CHEMICAL ENGINEERING

2 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

From the steam tables, in Appendix A2, Volume 1, total enthalpy of steam at 1825 kN/m^2 D
2798 kJ/kg.

∴ enthalpy of steamD 5. 2 ð 2798 D 14 ,550 kW

Neglecting the enthalpy of the feed water, this must be derived from the coal. With an
efficiency of 75%, the heat provided by the coalD 14 , 550 ð 100 / 75 D 19 ,400 kW.
For a calorific value of 27,200 kJ/kg, rate of coal consumptionD 19 , 400 / 27 , 200 
D 0 .713 kg/s

or:  0. 713 ð 3600 ð 24 / 1000 D 61 .6 Mg/day

20% of the enthalpy in the steam is converted to power or:

 14 , 550 ð 20 / 100 D2910 kW or 2.91 MW say 3 MW

PROBLEM 1.

The power required by an agitator in a tank is a function of the following four variables:

(a) diameter of impeller,

(b) number of rotations of the impeller per unit time,

(c) viscosity of liquid,

(d) density of liquid.

From a dimensional analysis, obtain a relation between the power and the four variables.
The power consumption is found, experimentally, to be proportional to the square of
the speed of rotation. By what factor would the power be expected to increase if the
impeller diameter were doubled?

Solution

If the powerPDfDN
, then a typical form of the function isPDkDaNb c d,where
kis a constant. The dimensions of each parameter in terms ofM, L,andTare: power,
PDML^2 /T^3 , density, DM/L^3 , diameter,DDL, viscosity, DM/LT, and speed of
rotation,NDT^1
Equating dimensions:

M: 1 DcCd

L: 2 Da 3 cd

T: 3 Dbd

Solving in terms ofd:aD 5  2 d,bD 3 d,cD 1 d

∴ PDk

(

D^5

D^2 d

N^3

Nd

d

d

)

or: P/D^5 N^3 DkD^2 N
/d

that is: NPDkRem