Question 19.5: Both of the large vessels shown in Figure 19.10 are filled w...

a. For an isothermal system, dT = 0 and Eq. (19.50) gives

J_M = −\frac{ρk_o}{T} (\frac{dT}{dx}) − \frac{ρk_p}{μ} (\frac{dp}{dx})            (19.50)

J_M = \frac{\dot{m}}{A} = – \frac{ρk_p}{μ} (\frac{dp}{dx})

so

\dot{m} = – \frac{ρAk_p}{μ} (\frac{dp}{dx})

For saturated liquid water at 30.0°\text{C}, ρ = 996  \text{kg/m}^3 and μ = 891 × 10^{−6}  \text{kg/(s.m)}. Then,

\dot{m} = \frac{(996  \text{kg/m}^3) (π/4) (0.0100  \text{m})^2 (10^{−12})}{891× 10^{−6}  \text{kg/(s.m)}} (- \frac{1.00 × 10^4  \text{N/m}^2}{0.100  \text{m}}) = − 8.78 × 10^{−6}  \text{kg/s}

(Note that the mass flow rate is in the same direction as the pressure drop.)

b. We also have

J_Q \mid_{T = \text{constant}} = \frac{\dot{Q}_i}{A} =  − k_o (\frac{dp}{dx })

so that

k_o = – \frac{\dot{Q} / A}{(dp/dx)} = – \frac{(15.0  \text{J/s})/[(π/4) (0.0100  \text{m})^2]}{(−1.00 × 10^4  \text{N/m}^2)/ (0.100  \text{m})} = 1.91  \text{m}^2/\text{s}

c. From Eq. (19.56), we have

(\frac{dp}{dT}) \mid_{J_M =0} = – \frac{ρ\dot{Q}_i}{T\dot{m}} = – \frac{ρ\dot{S}_i}{\dot{m}}            (19.56)

\dot{S}_i = \frac{\dot{Q}_i}{T} = \frac{15.0  \text{J/s}}{(273.15 + 30.0  \text{K})} =  0.0500  \text{J/(s.K)}