Chapter 4.5, Problem 79P

The state of liquid water is changed from 50 psia and 50°F to 2000 psia and 100°F. Determine the change in the internal energy and enthalpy of water on the basis of the (a) compressed liquid tables, (b) incompressible substance approximation and property tables, and (c) specific-heat model.

To determine

The specific internal energy of water on the basis of the compressed liquid table.

The specific enthalpy of water on the basis of the compressed liquid table.

Answer to Problem 79P

The specific internal energy of water on the basis of the compressed liquid table is $\underset{\_}{49.29\text{Btu}/\text{lbm}}$.

The specific enthalpy of water on the basis of the compressed liquid table is $\underset{\_}{55.08\text{Btu}/\text{lbm}}$.

Explanation of Solution

Determine the change in an initial specific enthalpy of the liquid water.

$h_1=h_{f@50°\text{F}}+v_f\left(P-P_{\text{sat@T}}\right)$ (I)

Here, the specific enthalpy at 50 F temperature is $h_{f@50°\text{F}}$, the specific volume of saturated liquid is $v_f$, the pressure of liquid water is P, and the pressure at saturated temperature is $P_{\text{sat@T}}$.

Determine the specific internal energy of water on the basis of the compressed liquid table.

$\Delta u=u_2-u_1$ (II).

Here, the initial specific internal energy of water is $u_1$ and the final specific internal energy of water is $u_2$.

Determine the specific enthalpy of water on the basis of the compressed liquid table.

$\Delta h=h_2-h_1$ (III).

Here, the initial specific enthalpy energy of water is $h_1$ and the final specific enthalpy of water is $h_2$.

Conclusion:

From the Table A-4E, “Saturated water”, obtain the value of initial internal energy and enthalpy at temperature 50 F.

$\begin{array}{l}{h}_f=18.07\text{Btu}/\text{lbm}\\ v_f=0.01602{\text{ft}}^{\text{3}}/\text{lbm}\\ P_{\text{sat@T}}=0.17812\text{psia}\end{array}$.

Substitute $18.07\text{Btu}/\text{lbm}$ for $h_f$, $0.01602{\text{ft}}^{\text{3}}/\text{lbm}$ for $v_f$, $0.17812\text{psia}$ for $P_{\text{sat@T}}$, and 50 psia for P in Equation (I).

$\begin{array}{c}{h}_1=\left(18.07\text{Btu}/\text{lbm}\right)+\left(0.01602{\text{ft}}^{\text{3}}/\text{lbm}\right)\left(50-0.17812\right)\text{psia}\\ =\left(18.07\text{Btu}/\text{lbm}\right)+\left(0.01602{\text{ft}}^{\text{3}}/\text{lbm}\right)\left(49.82\text{psia}\right)\times \left(\frac{1\text{}\text{Btu}}{5.404\text{psia}\cdot {\text{ft}}^{\text{3}}}\right)\\ =\left(18.07\text{Btu}/\text{lbm}\right)+\left(\text{0}\text{.147696}\text{Btu}/\text{lbm}\right)\\ =\text{18}\text{.21}\text{Btu}/\text{lbm}\end{array}$

From the Table A-7E, “Compressed liquid water”, obtain the value of final internal energy and enthalpy at pressure 2000 psia and temperature 100 F.

$\begin{array}{l}{u}_2=67.36\text{Btu}/\text{lbm}\\ h_2=73.30\text{Btu}/\text{lbm}\end{array}$

Substitute $67.36\text{Btu}/\text{lbm}$ for $u_2$ and $18.07\text{Btu}/\text{lbm}$ for $u_1$ in Equation (II).

$\begin{array}{c}\Delta u=\left(67.36-18.07\right)\text{Btu}/\text{lbm}\\ =49.29\text{Btu}/\text{lbm}\end{array}$

Thus, the specific internal energy of water on the basis of the compressed liquid table is $\underset{\_}{49.29\text{Btu}/\text{lbm}}$.

Substitute $73.30\text{Btu}/\text{lbm}$ for $h_2$ and $18.21\text{Btu}/\text{lbm}$ for $h_1$ in Equation (II).

