. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298 K and 20 bars to 200 bars. Compute the heat removal and work required.

Answers

Answer 1

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

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The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?
How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h
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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38 wt% at a point 2.3 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 5.3 mm position for an identical steel and at the same carburizing temperature.

Answers

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.

Answers

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Calculate the amount of power (in Watts) required to move an object weighing 762 N from point A to point B within 29 seconds. Distance between point A and point B is 5.0 meters and they are at the same level. (Ignore the frictional losses)

Answers

Answer:

0.556 Watts

Explanation:

w = Weight of object = 762 N

s = Distance = 5 m

t = Time taken = 29 seconds

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$s=ut+(1)/(2)at^2\n\Rightarrow a=(2* (s-ut))/(t^2)\n\Rightarrow a=(2* (5-0))/(29^2)=(10)/(481)$

Mass of the body

$m=(w)/(g)=(762)/(9.81)$

Force required to move the body

$F=ma\n\Righarrow F=(762)/(9.81)* (10)/(481)$

Velocity of object

$v=u+at\n\Rightarrow v=0+(10)/(481)* 29\n\Rightarrow v=(10)/(29)$

Power

$P=Fv\n\Rightarrow P=(762)/(9.81)* (10)/(481)* (10)/(29)=0.556\ W$

∴ Amount of power required to move the object is 0.556 Watts

A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

Efficiency = 38%
Power output = 800 MW

Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

Total heat = 2105.26 MW
Heat used by plant = 800 MW
Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

brainly.com/question/2009210

Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

$\eta = (Wn)/(Qin)$ ......................1

put here value we get

38% = $(800)/(Qin)$

Qin = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

Answers

Answer:

$\dot W_(out) = 3374.289\,(BTU)/(s)$

Explanation:

The model for the turbine is given by the First Law of Thermodynamics:

$- \dot W_(out) + \dot m \cdot (h_(in) - h_(out)) = 0$

The turbine power output is:

$\dot W_(out) = \dot m\cdot (h_(in)-h_(out))$

The volumetric flow is:

$\dot V = (\pi)/(4) \cdot \left( (2)/(12)\,ft \right)^(2)\cdot (620\,(ft)/(s) )$

$\dot V \approx 13.526\,(ft^(3))/(s)$

The specific volume of steam at inlet is:

State 1 (Superheated Steam)

$\nu = 1.33490\,(ft^(3))/(lbm)$

The mass flow is:

$\dot m = (\dot V)/(\nu)$

$\dot m = (13.526\,(ft^(3))/(s) )/(1.33490\,(ft^(3))/(lbm) )$

$\dot m = 10.133\,(lbm)/(s)$

Specific enthalpies at inlet and outlet are, respectively:

State 1 (Superheated Steam)

$h = 1479.74\,(BTU)/(lbm)$

State 2 (Saturated Vapor)

$h = 1146.1\,(BTU)/(lbm)$

The turbine power output is:

$\dot W_(out) = (10.133\,(lbm)/(s) )\cdot (1479.1\,(BTU)/(lbm)-1146.1\,(BTU)/(lbm))$

$\dot W_(out) = 3374.289\,(BTU)/(s)$

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Answers

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

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