Chapter 10, Problem 10.1P

A solution containing hydrogen peroxide with a mass fraction . $x_p{}_0\left(kg{H}_2{O}_2/kgsolution\right)$is added to a storage tank at a steady rate ${\stackrel{\dot{}}{m}}_0$(kg/h). During this process, the liquid level reaches a corroded spot in the tank wall and a leak develops. As the filling continues, the leak rate ${\stackrel{\dot{}}{m}}_0$(kg/h) becomes progressively worse. Moreover, once it is in the tank the peroxide begins to decompose at a rate

   $r_d\left(kg/h\right)=k{M}_p$

where $M_p$(kg) is the mass of peroxide in the tank. The tank contents are well mixed, so that the peroxide concentration is the same at all positions. At a time t = 0 the liquid level reaches the corroded spot. Let $M_0$and $M_{P0}$be the total liquid mass and mass of peroxide, respectively, in the tank at t = 0, and let M(t) be the total mass of liquid in the tank at any time thereafter.

Show' that the leakage rate of hydrogen peroxide at any time is ${\stackrel{\dot{}}{m}}_1{M}_P/M$.

1. Write differential balances on the total tank contents and on the peroxide in the tank, and provide initial conditions. Your solution should involve only the quantities ${\stackrel{\dot{}}{m}}_0$, ${\stackrel{\dot{}}{m}}_1$, $x_p{}_0$, k, M, $M_0$, $M_P$, $M_{P0}$, and t.

Interpretation Introduction

(a)

Interpretation:

Prove that leakage rate of hydrogen peroxide at any time is $m_1{M}_p/M$.

Concept introduction:

The General Balance Equation is given as:

Accumulation = Input − Output + Generation − Consumption

Answer to Problem 10.1P

$\text{Leakage}\text{rate}=m_1{M}_p/M$

Explanation of Solution

The peroxide mass fraction in the liquid equals that in the contents of the tank. We can write the following equation,

$x_p=\frac{M_P}{M}$

Where M is the mass of all the contents in the tank.

The leakage rate of the hydrogen peroxide is given by the product of the peroxide mass fraction and the leak rate,

$\text{Leakage}\text{rate}=m_1{M}_p/M$

Interpretation Introduction

(b)

Interpretation:

Write differential balances on the total tank contents and on the peroxide in the tank, provide initial conditions.

Concept introduction:

The General Balance Equation is given as:

Accumulation = Input − Output + Generation − Consumption

Answer to Problem 10.1P

The derived equations are,

$\begin{array}{l}\frac{dM}{dt}=m_0-m_1\\ whent=0,M=M_0\left(initialcondition\right)\end{array}$

And

$\begin{array}{l}\frac{d{M}_p}{dt}=m_0{x}_{P_0}-m_1\left(\frac{M_P}{M}\right)-k{M}_p\\ whent=0,M_P=M_{P_0}\left(initialcondition\right)\end{array}$

Explanation of Solution

The total accumulation in the tank is equal to the difference in the mass input and the mass output.

First perform mass balance as,

$\begin{array}{l}Accumulation=input-output\\ \frac{dM}{dt}=m_0-m_1\\ whent=0,M=M_0\left(initialcondition\right)\end{array}$

Perform H₂ O₂ balance as,

$\begin{array}{l}Accumulation=input-output-consumption\\ \frac{d{M}_p}{dt}=m_0{x}_{P_0}-m_1\left(\frac{M_P}{M}\right)-k{M}_p\\ whent=0,M_P=M_{P_0}\left(initialcondition\right)\end{array}$