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3 years ago

12

Plateau Creek carries 5.0 m^3 /s of water with a selenium (Se) concentration of 0.0015 mg/L. A farmer withdraws water at a certa

in flowrate (m3 /s) from the creek for irrigation. After using the water, half of the water withdrawn returns to the creek and contains 1.00 mg/L of Se. Fish in the creek are sensitive to Se levels over 0.04 mg/L.
Required:

How much water (m3 /s) can the farmer withdraw from the stream to maintain Se at 0.04 mg/L after the contaminated water is returned to the creek?

1 answer:

Bond [772]3 years ago

7 0

Answer:

The correct answer is "4.8137 m³". The further explanation is given below.

Explanation:

Firstly we have to calculate the concentration of Se:

$C = 0.0015 \ mg/L\times \frac{1g}{1000 mg}\times \frac{1 \ mol}{79 \ g}$

$=1.9\times 10^{-8} \ mol/L$

Concentration the fish can take:

$=0.04 \ mg/L\times \frac{1 \ g}{1000mg}\times \frac{1 \ mol}{79 \ g}$

According to the general dilution principle will be:

⇒ $M_1V_1 = M_2V_2$

The volume that can take the farmer will be:

$V_2 = 1.9\times 10^{-8} M\times \frac{5\times 10^3 \ L}{5.1\times 10-7 M}$

$=186.27 \ L$

On converting this into m³, we get

= $0.18627 \ m^3$

Finally the volume the farmer can remove would be:

$V = 5-0.18627$

$= 4.8137 \ m^3$

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3 years ago

A beam has a rectangular cross section that is 5 inches wide and 1.5 inches tall. The supports are 60 inches apart and with a 12

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Answer:

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Load = 125 × 32 = 4000 $\frac{ft lbm}{s^{2} }$

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2 years ago

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tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

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u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

0.014m³ / 0.2066 m³/kg

m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

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For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ = 0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524 . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0

3 years ago

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