Here is the missing part of the question
To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C
Answer:
Explanation:
The image attached below shows the process on T - v diagram
At State 1:
The first step is to find the vapor pressure
[tex]P_{v1} = \rho_1 P_g_1[/tex]
[tex]= \phi_1 P_{x \ at \ 125^0C}[/tex]
= 0.5 × 232 kPa
= 116 kPa
The initial specific volume of the vapor is:
[tex]P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1[/tex]
[tex]116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)[/tex]
[tex]116 \times 10^3 \times v_{v_1} = 183831.7778[/tex]
[tex]v_{v_1} = 1.584 \ m^3/kg[/tex]
At State 1:
The next step is to determine the mass of water vapor pressure.
[tex]m_{v1} = \dfrac{V}{v_{v1}}[/tex]
[tex]= \dfrac{2.5}{1.584}[/tex]
= 1.578 kg
Using the ideal gas equation to estimate the mass of the dry air [tex]m_a[/tex][tex]P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1[/tex]
[tex](P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1[/tex]
[tex](4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)[/tex]
[tex]710000= m_a \times 114220.642[/tex]
[tex]m_a = \dfrac{710000}{114220.642}[/tex]
[tex]m_a = 6.216 \ kg[/tex]
For the specific volume [tex]v_{v_1} = 1.584 \ m^3/kg[/tex] , we get the identical value of saturation temperature
[tex]T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)[/tex]
[tex]T_{sat} =101.92 ^0\ C[/tex]
Thus, at [tex]T_{sat} =101.92 ^0\ C[/tex], condensation needs to begin.
However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.
Heat can now be determined by using the formula
Q = ΔU + W
Recall that: For a rigid tank, W = 0
Q = ΔU + 0
Q = ΔU
Q = U₂ - U₁
Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.
At State 1;
The internal energy is calculated as:
[tex]U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )[/tex]
At [tex]T_1[/tex] = 125° C, we obtain the specific internal energy of air
SO;
[tex]U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )[/tex]
[tex]=278.93 + ( 7.23) (\dfrac{8}{10} )[/tex]
[tex]= 284.714 \ kJ/kg\\[/tex]
At [tex]T_1[/tex] = 125° C, we obtain the specific internal energy of water vapor
[tex]U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg[/tex]
[tex]U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )[/tex]
= 6.216 × 284.714 + 1.578 × 2534.5
= 5768.716 kJ
At State 2:
The internal energy is calculated as:
[tex]U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )[/tex]
At temperature 110° C, we obtain the specific internal energy of air
SO;
[tex]U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )[/tex]
[tex]271.69+ (7.24) (0.3)[/tex]
[tex]= 273.862 \ kJ/kg\\[/tex]
At temperature 110° C, we obtain the specific internal energy of water vapor
[tex]U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg[/tex]
[tex]U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )[/tex]
= 6.216 × 273.862 + 1.578 × 2517.9
= 5675.57 kJ
Finally, the heat transfer during the process is
Q = U₂ - U₁
Q = (5675.57 - 5768.716 ) kJ
Q = -93.146 kJ
with the negative sign, this indicates that heat is lost from the system.
