Solutions%20to%20Homework#4

Solutions%20to%20Homework#4 - Assignment Solution 4...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment Solution 4 Chemical Engineering 2800 (Summer 2008) Fundamentals of Process Engineering University of Utah Dr.Mataz Alcoutlabi Problem 1 The specific heat of a substance is 5320 J/(kg°C) and its molecular weight is 37.4. Find its specific heat in Solution C = 5320 J/kg°C a) 5320 J/(kg K) b) 1 Btu = 1055.056 J 1 kg = 2.2046 lbm 1.8Δ°F = 1.0 Δ°C F lbm Btu 1270 lbm 2.2046 1kg 1055.1J 1Btu 8 . 1 1 kg J 5320 ° = ° ° ° F C C c) K kgmol kJ 199 1000J kJ kgmol 37.4kg kg J 5320 = ° C Problem 2 A Bourden gage that is attached to a nitrogen cylinder reads 400 psig. The barometric pressure is 650 mmHg. What is the absolute pressure in the tank (psia)? Solution P = 400 psig psia mmHg psi mmHg P atm 57 . 12 760 696 . 14 650 = = P abs = P gage +P atm = 413 psia Problem 3 The composition of a Power River Basin coal is given below (weight percent). Calculate its composition (weight percent) on a dry basis C 51.19 H 3.64 O 12.29 N 0.72 S 0.32 ash 6.03 moisture 25.81 Total 100 Solution Basis: 100 kg net coal Based on dry basis Sum of dry components = 100 - 25.81 = 74.19 kg
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
C 68.99852 H 4.906322 O 16.56557 N 0.970481 S 0.431325 ash 8.12778 Total 100 Problem 4 The specific heat of N 2 is given by 3 2 dT cT bT a c P + + + = where Cp is in kJ/(kmol K) and T is in K. The values of the constants are a = 28.90, b = -0.1571x10 -2 , c = 0.8081x10 -5 and d = -2.873x10 -9 . Suppose we want to use our formula with T in °C rather than K. What are the new values of a, b, c, d? Solution ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 K T K T K T c P 3 9 2 5 2 10 873 . 2 10 8081 . 0 10 1571 . 0 9 . 28 - - - × - × + × - = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 3 9 2 5 2 15 . 273 10 873 . 2 15 . 273 10 8081 . 0 15 . 273 10 1571 . 0 9 . 28 + ° × - + ° × + + ° × - = - - - C T C T C T c P ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 C T C T C T c P ° × - ° × + ° × + = - - - 3 9 2 5 2 10 873 . 2 10 57267 . 0 10 22006 .. 0 01 . 29
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/07/2009 for the course CHEN 2800 taught by Professor Process during the Summer '08 term at University of Utah.

Page1 / 8

Solutions%20to%20Homework#4 - Assignment Solution 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online