Answer:
see explain
Explanation:
Given that,
Water pressure  P 1 = 20 psia
Water temperature  T 1 = 80F
Steam pressure  P 2 = 20 psia
Calculating the enthalpy of steam and water at given pressure and temperature by using a steam table
h 1 = 48.07 BTU/lb
h 2 = 1156.20 Â BTU/lb
The enthalpy of water is determine using the saturated liquid approximation for the given temperature with the data from A -4E. the enthalpy of the vapor is determine from A - 5E for given pressure .the enthalpy of the mixture is determine from the energy balance.
[tex]m_1h_1 + m_2h_2 = m_3h_3\\m_1h_1 + m_2h_2 =2mh_3\\h_3 = \frac{h_1 + h_2}{2} \\= \frac{48.07 + 1156.2}{2} \\602.135\frac{btu}{ibm} \\[/tex]
the quality of the mixture is determine from the total enthalpy and the enthaipies of the constituent at the given pressure obtained from A - 5E
[tex]q = \frac{h_3 - h_l_i_q_2_0}{h_e_v_a_p_2_0} \\ = \frac{602.135 - 196.27}{959.93} \\ = 0.423[/tex]
≅ 0.4
the mixture temperature is simply the saturation temperature for the given pressure obtain from A - 5E
T₃ = 227.92°F
≅ 230°F
