A machine which continuously produces work from nothing is called a perpetual motion machine of the first kind. As per Fig. 4.4, W (work) is produced without taking any energy from surroundings. It creates its own energy. This concept violates the first law of thermodynamic. All efforts to create such PMM-1 have failed, thus providing the experimental proof of the validity of the first law of thermodynamics.

**SOLVED EXAMPLES**

**Example 4.1.** An engine is tested by means of a water brake at 1100 rpm. The torque of the engine is 11000 Nm and the water consumption of the brake is 1 m3/S, its inlet tamperature being 25°C. Calculate the water temperature at exit, assuming that the whole of the engine power is ultimately transformed into heat which is absorbed by the cooling water (specific heat of water= cP = 4.187 KJ/KgK).

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**Solution.** Given T = 11000 Nm

N = 1100 rpm

t1 = 25°C, t2 =?

Power , p = tw = t x 2πN / 60

P = 11000 x 2π x 1100 / 60 j /s

Work = heat = p x t

m cp ( t_{2} – t_{1}) = p x t

m / t cp ( t_{2} – t_{1}) = p,

or 1 x 10^{3 }x 4.187 x 10^{3} x ( t_{2} -2_{1}) = 11000 x 2π x 1100 / 60

t_{2} – t_{1} = 11 x 2π x 11 / 60 x 41.87 = 0.3

t_{2} = ( t_{1} + 0.3) = 25.3 c.

**Example 4.2.** In a cyclic process, heat transfers are + 15 KJ, – 26 KJ, – 4.56 KJ and + 32.56

KJ. Calculate the net work transfer.

**Solution.**

In a cyclic operation,

ΣQ _{cycle} = ΣW _{cycle}

or ( + 15 – 26 – 4.56 + 32.56) KJ = ΣW _{cycle}

17 KJ = Σ W _{cycle }= net work transfer.

**Example 4.3.** A slow chemical reaction takes place in a fluid at the constant pressure of 1cf N/m2. The fluid is surrounded by adiabatic wall during the reaction which begins at state A and ends at state B. The adiabatic wall is then removed and 106 KJ of heat flows to the surroundings as the fluid goes to state C. The following data are given.

State V(m3) t( c)

A 0.004 20

B 0.4 370

C 0.08 20

Calculate E8 and Ec if EA = 0 for the fluid system.

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**Solution.** For states A to B

We have dQ = dE + dW

or integrating 1-2

Q _{A-B} = EB – EA + PDV

= E_{2} – E_{1} + P ( V_{2} – V_{1})

Q_{A-B} = 0 and EA =0 gien

O = E_{B} – 0 + 105 ( 0.5 – 0.004)

E_{B} = – 105 ( 0.4 – 0.004) Joule

= – 39.6 x 103 joule = – 39.6 kj.

For state b to c

Q_{B-C }= E_{C} – E_{B }+ P ( V_{3} – V_{2})

-106 x 10^{3} = E_{C} + 39.6 X 10^{3} + 10^{5} ( 0.08 – 0.4)

[Q8c is heat out from system so it is negative]

or Ec = -113.6 x 10^{3} Joule

or Ec = -113.6 KJ. Ans.

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**Example 4.4.** During a certain period a 11Ulchine consumes 1 KWh of energy and the internal energy of the system drops by 5000 KJ. Fin,d out the net heat transfer for the system.

**Solution.** Work W1- 2 is done on the system.

W_{1-2} = – 1 KWh = – 3600 kj

Internet energy drops i.e , E_{2 }–E_{1} = – 5000 KJ

We known

Q_{1-2} = E_{2} –E_{1} + w_{1-2 }

= -5000 kj – 3600 kj

= – 8600 kj

= – 8.6 mj.