Chapter 3, Problem 3.24P

Interpretation Introduction

Interpretation:

It is to determine heat loss from the tank of volume $\text{0}\text{.1}{\text{m}}^{\text{3}}$ contains air at ${\text{25}}^{\text{o}}\text{C}$ and $\text{101}\text{.33}\text{kPa}$ to the compressed air line connected to tank, supplies air at constant conditions of ${\text{45}}^{\text{o}}\text{C}$ and $\text{1500}\text{kPa}$ as a result of crack of valve in gas line so that air flows slowly from air line to the tank until both the pressure will be same keeping tank temperature constant at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

To solve the problem, it is needed to find a relation which relates heat transfer of the tank to quantity of air supplies to it and its temperaturechange. Hence,

  $n_1=\frac{P_1{V}^t}{RT}$

And

  $n_2=\frac{P_2{V}^t}{RT}$

Where $V^t$ is the total volume of the tank, $n_1$ and $n_2$ are the number of moles at different pressure and T is constant temperature.

So, the quantity of air supplies to tank is

  $n^{\text{'}}=n_2-n_1=\frac{(P_2-P_1)V^t}{RT}$.....(1)

For non adiabatic, no shaft work steady state neglecting kinetic and potential terms, the resultant energy balance shall be

  $\stackrel{•}{Q}=-n^{\text{'}}{H}^{\text{'}}+\frac{d(nU)}{dt}$

Multiplying by $dt$ and integrating with respect to time implies

  $\begin{array}{l}Q=n_2{U}_2-n_1{U}_1-n^{\text{'}}{H}^{\text{'}}\\ Q=n_2(U_2-H^{\text{'}})-n_1(U_1-H^{\text{'}})\\ Q=n_2(H_2-RT-H^{\text{'}})-n_1(H_1-RT-H^{\text{'}})\\ Q=n_2(C_P(T-T^{\text{'}})-RT)-n_1(C_P(T-T^{\text{'}})-RT)\end{array}$

Hence

  $Q=n^{\text{'}}(C_P(T-T^{\text{'}})-RT)$.....(2)