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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 an ideal Hampson-Linde cycle in which no heat is lost to the environment.

(a) Argue that the combination of the throttling valve and the heat exchanger is a constant-enthalpy device, so that the total enthalpy of the fluid coming out of this combination is the same as the enthalpy of the fluid going in.

(b) Let xbe the fraction of the fluid that liquefies on each pass through the cycle. Show that

x=Hout-HinHout-Hliq,

where Hinis the enthalpy of each mole of compressed gas that goes into the heat exchanger, Houtis the enthalpy of each mole of low-pressure gas that comes out of the heat exchanger, and Hliqis the enthalpy of each mole of liquid produced.

(c) Use the data in Table 4.5to calculate the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle operating between 1 bar and 100 bars, with an input temperature of 300K. Assume that the heat exchanger works perfectly, so the temperature of the low-pressure gas coming out of it is the same as the temperature of the high-pressure gas going in. Repeat the calculation for an input temperature of 200K.

Prove that if you had a refrigerator whose COP was better than the ideal value (4.9), you could hook it up to an ordinary Carnot engine to make an engine that produces no waste heat.

Can you cool off your kitchen by leaving the refrigerator door open? Explain.

The magnetic field created by a dipole has a strength of approximately μ0/4πμ/r3, where r is the distance from the dipole and μ0is the "permeability of free space," equal to exactly 4π×10-7in SI units. (In the formula I'm neglecting the variation of field strength with angle, which is at most a factor of 2.) Consider a paramagnetic salt like iron ammonium alum, in which the magnetic moment μof each dipole is approximately one Bohr magneton 9×10-24J/T, with the dipoles separated by a distance of 1nm. Assume that the dipoles interact only via ordinary magnetic forces.

(a) Estimate the strength of the magnetic field at the location of a dipole, due to its neighboring dipoles. This is the effective field strength even when there is no externally applied field.

(b) If a magnetic cooling experiment using this material begins with an external field strength of 1T, by about what factor will the temperature decrease when the external field is turned off?

(c) Estimate the temperature at which the entropy of this material rises most steeply as a function of temperature, in the absence of an externally applied field.

(d) If the final temperature in a cooling experiment is significantly less than the temperature you found in part (c), the material ends up in a state where ∂S/∂Tis very small and therefore its heat capacity is very small. Explain why it would be impractical to try to reach such a low temperature with this material.

Suppose that the throttling valve in the refrigerator of the previous problem is replaced with a small turbine-generator in which the fluid expands adiabatically, doing work that contributes to powering the compressor. Will this change affect the COP of the refrigerator? If so, by how much? Why do you suppose real refrigerators use a throttle instead of a turbine?

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