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Igoryamba
4 years ago
7

A cylinder with a piston restrained by a linear spring contains 2 kg of carbon dioxide at 500 kPa and 400°C. It is cooled to 40°

C and 300 kPa. Calculate the heat transfer for the process.
Engineering
1 answer:
leonid [27]4 years ago
7 0

Answer:

heat transfer for the process is - 643.3 kJ

Explanation:

given data

mass m = 2 kg

pressure p1 = 500 kPa

temperature t1 = 400°C = 673.15 K

temperature t2 = 40°C = 313.15 K

pressure p2 = 300 kPa

to find out

heat transfer for the process

solution

we know here mass is constant so

m1 = m2

so by energy equation

m ( u2 - u1 ) = Q - W

Q is heat transfer

and in process  P =  A+ N that is linear spring

so

W = ∫PdV

= 0.5 ( P1+P2) ( V1 - V2)

so for case 1

P1V1 = mRT

put here value

500 V1 = 2 (0.18892) (673.15)

V1 = 0.5087 m³

and

for case 2

P2V2 = nRT

300 V2 = 2 (0.18892) (313.15)

V2 = 0.3944 m³

and

here W will be

W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )

W = -45.72 kJ

and

Q is here for Cv = 0.83 from ideal gas table

Q = mCv ( T2-T1 ) + W

Q = 2 × 0.83 ( 40 - 400 ) - 45.72

Q = - 643.3 kJ

heat transfer for the process is - 643.3 kJ

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A 1 phase load operates at 600 V and consumes 75 kW with a 0.85 lagging power factor. Compute the complex power consumed by the
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Answer:

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4 years ago
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A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.
mart [117]

Given :

Capacitor , C = 55 μF .

Energy is given by :

\omega(t)=10cos^2 (377t)\ J .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)

Now , current i is given by :

i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .

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3 years ago
Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo
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Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

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