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Modern Engineering Thermodynamics

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138 CHAPTER 4: The First Law of <strong>Thermodynamics</strong> and Energy Transport Mechanisms<br />

Now, we incorporate our earlier results that 1 W 2 = 0andu 2 − u 1 = c v (T 2 − T 1 ) = 0, because T 2 = T 1 here (the<br />

process is also isothermal). Our latest rendition of the problem statement makes it clear that V 2 = V 1 ,andwhen<br />

these conditions are incorporated into the energy balance, we obtain our final equation for the heat transfer as<br />

<br />

1Q 2 = m 0 + 0 + gðZ <br />

2 − Z 1 Þ<br />

g<br />

+ 0 = mg<br />

c<br />

g<br />

ðZ 2 − Z 1 Þ<br />

system<br />

c<br />

and, from the equation of state, we can determine the solution to part (b) as<br />

since T 2 = T 1 and v 2 = v 1 .<br />

p 2 = RT 2 /v 2 = RT 1 /v 2 = RT 1 /v 1 = p 1<br />

Step 7 allows us to calculate the values of the unknowns:<br />

ðaÞ 1 Q 2 = ð5:00 kgÞð9:81 m/s2 Þ<br />

ð0 − 50:0mÞ = −2450 kg.m 2 /s 2<br />

1<br />

= −2450 N ⋅ m = −2450 J = −2:45 kJ<br />

and<br />

ðbÞ p 2 = p 1 = 0:300 MPa<br />

The negative sign in the answer for part (a) tells us that the heat transfer is out of the system. Note that the<br />

answer in part (b) was not the result of a complex calculation. However, it did result from a rather complex analysis<br />

and, therefore, is not trivial. Also note that we did not need the value of v 2 in the solution of the problem,<br />

so it would have been a waste of time to have calculated it early in the solution.<br />

In step 8, since the solution now seems to work well, the problem statement is complete and accurate. We<br />

should now check all the algebra, units, and calculations before creating and solving additional problems with<br />

similar or different scenarios.<br />

Exercises for the problem solved in Steps 1–8<br />

1. Rewrite this problem and make the barrel insulated but not isothermal. (Can it be both insulated and<br />

isothermal?) Resolve the problem with these new conditions. Is any additional information needed to find<br />

T 2 and p 2 ?<br />

2. Write a thermodynamics problem about a computer chip. Look up the steady state voltage and current<br />

required by a typical computer chip in a handbook and supply these values in the problem statement. This<br />

is a closed system, and the chip cannot be insulated (otherwise, it would overheat). Use the energy rate<br />

balance in the formulation of your problem scenario.<br />

3. Write a thermodynamics problem about an electrical generator. Use the closed system energy rate balance.<br />

Make the process steady state. You may have the generator insulated or uninsulated. Note that there are two<br />

work modes here, shaft work and electrical work.<br />

4. Write a thermodynamics problem about an airplane. Make it a closed system and have it change altitude and<br />

speed. Choose an appropriate unknown and provide all the necessary values for the remaining variables.<br />

SUMMARY<br />

In this chapter, we discover that the first law of thermodynamics is simply the conservation of energy principle.<br />

Since energy is conserved in all actions, the change in a system’s energy can be equated to the net transport of<br />

energy into the system. Only three possible energy transport mechanisms are available to us: (1) heat transport<br />

of energy (commonly called heat transfer), (2) work transport of energy (commonly called work), and (3) energy<br />

transported with a mass flowing across a system’s boundary. This information produced the very powerful<br />

energy balance and energy rate balance equations.<br />

The general closed system energy balance:<br />

The general closed system energy rate balance:<br />

1Q 2 − 1 W 2 = ðE 2 − E 1 Þ system<br />

= m½ðu 2 − u 1 Þ + ðV 2 2 − V2 1 Þ/ð2g cÞ + ðZ 2 − Z 1 Þg/g c Š system<br />

_Q − _W = ðdE/dtÞ system<br />

= ðm _u + mV _V /g c + mg _Z/g c Þ system

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