Chapter 3, Problem 3.25P

(a)

Interpretation Introduction

Interpretation:

A closed tank with a valve which maintain gas flow from supply line contains gas at a lower pressure than same gas in supply line at constant $T$ and $P$ . After allowing the flow of some gas through value and then shut down, Derive a general equations of energy balance between $n_1\text{and }{n}_2,U_1\text{and}{U}_2,H^{\text{'}}\text{and }Q$ .

Concept Introduction:

The general mass balance across any system is,

$\frac{$ \stackrel{•}{Q}+\stackrel{•}{W}=\Delta [(H+\frac{1}{2}{u}^2+gz)\stackrel{•}{m}]+\frac{d(mU)}{dt}$}{}$

dmdt+Δm=0

.....(1)

The energy balance for a system is:

.....(2)

(b)

Interpretation Introduction

Interpretation:

Reduce the general form of energy balance found in part (a) to its simplest form for the ideal gas with constant heat capacities.

Concept Introduction:

For constant specific heat

  $\Delta U=C_v\Delta T$

And for ideal gas,

$R=C_P-C_V$ and $\frac{C_P}{C_V}=R$

(c)

Interpretation Introduction

Interpretation:

Reduce the general form of energy balance form for the ideal gas with constant heat capacities found in part (b) to its simplest form for case of $n_1=0$ .

Concept Introduction:

If it is given that $n_1=0$ this means that number of moles present in tank at the beginning is zero. Hence there is no gas present is gas or this is case of vacuum inside tank and gas is filling from supply line.

(d)

Interpretation Introduction

Interpretation:

Reduce the general form of energy balance form for the ideal gas with constant heat capacities for case of $n_1=0$ found in part (c) to its simplest form for case of $Q=0$ .

Concept Introduction:

If it is given that $Q=0$ this means that there is no heat transfer and process is completely adiabatic.

And for ideal gas,

$R=C_P-C_V$ and $\frac{C_P}{C_V}=\gamma$

(e)

Interpretation Introduction

Interpretation:

Assume nitrogen as an ideal gas with $C_P=\frac{7}{2}R$supplying to the closed tank with volume $\text{4}{\text{m}}^{\text{3}}$ at $\text{298}\text{.15}{\text{(25}}^{\text{o}}\text{C)}$ and $\text{3}\text{bar}$ .Determine number of moles of nitrogen that flow into the tank to equalize the pressures for two cases: 1. No heat loss from the tank 2. A insulated tank with weigh $\text{400}\text{kg}$ has initial temperature of $\text{298}\text{.15}{\text{(25}}^{\text{o}}\text{C)}$ and specific heat $\text{0}\text{.46}\text{kJ}{\text{kg}}^{-1}{\text{K}}^{-1}$ is heated by gas so as always to be at temperature of the gas in the tank.

Concept Introduction:

Number of moles of nitrogen that flow into the tank to equalize the pressures for two cases calculated by:

$n_2=\frac{P_2{V}^t}{R{T}_2}$  ............(1)

Where $n_2$ = number of moles of gas in the tank at end of process which we need to calculate. $P_2\text{and }{T}_2$ = Pressure and temperature in the tank at end of process and $V^t$ is the total volume of the tank.

The heat transfer between the supply line gas with temperature $T^{\text{'}}$and tank $T_2$ is given by:

  $Q=-m_{\text{tank}}C(T_2-T^{\text{'}})$