Download e-book for kindle: Chemical Engineering: Solutions to the Problems in Volume 1 by J R Backhurst, J H Harker, J. F. Richardson

By J R Backhurst, J H Harker, J. F. Richardson

ISBN-10: 075064950X

ISBN-13: 9780750649506

This quantity within the Coulson and Richardson sequence in chemical engineering includes complete labored ideas to the issues posed in quantity 1. when the most quantity comprises illustrative labored examples through the textual content, this e-book comprises solutions to the more difficult questions posed on the finish of every bankruptcy of the most text.

those questions are of either a customary and non-standard nature, and so will turn out to be of curiosity to either educational employees educating classes during this sector and to the prepared scholar. Chemical engineers in who're searching for a typical technique to a real-life challenge also will locate the e-book of substantial curiosity.

* a useful resource of data for the coed learning the cloth contained in Chemical Engineering quantity 1
* A beneficial approach to studying - solutions are defined in complete

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Additional info for Chemical Engineering: Solutions to the Problems in Volume 1

Example text

074 m3 /s. 36 m/s. 44 m/s. 6 ð 105 . 32 ð 106 . 0005 respectively. From Fig. 6 kN/m2 In addition, there will be a small pressure drop at the junction of the two pipes although this has been neglected in this solution. Thus the existing pump is satisfactory for this duty. 11 Explain the phenomenon of hydraulic jump which occurs during the flow of a liquid in an open channel. 5 m/s and a depth of 75 mm. Calculate, from first principles, the corresponding velocity and depth after the jump. 9. 12 What is a non-Newtonian fluid?

5 ð 10 1 ð 10 7 7 6 6 6 5 4 Suggest a suitable model to describe the fluid properties. Solution Inspection of the data shows that the pressure difference increases less rapidly than the flowrate. Taking the first and the last entries in the table, it is seen that when the flowrate increases from 1 ð 10 7 to 1 ð 10 4 m3 /s, that is by a factor of 1000, the pressure difference increases from 1 ð 103 to 1 ð 105 N/m2 that is by a factor of only 100. In this way, the fluid appears to be shear-thinning and the simplest model, the power-law model, will be tried.

182 D 9750 ∴ From Fig. 0037 which does not agree with the original assumption. 0022. 175 m. 24 On the assumption that the velocity profile in a fluid in turbulent flow is given by the Prandtl one-seventh power law, calculate the radius at which the flow between it and the centre is equal to that between it and the wall, for a pipe 100 mm in diameter. 5. 4 m3 /s. 5 mm, what is the pressure drop? 02 Ns/m2 ? 04 m/s. 001 and from Fig. 6 ð 106 From Fig. 26 Water (density 1000 kg/m3 , viscosity 1 mNs/m2 ) is pumped through a 50 mm diameter pipeline at 4 kg/s and the pressure drop is 1 MN/m2 .

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Chemical Engineering: Solutions to the Problems in Volume 1 by J R Backhurst, J H Harker, J. F. Richardson


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