A vertical piston cylinder assembly contains 10.0kg of a saturated liquid-vapor water mixture with initial quality of 0.85 The water receives energy by heat transfer until the temperature reaches 320*C. The piston has a mass of 204kg and area of 0.005m2. Atmospheric pressure of 100kPa acts on the top side of the piston. Local gravitational acceleration is 9.81m/s2 Calculate the amount of heat transfer between the water and the surroundings in kJ. Enter a numeric value only. 6735.66

Answers

Answer 1

Answer:

Explanation:

From the given information:

At state 1:

Initial Quality [tex]= x_1 = 0.85[/tex]

mass = 10.0 kg

At state 2:

Temperature [tex]T_2 = 320^0[/tex]

mass of the piston [tex]m_p = 204 \ kg[/tex]

area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]

Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]

Gravitational acceleration = 9.81 m/s²

[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.

To calculate the applying force balance over the piston by using force balance in the vertical direction:

[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]

(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05

P = 500248 Pa

P = 500.25 kPa

At state 1:

[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]

[tex]x_1 = 0.85[/tex]

Hence, this is a saturated mixture of liquid and vapor

Using the steam tables at 500.25 kPa

[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]

Specific volume at state 1 is given as:

[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]

volume at state 1 is given by:

[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]

Similarly, the specific internal energy is:

[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]

[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]

[tex]U_1 = 2272.57 \ kJ/kg[/tex]

At state 2:

[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]

Using steam tables at P = 500.25 kPa and T = 320° C

[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]

[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]

[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]

[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]


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