Chapter 4, Problem 4.40P

Interpretation Introduction

Interpretation:

To find the amount of water evaporated in the dryer per lbm of oil burnt

Concept Introduction:

By definition of rate of change of enthalpy,

  $\Delta H_R=R\times MCPH(T_0,T,A,B,C,D)\times (T-T_0)$

  $MCPH=\frac{{(C_p)}_H}{R}=A+\frac{B}{2}{T}_0(\tau +1)+\frac{C}{3}{T}_0^2\left({\tau }^2+\tau +1\right)+\frac{D}{\tau T_0^2}$

A,B,C,D are constants

T₀ = Initial temperature

T = Final temperature

R = Universal Gas Constant = 8.314 J/mol K = 1.986 BTU/lbmol.R

  $\tau =\frac{T}{T_O}$

  $\Delta H_R$ = Change of enthalpy of Reactants stream

  $\Delta H_{298}$ = Standard enthalpy at 298K

  $\Delta H_P$ = Change of enthalpy of Products stream

Q = Net heating value of oil = 19000 BTU/lbm

Taking energy balance

  $Q=(\Delta H_R+\Delta H_{298}+\Delta H_P)$

Find the standard heat of the reactions taking place in the drier.

Also find the standard heat of reaction at 298 K