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nika2105 [10]
3 years ago
8

Two kilograms of air in a piston-cylinder assembly undergoes an isothermal process from an initial state of 200K, 300kPa to 600k

Pa. a) How much entropy is generated in the process, in kJ/K. b) Calculate the amount of lost work.
Engineering
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

a) 0.39795 kJ/K

b) 79.589.37 kJ

Explanation:

m = Mass of air = 2 kg

Temperature = 200 K

P₁ = Initial pressure = 300 kPa

P₂ = Final pressure = 600 kPa

R = mass-specific gas constant for air = 287.058 J/kgK

a) For isentropic process

\Delta S=mRln\frac{P_1}{P_2}\\\Rightarrow \Delta S=2\times 287.058ln\frac{300}{600}\\\Rightarrow \Delta S=-397.95\ J/K

∴ Entropy is generated in the process is 0.39795 kJ/K

b)

W=mRTln\frac{P_1}{P_2}\\\Rightarrow W=2\times 287.058\times 200ln\frac{300}{600}\\\Rightarrow W=-79589.37\ J

∴ Amount of lost work is 79.589.37 kJ

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Answer:

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Explanation:

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Deductions;                                        |                       Deduct

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8 0
3 years ago
An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �
Pani-rosa [81]

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

5 0
3 years ago
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