The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?

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Answer 1

Answer:

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Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).After finishing computation, the app sends some data to the cloud, regardless of the mode itâ€™s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.

Required:
What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?

Answers

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Why reducing leads to increasing wages?

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

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A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. The transmitted tangential load is 22 kN. Conditions are such that Km = 1.7. The teeth are standard 20-degree, full-depth. The module is 5 and the face width 62 mm. Determine the bending stress when the mesh is at the highest point of single tooth contact.

Answers

Answer:

The bending stress of the face tooth is $\sigma _(bg) = 502.82 MPa$

Explanation:

From the question we are told that

The number of tooth of the pinion is $N_t = 26 \ tooth$

The velocity of rotation is given as $\omega_p = 1800 rpm$

The number of tooth is of the gear is $N_g = 55 \ tooth$

The quality level is $Q_r = 10$

The transmitted tangential load is $F_T = 22\ kN$ = $22 KN * (1000N)/(1KN) = 22*10^3 N$

$k_m = 1.7$

The angle of the teeth is $\theta_t = 20^o$

The module is $M= 5$

The face width is $W_f = 62mm$

The diameter of the pinion is mathematically represented as

$d_p = M * N_t$

Substituting the values

$d_p = 5 *26$

$= 130 mm = (130)/(1000) = 0.130m$

The pitch line velocity is mathematically represented as

$V_t = (d_p )/(2) (2 \pi \omega_p)/(60)$

Substituting values

$= (0.130)/(2) * (2 * 3.142 * 1800 )/(60)$

$= 12.25\ m/s$

Generally the dynamic factor is mathematically represented as

$K_v = [(A)/(A +√(200V_t) ) ]^B$

Now B is a constant that is mathematically represented as

$B = ((12 -Q_r )^(2/3))/(4)$

substituting values

$= ((12- 10 )^(2/3))/(4)$

$=0.3968$

A is also a constant that is mathematically represented as

$A = 50 + 56(1 -B)$

Substituting values

$= 50 +56 (1- 0.3968)$

$= 83.779$

Substituting these value into the equation for dynamic factor we have

$K_v = [(83.779)/(83.779 + √(200 * 12.25) ) ]^(0.3968)$

$= 0.831$

The geometric bending factor for a 20° profile from table

"AGMA Bending Geometry Factor J for 20°, Full -Depth Teeth with HPSTC Loading , Table 2-9"

That corresponds to 55 tooth gear meshing with 26 pinion is

$J_g = 0.41$

the diameter pitch can be mathematically represented as

$p_d = (1)/(M)$

Substituting values

$p_d = (1)/(5)$

$=0.2mm^(-1)$

The mathematically representation for gear tooth bending stress in the teeth face is as follows

$\sigma_(bg) = (F_T \cdot p_d )/(W_f * J_g)(K_a K_(dt) )/(K_v) K_s K_B K_t ----(1)$

Where $W_t$ is the tangential load

$W_f$ is the face width

$K_a$ is the application factor this is obtained from table "Application Factors, Table 12-17 " and the value is $K_a$ = 1

$K_(dt)$ is the load distributed factor

$K_s$ is the size factor

$K_B$ is the rim thickness factor which is obtained for M which has a value 1

$K_t$ is the idler

Substituting values into equation 1

$\sigma_(bg) = (22*10^3 *0.2)/(62 * 0.41) * (1 * 1.7 )/(0.831) * 1 *1 *1.42$

$= 502.82 N/mm^2$

$= 502.82 * 1000 * (N)/(m^2)$

$= 502.82 MPa$

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks

Answers

Answer:

branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
branch 2: i = 21.466∠63.435°; pf = 0.447 leading
total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

$X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega$

The capacitive reactance is ...

$X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega$

Branch 1

The impedance of branch 1 is ...

Z1 = 8 +j4.99984 Ω

so the current is ...

I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answers

Answer:

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Explanation:

From Open Channel Theory we know that critical depth of the rectangular channel ( $y_(c)$ ), measured in meters, is calculated by using this equation:

$y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }$ (Eq. 1)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$b$ - Channel width, measured in meters.

If we know that $\dot V = 15\,(m^(3))/(s)$ , $g = 9.807\,(m)/(s^(2))$ and $b = 2\,m$ , then the critical depth is:

$y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }$

$y_(c) \approx 1.790\,m$

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity ( $v$ ), measured in meters, is obtained from this formula:

$v = (\dot V)/(b\cdot y_(c))$ (Eq. 2)

If we know that $\dot V = 15\,(m^(3))/(s)$ , $b = 2\,m$ and $y_(c) = 1.790\,m$ , then the flow velocity in the rectangular channel is:

$v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))$

$v = 4.190\,(m)/(s)$

The flow velocity in the rectangular channel is 4.190 meters per second.

Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur

Answers

Answer:

While air is compressed in a piston cylinder there are following types of irreversibilities

1.Due to finite temperature difference heat transfer take place between cylinder and surrounding.

2.Due friction force between cylinder and piston .

3.Compression process is so fast due to this ,it leads in the irreversibility of system.

Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

Answers

x + 2 * x is the correct option. The above-selected option demonstrates implicit conversion, which is an automated type of conversion. Thus, option B is correct.

The series of conversions are necessary to change the type of a function call's argument to that of the parameter with the same name in the function declaration is known as an implicit conversion sequence. For each parameter, the compiler tries to identify an implicit conversion sequence.

If both user-defined conversion sequences A and B contain the same user-defined conversion function or constructor, and if the second standard conversion sequence of A is superior to the second standard conversion sequence of B, then user-defined conversion sequence A is preferable to user-defined conversion sequence B.

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