Question 19.6: If the vessels in Example 19.5 were maintained isobaric at a...
If the vessels in Example 19.5 were maintained isobaric at a mean temperature of 30.0°\text{C} and the measured thermomechanical heat transfer rate was 8.70 \text{J/s}, then find the induced isobaric mass flow rate and the resulting temperature difference between the vessels.
From Eq. (19.59), we have
\dot{m}_i = \frac{ρ\dot{Q}}{Tk_t/k_o + μk_o/k_p}
where the values of μ, ko, and kp are the same as in Example 19.5; and for saturated liquid water at 30.0° \text{C}, we have k_t = 0.610 \text{W/(K.m)}. Then, Eq. (19.59) gives
\dot{m} = \frac{(996 \text{kg/m}^3) (8.70 \text{J/s})}{\frac{(303 \text{K})[0.610 \text{J/(s.K.m)}]}{1.91 \text{m}^2/\text{s}} + \frac{[891 × 10^{−6} \text{kg/(s.m)}] (1.91 \text{m}^2/\text{s})}{1.00 × 10^{−12} \text{m}^2}} = 5.10 × 10^{−6} \text{kg/s}
Then,
\frac{dT}{dx} = – \frac{T}{ρk_o} J_M \mid_{p = \text{constant}} = – \frac{T\dot{m}}{ρk_o} = – \frac{(303 \text{K}) (5.10 × 10^{−6} \text{kg/s})}{(996 \text{kg/m}^3) (1.91 \text{m}^2/\text{s})} = −8.11 × 10^{−7} \text{K/m}
and so
dT = ΔT = (−8.11 × 10^{−7} \text{K/m}) (0.100 \text{m}) = −8.11 × 10^{−8} \text{K}