Respuesta :
Explanation:
The given data is as follows.
Initial temperature of the system ([tex]T_{1}[/tex]) = [tex]-20^{o}C[/tex]
So, at state 1 quality of the system is ([tex]x_{t}[/tex]) = 0.5036
Pressure at state 2, ([tex]P_{2}[/tex]) = 6 bar
As, it is a rigid tank hence, the specific volumes at state 1 and state 2 are equal.
So, [tex]\nu_{2} = \nu_{t}[/tex]
Now, from the saturated refrigerant-22 table taking specific volume at [tex]-22^{o}C[/tex] is as follows.
[tex](\nu_{f})_{t} = 0.7427 \times 10^{-3} m^{3}/kg[/tex]
[tex](\nu_{g})_{t} = 0.0926 m^{3}/kg[/tex]
Thus, [tex]\nu_{2} = \nu_{t}[/tex]
= [tex](\nu_{f})_{t} + x_{t} ((\nu_{g})_{t} - (\nu_{f})_{t})[/tex]
= 0.0007427 + 0.5036 (0.0926 - 0.0007427)
= [tex]0.04700 m^{3}/kg[/tex]
Hence, [tex]\nu_{g}[/tex] = [tex]0.04700 m^{3}/kg[/tex]
Super heated refrigerant-22 tables take pressure at 6 bar. Interpolation method used to find the temperature is as follows.
[tex]\nu_{t} = 0.04628 m^{3}/kg[/tex] and [tex]T_{t} = 40^{o}C[/tex]
[tex]\nu_{t} = 0.04724 m^{3}/kg[/tex] and [tex]T_{2} = 45^{o}C[/tex]
[tex]T_{vap} = T_{t} + \frac{(\nu_{g} - \nu_{t})(T_{2} - T)_{t}}{\nu_{2} - \nu_{t}}[/tex]
= [tex]40^{o}C + \frac{(0.04700 - 0.04727) \times (45 - 40)}{0.04724 - 0.04628}[/tex]
= 40 + 3.75
= [tex]43.75^{o}C[/tex]
Saturated vapor refrigerant-22 takes specific volume at [tex]0.04700 m^{3}/kg[/tex].
Therefore, interpolation method used to find the temperature will be as follows.
[tex]\nu_{t} = 0.0492 m^{3}/kg[/tex] and [tex]T_{t} = -1.45^{o}C[/tex]
[tex]\nu_{2} = 0.0469 m^{3}/kg[/tex] and [tex]T_{2} = 0.12^{o}C[/tex]
[tex]T_{\text{saturated vapor}} = T_{t} + \frac{(\nu_{g} - \nu_{t})(T_{2} - T_{t}}{\nu_{2} - \nu_{t}}[/tex]
= [tex]-1.45^{o}C + \frac{(0.04700 - 0.0492) \times (0.12 - (-1.45))}{0.0469 - 0.0492}[/tex]
= -1.45 + 1.50
= [tex]0.051^{o}C[/tex]
Thus, we can conclude that at [tex]0.051^{o}C[/tex] tank contains only saturated vapor.
A saturated vapor is when a substance exists as vapor at its saturation temperature.
At 0.05 °C, the tank will contain only saturated vapor.
What is the energy transfer?
The transfer of energy from one form to other is called energy transfer.
Given,
Initial temperature is −20°C
The state quality of the system is 50.36%
The specific volume of state 1 and state 2 are equal
So, [tex]\bold{V_2 = V_t}[/tex]
Now, taking the specific volume is -22°C
[tex]\bold{(v_f)_t = 0.7427\times10^-^3 m^3/kg}[/tex]
[tex]\bold{(v_g)_t = 0.926 \; m^3/kg}[/tex]
Therefore,
[tex]\bold{V_2 = V_t}[/tex]
[tex]v_g = (V_f)_t+ x_t( (v_q)-(v_t)_t)\\\\v_g= 0.0007427+ 0.5036(0.0926-0.0007427)\\\\v_g= 0.04700\; m^3/kg\\\\v_g= 0.04700\; m^3/kg\\[/tex]
The pressure is 6 bar at Super heated refrigerant-22 table. Interpolation method will use here to find the temperature.
[tex]\bold{v_t = 0.04628\;m^3/kg\; and \; T_t= 40^\circ C}\\\\\bold{v_t = 0.04724\;m^3/kg\; and \; T_2= 45^\circ C}\\\\\\\bold{T_v_a_p= T_t + \dfrac{(v_q-v_t)(T_2-T_1)_t}{v_2-v_t}}\\\\\\\\\bold{T_v_a_p = 40^\circ c+ \dfrac{(0.04700-0.04727)(45-40)_t}{0.04724- 0.04628}}\\\\\\\\\bold{T_v_a_p = 40^\circ c + 3.75}\\\\\bold{T_v_a_p = 43.75^\circ c}[/tex]
The volume is [tex]v_g= 0.04700\; m^3/kg\\[/tex]
The interpolation method will be used to find the temperature
[tex]\bold{v_t = 0.0492\;m^3/kg\; and \; T_t= -1.45^\circ C}\\\\\bold{v_t = 0.0469\;m^3/kg\; and \; T_2= 0.12^\circ C}\\\\\\\bold{T _s_/_v= T_t + \dfrac{(v_q-v_t)(T_2-T_1)_t}{v_2-v_t}}\\\\\\\\\bold{T_v_a_p = 1.45^\circ c+ \dfrac{(0.04700-0.0492)(0.12-(-1.45)}{0.0469- 0.0492}}\\\\\\\\\bold{T_v_a_p = 1.45^\circ c + 1.50}\\\\\bold{T_v_a_p = 0.051^\circ c}[/tex]
Thus, At 0.05 °C, the tank will contain only saturated vapor.
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