Question

A rigid container of volume 10 m^3 stores 11.221 kg of water initially as vapor and liquid at (T_1) 85 degree C. What is the pressure (P_1) of the system? What is the quality (q_1) of the system? When the temperature of the container is raised to T_2 = 105 degree C, what is the quality of the system (q_2), and what is the pressure (P_2)? What are the changes of enthalpy (Delta H) and changes of internal energy (Delta U) of the mixture at this point relative to the initial state? As the temperature is kept increasing from T_2 = 105 degree C, at what temperature (T_3) and pressure (P_3) does the container contain only saturated vapor? What are the changes of enthalpy (Delta H) and changes of internal energy (Delta U) of the mixture at this point relative to the initial state (i.e. T_1- 85 degree C)? Make a qualitative sketch of parts (a) through (d) on a P-V diagram, showing the phase envelope.

Answer 1

Consider a rigid container of volume 10 m'. Mass water in the container is 11.221 kg. The initial temperature of the water is 85°C. (a) Calculate the pressure of the system as follows: From steam tables, at 85°C pressure obtained is 0.5784 bar. Therefore, pressure of the system is 0.5784 bar . (b) Calculate the quality of the system as follows: At 85°C specific volume of vapor and liquid are as follows: Specific volume of saturated vapor, V, = 2.8 Specific volume of saturated liquid, V, = 0.001032 Consider the following equation. Vi =V, +q(V, -V.) V, is calculated as follows:

10 m 11.221 kg = 0.8911 m kg Substitute the data in the above equation as follows: Vi =V+9(V, -VL) 0.8912 = 0.001032 m +9(2.825 -0.001032) kg q=0.3151 Hence, quality of the system is 31.51% vapor. The system temperature has been raised to 105º C. Calculate the quality of the system and pressure of saturated vapor. From steam tables, note the pressure at 105 C. Pressure of the system at 105° C is 1.209 bar. Calculate the quality of the system as follows:

At 105 C, note the specific volume of saturated vapor and liquid as follows: Specific volume of saturated vapor, V, = 1.4184 - Specific volume of saturated liquid, V, = 0.0010474 Vi =V, +9(V, -V;) 0.8911 m.* = 0.0010474 m*+9(1.4184 – 0.0010474)m. kg kg q=0.6279 Hence, quality of the system when temperature raised to 105° C is 62.79. Calculate the changes in internal energy and enthalpy as follows: Consider internal energy of vapor and liquid at 85°C and 105° C as follows:

At 85°C, Specific enthalpy of saturated vapor = 2651.33 kJ Specific enthalpy of saturated liquid = 355.946 Hence, H (85°C) =(1-x)H, + xH, =(1-0.3151)35.946 k + (0.3151)261.33 =1079.22 kJ At 85°C, Specific internal energy of saturated vapor = 2488 Specific internal energy of saturated liquid = 355.8 U(85°C)=(1-x)U + XU = (1-0.3151)355.8 +(0.3151) 2488 kW =1027.65 kl

At 105°C, Specific enthalpy of saturated vapor = 2683.39 Specific enthalpy of saturated liquid = 440.213 H (105°C)=(1-x)Hz + xH, = (1+0.6279) 440.213 ) +(0.6279)2683.39 =1848.70 kJ At 105°C, Specific internal energy of saturated vapor = 2512.33 Specific intemal energy of saturated liquid = 440.061 U(105°C)=(1-x)U,+ XU, =(1-0.6279)440.061 od +(0.6279) 25 12.33 km =1741.23 La kg Calculate internal energy as follows:

U = m(U (105°C)-U (85C)) =11.221 kg[1741.23–1027.65]k! = 8007.17 kJ Hence, change internal energy is 8007.17 kJ. Calculate enthalpy as follows: H = m(H (105°C) - H (85°C)) =11.221 kg [1848.70–1079.22] = 8634.33 kJ Hence, change in enthalpy is 8634.33 kJ. (d) As the temperature increasing from 105° C, calculate the temperature and pressure at which at which the system contains only saturated vapor. Since only saturated vapor present, V =Vv. Hence, V= 0.89111 From steam tables, take the pressure and temperature at V, = 0.8911 Hence,

From interpolation, pressure at V, = 0.8911 – m? kg is 1.986 bar. Temperature at saturated vapor is 119.978°C. At 119.978°C, Specific enthalpy of saturated vapor is 2705.9 Specific internal energy is of saturated vapor is 2528.96 kg Calculate change in internal energy from initial state to final state as follows: U = m(U (119.978°C)-U (85°C)) =11.221 kg [2528.96 –1027.65] * =16846.199 kJ Hence, change in internal energy is 16846.199 kJ. Calculate change in enthalpy from initial state to final state as follows: H=m(H (105°C)-H (85°C)) =11.221 kg[2705.9 – 1079.22] kl =18252.97 kJ Hence, change in enthalpy is 18252.97 kJ .

PV-diagram for the above steps is given as follows: P(bar) Critical point 1.986 1.209 0.5784 119.978C -1050 85C Vapor-Liquid V = 0.891 VE)