Modern Engineering Thermodynamics

19.5 Thermomechanical Coupling 781
Solution
a. For an isothermal system, dT = 0 and Eq. (19.50) gives
J M =
_m A = − ρk p
μ
dp
dx
so
_m = − ρAk p
μ
dp
dx
For saturated liquid water at 30.0°C, ρ = 996 kg/m 3 and μ = 891 × 10 −6 kg/(s·m). Then,
_m = ð996 kg/m3 Þðπ/4Þð0:0100 mÞ 2 ð10 −12 Þ
891 × 10 −6 kg/ðs.mÞ
− 1:00 × 104 N/m 2
0:100 m
(Note that the mass flow rate is in the same direction as the pressure drop.)
b. We also have
so that
c. From Eq. (19.56), we have
J T
Q
Q = constant
= _
i
A = −k dp
o
dx
= −8:78 × 10 −6 kg/s
k o = − _ Q /A
ðdp/dxÞ = − ð15:0 J/sÞ/½ðπ/4Þð0:0100 mÞ2 Š
ð−1:00 × 10 4 N/m 2 Þ/ð0:100 mÞ = 1:91 m2 /s
Q
_S i = _ i
T = 15:0 J/s
ð273:15 + 30:0KÞ = 0:0500 J/ðs .KÞ
EXAMPLE 19.6
If the vessels in Example 19.5 were maintained isobaric at a mean temperature of 30.0°C and the measured thermomechanical
heat transfer rate was 8.70 J/s, then find the induced isobaric mass flow rate and the resulting temperature difference
between the vessels.
Solution
From Eq. (19.59), we have
ρ _Q
_m i =
Tk t /k o + μk o /k p
where the values of μ, k o ,andk p are the same as in Example 19.5; and for saturated liquid water at 30.0°C, we have
k t = 0.610 W/(K·m). Then, Eq. (19.59) gives
Then,
and so
_m =
ð303 KÞ½0:610 J/ðs.K.mÞŠ
1:91 m 2 /s
ð996 kg=m 3 Þð8:70 J=sÞ
+ ½891 × 10−6 kg/ðs.mÞŠð1:91 m 2 /sÞ
1:00 × 10 −12 m 2 = 5:10 × 10 −6 kg=s
dT
dx = − T
J p M
ρk = constant
= − T _m = − ð303 KÞð5:10 × 10−6 kg/sÞ
o ρk o ð996 kg/m 3 Þð1:91 m 2 /sÞ = −8:11 × 10−7 K/m
dT = ΔT = ð−8:11 × 10 −7 K=mÞð0:100 mÞ = −8:11 × 10 −8 K