Question E.1: From data in the steam tables, find: a. The specific volume ...

(a) The following table shows specific volumes from Table E.2 for superheated steam at conditions adjacent to those specified:

Substitution of values in Eq. (E.2) with M = V, X = t, and Y = P yields:

M=\left[\left(\frac{X_2-X}{X_2-X_1}\right) M_{1,1}+\left(\frac{X-X_1}{X_2-X_1}\right) M_{1,2}\right] \frac{Y_2-Y}{Y_2-Y_1}

+\left[\left(\frac{X_2-X}{X_2-X_1}\right) M_{2,1}+\left(\frac{X-X_1}{X_2-X_1}\right) M_{2,2}\right] \frac{Y-Y_1}{Y_2-Y_1}                 (E.2)

+\left[\frac{38}{50}(429.65)+\frac{12}{50}(458.10)\right] \frac{16}{25}=441.42 \mathrm{~cm}^3 \cdot \mathrm{g}^{-1}

(b) The following table shows enthalpy data from Table E.2 for superheated steam at conditions adjacent to those specified:

Here, the direct use of Eq. (E.2) is not convenient. Rather, for P = 2950 kPa, interpolate linearly at t1 = 350°C for H t 1 and at t2 = 375°C for H t 2 , applying Eq. (E.1) twice, first at t1 and second at t2, with M = H and X = P:

H_{t_1}=\frac{50}{100}(3119.7)+\frac{50}{100}(3117.5)=3118.6     (E.1)

H_{t_2}=\frac{50}{100}(3177.4)+\frac{50}{100}(3175.6)=3176.5

A third linear interpolation between these values with M = t and X = H in Eq. (E.1) yields:

t=\frac{3176.5-3150.6}{3176.5-3118.6}(350)+\frac{3150.6-3118.6}{3176.5-3118.6}(375)=363.82^{\circ} \mathrm{C}

Given this temperature, a table of entropy values can now be constructed:

Application of Eq. (E.2) with M = S, X = t, and Y = P yields:

S=\left[\frac{11.18}{25}(6.7654)+\frac{13.82}{25}(6.8563)\right] \frac{50}{100}

+\left[\frac{11.18}{25}(6.7471)+\frac{13.82}{25}(6.8385)\right] \frac{50}{100}=6.8066 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}

As a check, one can apply Eq. (E.2) with M = H, X = t, and Y = P, confirming that doing so produces H = 3150.6 kJ·kg^{−1}.