Chapter 7, Problem 7.17P

Interpretation Introduction

Interpretation:

The initial temperature and work produced per mole of nitrogen gas under the given conditions needs to be calculated

Concept Introduction:

• For an ideal gas under isentropic adiabatic conditions, the temperature and pressure are related as:

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}{\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)}^{\text{γ-1/γ}}---(1)\\ {\text{where, γ = C}}_{\text{p}}{\text{/C}}_{\text{v}}\end{array}$

• The work done under isentropic conditions is:

  $\text{W = }\frac{{\text{nR(T}}_{\text{1}}{\text{-T}}_{\text{2}}\text{)}}{\text{γ-1}}\text{ -----(2)}$

The initial temperature = $276.2\text{C}$

Work done/mole of nitrogen = 2623.9 J

Given Information:

Inlet pressure of steam, P₁ = 8.5 bar

Outlet Temperature of steam T₂= $150\text{C}$

Outlet pressure of steam, P₂ = 1.0 bar

Explanation:

In this case, nitrogen has been assumed to behave as an ideal gas. The initial temperature can be deduced from equation (1) and the work done per mole of nitrogen can be calculated from equation (2).

The heat capacity ratio for nitrogen, i.e. C_p/C_v = 1.4

Calculation:

Step 1:

Calculate the initial temperature, T₁

Based on equation (1) we have:

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}{\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)}^{\text{γ-1/γ}}\\ \frac{\text{150}}{{\text{T}}_{\text{1}}}={\left(\frac{1}{8.5}\right)}^{\text{1}\text{.4-1/1}\text{.4 }}=0.543\\ {\mathrm{T}}_1=276.2C\end{array}$

Step 2:

Calculate the work done

Based on equation (2) for 1 mole of nitrogen we have:

  $\text{W = }\frac{\text{1}\times \text{8}\text{.314}\times \text{(276}\text{.2-150)}}{\text{1}\text{.4-1}}\text{ = 2623}\text{.9 J}$

Thus, the initial temperature = $276.2\text{C}$

Work done/mole of nitrogen = 2623.9 J