:

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Explanation:

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4 0
3 years ago
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
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Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

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2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

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• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the
Rudik [331]

Answer:

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Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

\sigma a =  \sqrt{(12 \times 1)^2+3\times (0\times 1)^2}  

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and

\sigma m = \sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}     ...........2

put here value and we get

\sigma m = \sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}  

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now we apply here goodman line equation here that is

\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

\frac{17.34}{63.800} +  \frac{12}{31.9} = \frac{1}{FOS}  

solve it we get

FOS = 1.5432

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average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.

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