Chapter 7, Problem 7.10P

Interpretation Introduction

Interpretation:

The state of steam at the nozzle exit and the ${\text{S<mo>˙</mo>}}_{\text{G}}$ under the given conditions needs to be calculated

Concept Introduction:

• The energy balance equation that relates the inlet and outlet enthalpies and velocities is given based on the first law of thermodynamics as:

$\begin{array}{l}{\left({\text{H}}_{\text{1}}\text{+}\frac{{\text{u}}_{\text{1}}^{\text{2}}}{\text{2}}\right)}_{\text{inlet}}\text{= }{\left({\text{H}}_{\text{2}}\text{+}\frac{{\text{u}}_{\text{2}}^{\text{2}}}{\text{2}}\right)}_{\text{outlet}}\\ \text{(or)}\\ \text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ = 0 ----(1)}\end{array}$

• For an adiabatic process, there is no loss or gain of heat i.e. q = 0

  $\begin{array}{l}\text{ΔS =}\frac{\text{q}}{\text{T}}{\text{ + S}}_{\text{G}}\\ \text{ΔS}={\text{S}}_{\text{2}}{\text{-S}}_{\text{1}}{\text{ = S}}_{\text{G}}\text{-----(2)}\end{array}$

$$

The state of steam at exit is saturated steam with ${\text{S}}_{\text{G}}\text{ = }0.0616\text{ Btu/lbm-R}$

Given Information:

Inlet pressure of steam, P₁ = 130 psi

Inlet Temperature of steam T₁= $420\text{F}$

Inlet velocity, u₁ = 230 ft/s

Outlet pressure of steam, P₂ = 35 psi

Inlet velocity, u₂ = 2000 ft/s

Explanation:

Based on the steam tables, for inlet conditions: P₁ = 130 psi, T₁= $420\text{F}$

Specific enthalpy H₁= 1233.6 Btu/lbm

Specific entropy, S₁=1.6310 Btu/lbm-R

Step 1:

Calculate the enthalpy at exit, H₂

Based on equation (1) we have:

$\begin{array}{l}{\text{H}}_{\text{2}}-{\text{H}}_{\text{1}}\text{ + }\frac{{\text{u}}_2^2-{\text{u}}_1^2}{\text{2}}\text{ = 0 }\\ ({\text{H}}_{\text{2}}-1233.6\text{ Btu/lbm})\text{ = }-\frac{{(2000)}^2-{(230)}^2}{\text{2}}{\text{ft}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{H}}_{\text{2}}-1233.6\text{ Btu/lbm = }-\frac{1973550{\text{ ft}}^{\text{2}}{\text{/s}}^{\text{2}}\times 1\text{ Btu/lbm}}{25,037{\text{ft}}^{\text{2}}{\text{/s}}^{\text{2}}}\\ {\text{H}}_{\text{2}}\text{ = 1154}\text{.7 Btu/lbm}\end{array}$

Step 2:

Calculate the final state entropy, S₂

The exit temperature, T₂corresponding to the exit pressure P₂ = 35 psi and ${\text{H}}_{\text{2}}\text{ = 1154}\text{.7 Btu/lbm}$, can be calculated to be around $235\text{F}$ .

The entropy at the outlet = S₂ = 1.6926 Btu/lbm-R

Step 3: Calculate ${\text{S<mo>˙</mo>}}_{\text{G}}$

Based on equation (2) we have:

${\text{S}}_{\text{G}}{\text{ = S}}_{\text{2}}{\text{-S}}_{\text{1}}=1.6926-1.6310=0.0616\text{ Btu/lbm-R}$

Thus, the state of steam at exit is saturated steam with ${\text{S}}_{\text{G}}\text{ = }0.0616\text{ Btu/lbm-R}$