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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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3 years ago
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
I need answers for this sheet please.
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7 0
4 years ago
A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal
nirvana33 [79]

Answer:

a)  the temperature to which the pin must be cooled for assembly is T_2 = -101.89^ \ ^0}C

b) the radial pressure at room temperature after assembly is P_f = 62.8 \ MPa

c) the  safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion \alpha = 12.3*10^{-6} \  ^0 C

Yield strength \sigma_y = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature T_1 = 20°C

outer diameter of the collar D_o = 95 \ mm

inner diameter of the collarD_i = 60 \ mm

pin diameter D_p = 60.03 \ mm

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

(D_i - c )-D_p = \alpha * D_p(T_2-T_1)

(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}T_2  \ \ - \ \ 0.01476738

-0.09 +  0.01476738 = 7.38369*10^{-4}T_2

−0.07523262 =7.38369*10^{-4}T_2

T_2 = \frac{-0.07523262}{7.38369*10^{-4}}

T_2 = -101.89^ \ ^0}C

b)

To determine the radial pressure at room temperature after assembly ;we have:

P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\  P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2}  \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa

c)  the safety factor of the resulting assembly is calculated as:

safety factor =  \frac{Yield \ strength }{walking \ stress}

safety factor =  \frac{400}{62.8}

safety factor = 6.4

Thus, the  safety factor in the resulting assembly = 6.4

4 0
3 years ago
In 2002 Acme Chemical purchased a large pump for
Anestetic [448]

Answer:

$151094

Explanation:

Solution

Recall that:

Acme Chemical in 2002 purchased a large pump worth of = 112,000

The estimation for the pump to the industrial pump index is =100

The index in 2002= 212

The current index is = 286

k = is the  reference year for which cost or price is known.

n =  the year for which cost or price is to be estimated (n>k).

Cn = the estimated cost or price of item in year n.

Ck =  the cost or price of item in reference year k.

Cn = Ck * (In / Ik )

Now,

We find the estimated cost of the new pump which is stated as follows:

Cn = (112,000 * 286) /212

=32032000/212

=$151094

Therefore, the estimated cost of the new pump is $151094

7 0
3 years ago
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