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3 years ago

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A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr

ing sodium chloride concentration of 20 mg/L. If the storm sewer flow rate is 2000 L/min and the stream flow rate is 2 m3/s, what is the concentration of salt in the stream after the discharge point? Assume that the flow is completely mixed, the salt is nonreactive, and the system is at steady state

1 answer:

galina1969 [7]3 years ago

6 0

Answer:

Given Data:

concentration of sewer Csewer = 1.2 g/L

converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

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A series R-L circuit is given. Circuit is connected to an AC voltage generator. a) Derive equations for magnitude and phase of c

igomit [66]

Answer:

The equations for magnitude and phase of current and voltages on resistor and inductor are:

$I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

$V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

$V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}$

Explanation:

The first step is to find the impedances of the resistance (R) and the inductor (L).

The impedance of the resistor is:

Rectangular form: $Z_R=R$
Polar form: $Z_R=R\angle 0^{\circ}$

The impedance of the inductor is:

Rectangular form: $Z_L=j\omega L$

Polar form: $Z_L=\omega L \angle 90^{\circ}$

Where $\omega$ is the angular frequency of the source, and the angle is $90^{\circ}$ because a pure imaginary number is on the imaginary axis (y-axis).

The next step is to find the current expression. It is the same for the resistor and inductor because they are in series. The total impedance equals the sum of each one.

$I=\frac{V}{Z_R+Z_L}$

It is said that $V=V_m\angle \theta$ , so, the current would be:

$I=\frac{V_m\angle \theta }{R+j\omega L}$

The numerator must be converted to polar form by calculating the magnitude and the angle:

The magnitude is $\sqrt{R^2+(\omega L)^2}$
The angle is $tan^{-1}(\omega L / R)$

The current expression would be as follows:

$I=\frac{V_m\angle \theta }{\sqrt{R^2+(\omega L)^2}\, \angle tan^{-1}(\omega L / R)}$

When dividing, the angles are subtracted from each other.

The final current expression is:

$I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

The last step is calculating the voltage on the resistor $V_R$ and the voltage on the inductor $V_L$ . In this step the polar form of the impedances could be used. Remember that $V=I\cdot Z$ .

(Also remember that when multiplying, the angles are added from each other)

Voltage on the resistor $V_R$

$V_R=I\cdot Z_R=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (R\angle 0^{\circ})$

The final resistor voltage expression is:

$V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

Voltage on the inductor $V_L$

$V_L=I\cdot Z_L=\bigg( \frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)\bigg) \cdot (\omega L \angle 90^{\circ})$

The final inductor voltage expression is:

$V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}$

Summary: the final equations for magnitude and phase of current and voltages on resistor and inductor are:

$I=\frac{V_m}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

$V_R=I\cdot Z_R=\frac{V_m \cdot R}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)$

$V_L=I\cdot Z_L=\frac{V_m \cdot (\omega L)}{\sqrt{R^2+(\omega L)^2}}\angle \theta - tan^{-1}(\omega L / R)+90^{\circ}$

6 0

4 years ago

What is the energy change when the temperature of 15.0 grams of solid silver is decreased from 37.3 °C to 20.5 °C ?

Gala2k [10]

Answer:

Q=58.716 W

Explanation:

Given that

mass ,m = 15 g

The decrease in the temperature ,ΔT = 37.3 - 20.5 °C

ΔT = 16.8°C

We know that specific heat of silver ,Cp = 0.233 J/g°C

The energy change of the silver is given as

Q = m Cp ΔT

Now by putting the values we get

$Q= 15 \times 0.233\times 16.8\ W$

Q=58.716 W

Therefore the change in the energy will be 58.716 W.

8 0

3 years ago

A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional s

nlexa [21]

Answer:

(a) Modified Goodman criterion:

Factor of safety against fatigue failure = 1.0529

(b) Gerber criterion:

Factor of safety against fatigue failure = 1.31

(c) ASME-elliptic criterion:

Factor of safety against fatigue failure = 1.315

Explanation:

See the attached file for the calculation.

5 0

3 years ago

A fluid flows steadily through a pipe with a uniform cross sectional area. The density of the fluid decreases to half its initia

Vikentia [17]

Answer:

c. V2 equals V1

Explanation:

We can answer this question by using the continuity equation, which states that:

$A_1 v_1 = A_2 v_2$ (1)

where

A1 is the cross-sectional area in the first section of the pipe

A2 is the cross-sectional area in the second section of the pipe

v1 is the velocity of the fluid in the first section of the pipe

v2 is the velocity of the fluid in the second section of the pipe

In this problem, we are told that the pipe has a uniform cross sectional area, so:

A1 = A2

As a consequence, according to eq.(1), this means that

v1 = v2

so, the velocity of the fluid in the pipe does not change.

5 0

3 years ago

The dry unit weight of a soil sample is 14.8 kN/m3.

Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(c) Degree of saturation

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0

4 years ago