Question

Ammonia, 2 kg, at 400 kPa, 40'C is in a piston/cylinder together with an unknorvn mass of sat-urated liquid ammonia at 400 kPa. The pistonis loaded, so it maintains constant pressure, andthe two masses are allowed to mix to a final uni-form state ofsaturated vapor u'ithout external heattransfer. Find the total exergy destruction in theprocess.

Answer 1

Step-by-step explanation:

Given the following information

Initial Pressure P1 = 400kPa

Initial temperature T1 = 40°C

Mass m= 2kg

T(ambient) Ta = 298k

From superheated Ammonia table, corresponding to P1 =400kPa and T1 = 40°C, we can find the first stage entropy and enthalpy

h1 = 1543.6 KJ/Kg

s1 = 5.7111 KJ/KgK

Using interpolation method to find the second stage enthalpy and entropy

From saturated ammonia table,

at P = 354.9KPa, the liquid specific entropy and enthalpy is.

sf = 0.6266 KJ/KgK

hf = 157.31 KJ/Kg

Also, From saturated ammonia table, at P' = 429.6KPa, the liquid specific entropy and enthalpy is.

sf' = 0.7114 KJ/KgK

hf' = 180.36 KJ/Kg

So, using interpolation, stage 2 specific enthalpy and entropy is

s2 = sf + (P1 - P) / (P' - P) × (sf' - sf)

s2 = 0.6266 + (400-354.9) / (429.6-354.9) × (0.7114-0.6266)

s2 = 0.6266 + (45.1 / 74.7) × 0.0848

s2 = 0.6266 + 0.0512

s2 = 0.6778 KJ/KgK

h2 = hf + (P1 - P) / (P' - P) × (hf' - hf)

h2 = 157.31 + (400-354.9) / (429.6-354.9) × (180.36-157.31)

h2 = 157.31 + (45.1 / 74.7) × 23.05

h2 = 157.31 + 13.916

h2 = 171.23 KJ/Kg

At superheated ammonia table corresponding to P3 = 400Kpa, we can find the final stage specific entropy and enthalpy

s3 = 5.3559 KJ/KgK

h3 = 1440.2 KJ/Kg

Using equation of continuity

m1 + m2 = m3

We can now calculate the stage two mass

m1•h1 + m2•h2 = (m1+m2)•h3

Make m2 subject of the formula

m2•h2 - m2•h3 = m1•h3 - m1•h1

m2(h2 - h3) = m1 ( h3 - h1)

m2 = m1•(h3 - h1) / (h2 - h3)

m2 = 2 × (1440.2 - 1543.6) / (171.23 - 1440.2)

m2 = 2 × -103.4 / -1268.97

m2 = 0.163 kg

Then, the final mass is equal to

m3 = m1 + m2

m3 = 2 + 0.163

m3 = 2.163 kg

Calculating the irreversible process

I = Ta(m3•s3 - m2•s2 - m1•s1)

I = 298( 2.163 × 5.3559 - 0.163 × 0.6778 - 2 × 5.7111

I = 298(11.5848 - 0.1105 - 11.4222)

I = 298 × 0.0521

I = 15.5258 KJ

I ≈ 15.53 KJ