Physics, 18.09.2019 22:30, gustinya000

Aclosed, rigid tank contains a two-phase liquid–vapor mixture of refrigerant 22 initially at 2208c with a quality of 50.36%. energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. determine the final temperature, in 8c. if the final state is in the superheated vapor region, at what temperature, in 8c, does the tank contain only saturated vapor?

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The final temperature, when the pressure is 6 bar, is 43.75 C. And it is indeed in the superheated vapor region.The tank only would contain saturated vapor at 0 C.

Explanation:

It is needed property tables for Refrigerant 22.It must be noted, that if the tank is rigid, then the specific volume would be constant in each state of the refrigerant.Then knowing the initial state is at -20 C and 50.36% quality, it is possible to calculate the specific volume at the initial state.The specific volume would be calculated using $v_{1}=v_{1f}+X_1*(v_{1g}-v_{1f})$ where: $v_{1}$ is the specific volume at the initial state, $v_{1f}$ is the specific volume of the saturated liquid state at the temperature of the initial state, $(v_{1g}$ is the specific volume of the saturated vapor state at the same temperature of the initial state, and X_1

is the quality at the initial state.If $v_{1f}$ at -20 C is $0.7427*10^{-3}\frac{m^3}{kg}$ , and v_1g

at -20 C is $0.0926\frac{m^3}{kg}$ , then: $v_1=0.7427*10^{-3}\frac{m^3}{kg}+0.5036(0.0926\frac{m3}{kg}-0.7427*10^{-3}\frac{m^3}{kg})=0.047\frac{m^3}{kg}$ As the specific volume stays constant, then the specific volume would be compared with the specific volume at saturated vapor state at 6 bar. At this pressure $v_g=0.0392\frac{m^3}{kg}$ . Then, as this value is lower than the specific volume previously calculated ( $0.047\frac{m^3}{kg}$ ), it means that the final state is in the superheated vapor region.Looking for the specific volume at 6 bar pressure, it is found that is between 45 C ( $0.04724\frac{m^3}{kg}$ ) and 40 C ( $0.04623\frac{m^3}{kg}$ . Interpolation gives that with a specific volume of $0.047 \frac{m^3}{kg}$ and a pressure of 6 bar, the temperature would be 43.75 C.To found the temperature when the tank only contains saturated vapor, only is required to look when the saturated vapor specific volume is equal to the specific volume calculated for the process ( $0.047\frac{m^3}{kg}$ ). This is found to be at 0 C.
Aclosed, rigid tank contains a two-phase liquid–vapor mixture of refrigerant 22 initially at 2208c w

Aclosed, rigid tank contains a two-phase liquid–vapor mixture of refrigerant 22 initially at 2208c w

answered: Guest

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.

answered: Guest

Mass / weighthope it !

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Aclosed, rigid tank contains a two-phase liquid–vapor mixture of refrigerant 22 initially at 2208c w...