Problems of Chapter 4

Problem 4.1

The residence time is tres = 3.33-10-5 s and the measurement time tmeas = tref « 1.67-10-4 s.

Problem 4.2

pSuth = 2.67-10-5 kg/ms, р1 = 3.51-10-5 kg/ms, p2 = 2.64-10-5 kg/ms.

Дрі = (p1 — (ASuth)/^Suth = 0.31, Др2 = (м2 — I^Suth)/^Suth = —0.01.

Problem 4.3

kHan = 0.036 W/mK, k1 = 0.048 W/mK, k2 = 0.037 W/mK. Дк1 =

(k1 kHan)/kHan 0.33, Дк2 (k2 kHan)/kHan 0.03. kEucken °.°37

W/mK. ДкHan (kHan kEucken)/kEucken 0.027.

Problem 4.4

Cp = 1,032.71 m2 /s2, K, у = 1.385, PrHan = 0.7576, PrEucken = 0.7421, Preq. (4.19) = 0.7418.

Problem 4.5

On page 99 it was mentioned, that in that case fluid mechanics and thermo­dynamics are decoupled. With zero Mach number the Eckert number E is zero, too. The compression work and the dissipation work terms disappear from the energy equation.

However, because the velocity v is non-zero, the speed of sound a = J~fRT must be very large if M = 0:

v v

M = — : а м-гО = yrj———— t °°-

a M |^o

Assuming that the gas properties are unchanged, this means that the tem­perature is infinitely high. Obviously fluid mechanics and thermodynamics are decoupled.