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Liquid HFC-134a at its boiling point at 12 bars pressure is throttled to 1 bar pressure. What is the final temperature? What fraction of the liquid vaporizes?

Table 4.3. Properties of the refrigerant HFC-134a under saturated conditions (at its boiling point for each pressure). All values are for 1kgof fluid, and are measured relative to an arbitrarily chosen reference state, the saturated liquid at -40°c. Excerpted from Moran and Shapiro (1995).

Short Answer

Expert verified
  • The liquid vaporization is0.465.
  • The final temperature is -26.4°C

Step by step solution

01

To find

The final temperature and the fraction of liquid that vaporizes

02

Explanation

Given:

Initial pressure, P=12 bars

Final pressure, P=1 bar

Formula: Hf=aHliquid+(1-a)Hgas

Where,a=is the fraction of the HFC-134a, which ends up as liquid.

Calculation:

Using table 4.3, the initial temperature and enthalpy of the HFC -134aat a pressure of 12.0 bar are as follows:

Ti=46.3°CHi=116kJ

At a pressure of 12.0 bar, the final temperature of HFC-134a is as follows:

Tf=-26.4°C=(-26.4+273)K=246.6K

Therefore, the final temperature of the liquidHFC-134ais-26.4°Cor246.6K

The enthalpy of the liquid phase of HFC-134a at the boiling point at a final pressure of 1.0 bar is 16kJ, while the gas phase is231kJ.

Hliquid=16kJHgas=231kJ

03

Further calculation

A throttling operation conserves the enthalpy. The initial enthalpy of liquid HFC-134a ranges from16kJto 231kJ. The boiling point of

HFC-134a is -26.4°C, which results in a mixture of liquid and gas.

Hf=aHliquid+(1-a)Hgas

SubstituteHliquid=16kJ

Hgas=231kJ

Hf=a(16kJ)+(1-a)(231kJ)=231kJ-(215kJ)a

HFC-134a has the same initial and ultimate enthalpies as water.

Hf=Hi

SubstituteHf=231kJ-(215kJ)aHi=116kJ

solving for a is :

a=231kJ-116kJ215kJ=0.535

Hence, the fraction of liquid vaporizes is as follows:

role="math" localid="1648690588799" 1-a=1-0.535=0.465

Thus, the liquid vaporization is 0.465 .

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Most popular questions from this chapter

Consider a household refrigerator that uses HFC-134a as the refrigerant, operating between the pressures of 1.0barand 10bars.

(a) The compression stage of the cycle begins with saturated vapor at 1 bar and ends at 10 bars. Assuming that the entropy is constant during compression, find the approximate temperature of the vapor after it is compressed. (You'll have to do an interpolation between the values given in Table 4.4.)

(b) Determine the enthalpy at each of the points 1,2,3 and 4 , and calculate the coefficient of performance. Compare to the COP of a Carnot refrigerator operating between the same extreme temperatures. Does this temperature range seem reasonable for a household refrigerator? Explain briefly.

(c) What fraction of the liquid vaporizes during the throttling step?

A power plant produces1GWof electricity, at an efficiency of 40%(typical of today's coal-fired plants).

(a) At what rate does this plant expel waste heat into its environment?

(b) Assume first that the cold reservoir for this plant is a river whose flow rate is 100m3/s.By how much will the temperature of the river increase?

(c) To avoid this "thermal pollution" of the river, the plant could instead be cooled by evaporation of river water. (This is more expensive, but in some areas it is environmentally preferable.) At what rate must the water evaporate? What fraction of the river must be evaporated?

An apparent limit on the temperature achievable by laser cooling is reached when an atom's recoil energy from absorbing or emitting a single photon is comparable to its total kinetic energy. Make a rough estimate of this limiting temperature for rubidium atoms that are cooled using laser light with a wavelength of 780 nm.

An apparent limit on the temperature achievable by laser cooling is reached when an atom's recoil energy from absorbing or emitting a single photon is comparable to its total kinetic energy. Make a rough estimate of this limiting temperature for rubidium atoms that are cooled using laser light with a wavelength of 780 nm.

Prove that if you had a heat engine whose efficiency was better than the ideal value (4.5), you could hook it up to an ordinary Carnot refrigerator to make a refrigerator that requires no work input.

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