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solong [7]
3 years ago
13

After being purged with nitrogen, a low-pressure tank used to store flammable liquids is at a total pressure of 0.03 psig. (a) I

f the purging process is done in the morning when the tank and its contents are at 55°F, what will be the pressure in the tank when it is at 85℉ in the afternoon? (b) If the maximum design gauge pressure of the tank is 8 inches of water, has the design pressure been exceeded? (c) Speculate on the purpose of purging the tank with nitrogen

Engineering
1 answer:
g100num [7]3 years ago
8 0

Answer:

See attached picture.

Explanation:

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A 600 MW coal-fired power plant has an overall thermal efficiency of 38%. It is burning coal that has a heating value of 12,000
velikii [3]

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day

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3 years ago
Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th
True [87]
Um I think the answer for this is 10l
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3 years ago
How is an orthographic drawing similar to or different from an isometric drawing?
evablogger [386]
An isometrical drawing is a nearly 3d drawing showing the object's width and depth in a complete image, from each curved plane of the orthhographic view, the viewpoint is at a 45 degree angle. From an observations point of view, isometric differs, since all longitudes are true.
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3 years ago
Read 2 more answers
Your boss asks you to classify some of the components in a temperature measurement system. The system consists of a thermocouple
lianna [129]

Answer:

Explanation:

1) Resolution and uncertainty of both ADC ranges

a) Resolution (for +/- 5 mV range) = 5 mV / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 mV / 256 = 0.0195 mV    ..................1

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*0.0195 mV = 0.00975 mV       ..........2

b) Resolution (for +/- 5 V range) = 5 V / 2n where n = number of bits of ADC

Resolution = 5 mV / 28 = 5 V / 256 = 0.0195 V = 19.5 mV    ..................3

uncertainty (for +/- 5 mV range) = +/- 0.5*resolution = +/- 0.5*19.5 mV = 9.75 mV       ..........4

2)Thermocouple sensitivity = ( Maximum output voltage - Minimum Output voltage) / (Maximum Temperature - Minimum Temperature)

Thermocouple sensitivity = (3.649mv - 0 ) / (70 - 0) = 0.0521 mV / Deg.C           ............5

This is the required Thermocouple sensitivity

3) Water bath temperature is given as 57 deg.C

Hence voltage read by Thermocouple = Sensitivity*57 = 0.0521*57 mV = 2.9697 mV    ........6

4)We need to use ADC with a range of +/- 5 mV range as ADC with +/- 5 V range can not do measurement as it's resolution is higher than output voltage.

ADC will measure voltage as 2.9695 mV                    ......................7

8 0
3 years ago
If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo
Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)} and considering that E(s)=D(s)-G_c(s)C(s) then

\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)

\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)

\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}

E(s)=\frac {D(s)}{1+G_c(s)G(s)}

For ramp disturbance d(t)=at

D(s)=\frac {a}{s^{2}} therefore, the steady state error is given by

e(\infty)= \lim_{s \to 0} s E(s)

e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]

e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]

Whenever G_c(s) has a double intergrator, the error e(\infty) becomes zero

3 0
3 years ago
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