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Modern Engineering Thermodynamics

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194 CHAPTER 6: First Law Open System Applications<br />

enthalpy (h out ≠ constant); therefore, the integral in Eq. (6.37) cannot be evaluated. However, we can devise a<br />

way of dealing with this by defining an average specific enthalpy for the discharge flow stream as<br />

h avg = ð1/m out Þ<br />

where, from the conservation of mass law, m out =(m 1 − m 2 ) tank . Approximating h avg as<br />

Eq. (6.37) becomes<br />

Z 2<br />

h avg = ½ðh 1 + h 2 Þ/2Š tank<br />

1<br />

h out dm out (6.38)<br />

1Q 2 − h 1 ðm 1 − m 2 Þ/2 − h 2 ðm 1 − m 2 Þ/2 = m 2 u 2 − m 1 u 1<br />

where the numerical subscripts refer exclusively to states inside the tank from this point on.<br />

The effect of the emptying process can be more easily seen by simplifying the analysis somewhat. Let us stipulate<br />

that the tank is insulated (so 1 Q 2 =0)andthatm 1 ≫ m 2 ≈ 0. Then, the previous equation can be rearranged<br />

to give<br />

h 2 = 2u 1 − h 1 (6.39)<br />

Inthecaseofemptyingapressurizedincompressible fluid, we can write u = cT and h = cT + vp, thenEq.(6.39)<br />

gives<br />

T 2<br />

<br />

emptying<br />

incomp:<br />

fluid<br />

= T 1 − ðv/cÞðp 1 + p 2 Þ (6.40)<br />

Equation (6.40) tells us that the expansion process that accompanies the emptying process of an incompressible<br />

fluid always causes the final temperature inside the tank to be less than the initial temperature inside the tank.<br />

In the case of emptying a pressurized ideal gas, we can write u = c v T and h = c p T, then Eq. (6.39) gives<br />

T 2<br />

<br />

emptying<br />

ideal<br />

gas<br />

= T 1 ½ð2/kÞ− 1Š (6.41)<br />

Since a continuous expansion process lowers the temperature of the remaining contents, this explains why pressurized<br />

cans of paint, deodorant, and the like become very cold when they are continuously discharged.<br />

EXAMPLE 6.9<br />

Immediately after the scuba tank in Example 6.8 is filled, the air is released into the atmosphere through an open valve<br />

on the top of the tank. Assuming that the emptying process is adiabatic and ignoring the mass of any air remaining in<br />

the tank when its pressure reaches atmospheric, determine the final temperature inside the tank immediately after the<br />

tank is empty.<br />

Solution<br />

First, draw a sketch of the system (Figure 6.18).<br />

Escaping air<br />

Air T Final = ?<br />

FIGURE 6.18<br />

Example 6.9.<br />

The unknown is the final temperature inside the tank immediately after the tank is emptied. The material is air.

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