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defon
2 years ago
12

The density of oxygen contained in a tank is 2.0 kg/m3 when the temperature is 25 °C. Determine the gage pressure of the gas if

the atmospheric pressure is 97 kPa.
Engineering
1 answer:
Verdich [7]2 years ago
8 0

Answer:

57.8408 kPa

Explanation:

Data provided in the question:

Density of oxygen contained, ρ = 2.0 kg/m³

Temperature, T = 25 °C = 25 +273 = 298 K

Atmospheric pressure = 97 kPa.

Now,

Pressure due to gas = ρRT

Here,

R is the gas constant = 259.8 J/kg.K

Thus,

Absolute Pressure due to oxygen = 2 × 259.8 × 298

= 154840.8 Pa

= 154840.8 × 10⁻³ kPa

= 154.84 kPa

Thus,

Gage pressure = Absolute Pressure - Atmospheric pressure

=  154.84 kPa - 97 kPa

= 57.8408 kPa

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a) 44.4%

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Explanation:

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3 years ago
A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
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Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

b) For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

Part 2

For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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3 years ago
What is the reason for the development of new construction of materials for human use? (Select all that apply.)
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Answer:

All 4 could be justified.

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7 0
2 years ago
You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from
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Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

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Explanation:

Given data:

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steady state speed = 200 rad/s

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<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

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