Question 11.17: Suppose 1.00 lbm of superheated steam at 1.00 psia and 200.°...

The unknowns here are T_2 and p_2, and since this is a closed system, the energy balance (neglecting any changes in system kinetic and potential energy) is

\underset{\underset{\overset{0}{(\text{Badiabatic})} }{\downarrow } }{_1Q_2} – \underset{\underset{\overset{0}{(\text{aergonic})} }{\downarrow } }{_1W_2} = m(u_2 − u_1) = U_2 − U_1

or U_2 = U_1, or u_2 = u_1 = U_1/m_1 = (m_Au_A + m_Bu_B)/(m_A + m_B). Also, for a closed, rigid system, the total volume and mass are constant, so

v_2 = v_1 = (m_Av_A + m_Bv_B)/(m_A + m_B)

This problem is very difficult to solve using the steam tables or the Mollier diagram. It requires the construction of lines of constant u and constant v for various combinations of pressure and temperature, then finding the intersection of the u = u_1 = u_2 and v = v_l = v_2 lines. However, the steam states given in the problem statement fall within the ideal gas region of Figure 11.12, so we can solve this problem with reasonable accuracy using the ideal gas equations of state. Since the reference state values ultimately cancel out here, we can simplify the algebra by taking T_o = 0  \text{R}  and u_o = 0  \text{Btu/lbm} , then we can write u_A = c_vT_A, u_B = c_vT_B, and u_2 = c_vT_2. Also, from the Boyle-Charles ideal gas equation, we have that v_A = RT_A/p_A, v_B = RT_B/p_B, and v_2 = RT_2/p_2. Then,

u_2 = c_vT_2 = u_1 = c_v(m_AT_A + m_BT_B)/(m_A + m_B)

or

T_2 = (m_AT_A + m_BT_B)/(m_A + m_B)

and

v_2 = \frac{RT_2}{p_2} = \frac{R(m_AT_A/p_A + m_BT_B/p_B)}{m_A + m_B}

or

p_2 = \frac{(m_A + m_B)T_2}{m_AT_A/p_A + m_BT_B/p_B}

For the values given in the problem statement, we get

T_2 = \frac{(1.00  \text{lbm})(659.67  \text{R}) + (5.00  \text{lbm})(859.67  \text{R})}{6.00  \text{lbm}} = 827  \text{R} = 367°\text{F}

and

p_2 =\frac{(6.00 \text{lbm})(827 \text{R})}{(1.00  \text{lbm})(659.67  \text{R})/(1.00  \text{psia}) + (5.00  \text{lbm})(859.67 \text{R})/(5.00  \text{psia})} = 3.26  \text{psia  }

Note that, since the specific heat and gas constant cancel out in the equations for the final temperature and pressure in Example 11.17, they are independent of the material being mixed. The following exercises illustrate the use of this material.