$\begin{array}{c}\Delta h=\left(73.30-18.21\right)\text{Btu}/\text{lbm}\\ =55.08\text{Btu}/\text{lbm}\end{array}$

Thus, the specific enthalpy of water on the basis of the compressed liquid table is $\underset{\_}{55.08\text{Btu}/\text{lbm}}$.

To determine

The specific internal energy of water on the basis of the incompressible substance and property tables.

The specific enthalpy of water on the basis of the incompressible substance and property tables.

Answer to Problem 79P

The specific internal energy of water on the basis of the incompressible substance and property tables is $\underset{\_}{49.96\text{Btu}/\text{lbm}}$.

The specific enthalpy of water on the basis of the incompressible substance and property tables is $\underset{\_}{49.96\text{Btu}/\text{lbm}}$.

Explanation of Solution

Determine the specific internal energy of water on the basis of the incompressible substance and property tables.

$\Delta u=u_2-u_1$ (IV).

Here, the initial specific internal energy of water is $u_1$ and the final specific internal energy of water is $u_2$.

Determine the specific enthalpy of water on the basis of the incompressible substance and property tables.

$\Delta h=h_2-h_1$ (V).

Here, the initial specific enthalpy energy of water is $h_1$ and the final specific enthalpy of water is $h_2$.

Conclusion:

From the Table A-4E, “Saturated water”, obtain the value of initial internal energy and enthalpy at temperature 50 F and 100 F.

$\begin{array}{l}{u}_1\cong u_{f@50°\text{F}}=18.07\text{Btu}/\text{lbm}\\ h_1\cong h_{f@50°\text{F}}=18.07\text{Btu}/\text{lbm}\\ u_2\cong u_{f@100°\text{F}}=68.03\text{Btu}/\text{lbm}\\ h_2\cong h_{f@100°\text{F}}=68.03\text{Btu}/\text{lbm}\end{array}$.

Substitute $68.03\text{Btu}/\text{lbm}$ for $u_2$ and $18.07\text{Btu}/\text{lbm}$ for $u_1$ in Equation (II).

$\begin{array}{c}\Delta u=\left(68.03-18.07\right)\text{Btu}/\text{lbm}\\ =49.96\text{Btu}/\text{lbm}\end{array}$

Thus, the specific internal energy of water on the basis of the incompressible substance and property tables is $\underset{\_}{49.96\text{Btu}/\text{lbm}}$.

Substitute $68.03\text{Btu}/\text{lbm}$ for $h_2$ and $18.07\text{Btu}/\text{lbm}$ for $h_1$ in Equation (II).

$\begin{array}{c}\Delta h=\left(68.03-18.07\right)\text{Btu}/\text{lbm}\\ =49.96\text{Btu}/\text{lbm}\end{array}$

Thus, the specific enthalpy of water on the basis of the incompressible substance and property tables is $\underset{\_}{49.96\text{Btu}/\text{lbm}}$.

To determine

The specific internal energy of water on the basis of the specific heat model.

Answer to Problem 79P

The specific internal energy of water on the basis of the specific heat model is $\underset{\_}{50\text{Btu}/\text{lbm}}$.

Explanation of Solution

Determine the specific heat of water.

$\begin{array}{c}\Delta h=\Delta u\\ =c_p\left(T_2-T_1\right)\end{array}$ (VI)

Here, the specific heat of water is $c_p$, the initial temperature of water is $T_1$, and the final temperature of water is $T_1$.

Conclusion:

From the Table A-3E(a), “Properties of common liquids, solids, and foods”, obtain the value of specific heat of water is $1.00\text{Btu}/\text{lbm}\cdot \text{R}$.

Substitute $1.00\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ 100 F for $T_2$ and 50 F for $T_1$ in Equation (VI).

$\begin{array}{c}\Delta h=\left(1.00\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(100°\text{F}-50°\text{F}\right)\\ =50\text{Btu}/\text{lbm}\end{array}$

Thus, the specific internal energy of water on the basis of the specific heat model is $\underset{\_}{50\text{Btu}/\text{lbm}}$.