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prohojiy [21]
3 years ago
11

Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 second

s. 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 it’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?
Engineering
1 answer:
lara [203]3 years ago
3 0
N didn’t do it for you toroeriot everyone wwas wowowowoww
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jeka94

Answer:

why you doin this

Explanation:

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5 0
3 years ago
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Explain why the following scenario fails to meet the criteria for proper reverse engineering.
avanturin [10]

Answer:

he must document or remember the order he took it apart so he put it back together

Explanation:

5 0
2 years ago
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati
Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0
3 years ago
An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine t
lesya692 [45]

Answer:

the maximum bending stress in the strap is 3.02 ksi

Explanation:

Given the data in the question;

steel strap thickness = 0.125 in

width = 2 in

circular arc radius = 600 in

we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;

Now, using simple theory of bending

1/p = M/EI

solve for M

Mp = EI

M = EI / p ----- let this be equation 1

The maximum bending stress in the strap is;

σ = Mc / I -------let this be equation 2

substitute equation 1 into 2

σ = ( EI / p)c / I

σ = ( c/p )E

so we substitute in our values

σ = ( (0.125/2) / 600 )29 × 10³

σ = 0.00010416666 × 29 × 10³

σ = 3.02 ksi

Therefore, the maximum bending stress in the strap is 3.02 ksi

3 0
3 years ago
There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T
dmitriy555 [2]

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

3 0
3 years ago
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