Mechanical - Engineering Thermodynamics - The Second Law of Thermodynamics

**1. Two kg of air at
500kPa, 80°C expands adiabatically in a closed system until its volume is
doubled and its temperature becomes equal to that of the surroundings which is
at 100kPa and 5°C.**

**For this process, determine**

**1) ****the
maximum work **

**2) ****the
change in a availability and **

**3) ****the
irreversibility. **

**Take, C _{v} = 0.718 KJ/kg K, R =
0.287 KJ/kg K.**

**2. ****A
reversible heat engine receives 3000 KJ of heat from a constant temperature
source at 650 K . If the surroundings is at 295 K, **

**determine **

**i) ****the
availability of heat energy **

**ii) ****Unavailable
heat. **

A
= Q_{1} –T_{0} (ΔS)

= 3000
–295 (3.17)

= 2064.85
KJ.

Unavailable heat (U.A) = T_{0} (ΔS)

= 295
(3.17)

= 935.15
KJ.

**Result**:

1) The
availability of heat energy (A) = 2064.85 KJ.

2) Unavailable
heat (U.A) = 935.15 KJ.

**3. Air in a closed
vessel of fixed volume 0.15 m ^{3} exerts pressure of 12 bar at 250 °C.
If the vessel is cooled so that the pressure falls to 3.5 bar , determine the
final pressure, heat transfer and change of entropy. **

**Given Data:**

V_{1} = 0.15 m^{3}

p_{1} = 12 bar
= 1200 KN/m^{2} p_{2} = 3.5 bar = 350 KN/m^{2}

T_{1} = 250°C = 273+250 = 523 K

**To find:**

1) The
final pressure,

2) Heat
transfer

3) Change
of entropy.

**Solution:**

**Result:**

1) The
final pressure, T_{2} = 152.54 K.

2) Heat
transfer, Q = - 446.78 KJ.

3) Change
of entropy,-1.06KJ/K. ΔS =

**4. A domestic food
freezer maintains a temperature of - 15°C. The ambient air is at 30°C. If the
heat leaks into the freezer at a continuous rate of 1.75 KJ/s, what is the
least power necessary to pump the heat out continuously?**

**Given Data:**

T_{L} = - 15°C = 273 –15 = 258 K

T_{H} = 30°C
= 273 + 30 = 303 K

Q_{S} = 1.75 KW

**To find:**

Least power, (W)

**Solution:**

**Result:**

Least power necessary to pump heat, W = 0.305 KW.

**5. A refrigerator
working on reversed Carnot cycle requires 0.5 KW per KW of cooling to maintain
a temperature of -15°C. Determine the following:**

**a) ****COP
of the refrigerator **

**b) ****Temperature
at which heat is rejected and **

**Amount of heat rejected to the
surroundings per KW of cooling.**

**Given Data:**

W = 0.5 KW

Q_{2} = 1 KW

T_{2} = -15°C = 273 –15 = 258 K.

**To find:**

1) COP
of the refrigerator (COP)

2) Temperature
at which heat is rejected (T_{1})

3) Amount
of heat rejected to the surroundings per KW of cooling (Q_{1})

**Solution:**

**Result:**

1) COP
of the refrigerator (COP) = 2

2) Temperature
at which heat is rejected (T_{1}) = 387 K

Amount of heat rejected to the surroundings per KW
of cooling (Q_{1}) = 1.5 KW.

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Mechanical : Engineering Thermodynamics : The Second Law of Thermodynamics : Solved Problems: Thermodynamics Second Law |

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