Unit Operations of Chemical Engineering
7th Edition
ISBN: 9780072848236
Author: Warren McCabe, Julian C. Smith, Peter Harriott
Publisher: McGraw-Hill Companies, The
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Chapter 4, Problem 4.9P
(a)
Interpretation Introduction
Interpretation:
The variation between the total momentum and the product of mass flow rate and average velocity is to be determined.
Concept Introduction:
The volumetric flow rate of a fluid is given as,
Q = Volumetric flow rate of the fluid
u = Fluid velocity
S = Total cross-section area
The mass flow rate is given as,
The average velocity if the fluid is,
The momentum of the fluid is the product of the mass flow rate and the velocity of the fluid.
M = Momentum of the fluid
(b)
Interpretation Introduction
Interpretation:
The situation at which momentum flow is equal to the product of mass flow and average velocity is to be determined.
Concept Introduction:
The volumetric flow rate of a fluid is given as,
Q = Volumetric flow rate of the fluid
u = Fluid velocity
S = Total cross-section area
The mass flow rate is given as,
The average velocity if the fluid is,
The momentum of the fluid is the product of the mass flow rate and the velocity of the fluid.
M = Momentum of the fluid
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Only focus on H(3), which the answer is minus 1.26 KJ/mol. This also has the ideal gas of nitrogen gas N2. Two enthalpies need to be calculated for this. The first enthalpy is H = (specific volume) times (pressure difference). For the specific volume of nitrogen, how was 12.089 x10^(-5) m^3/mol obtained? I understand the second enthalpy for the heat capacity for nitrogen gas.
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Unit Operations of Chemical Engineering
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