(a) The mass flow rate, for each velocity, is given by \dot{M}_{f}=\int_{A_{u}}{\rho _{f}u_{f}dA}=\rho _{f} \left\langle u\right\rangle _{f}A_{u}=A_{u}\dot{m}_{f} as
\dot{M}_{f}=A_{u}\rho _{f} \left\langle u_{f}\right\rangle
where for a circular tube
A_{u}=\pi \frac{D^{2}}{4}
From Table, at T = 330 K, we have
\rho_{f} = 986.8 kg/m^{3} Table
c_{p,f} = 4,183 J/kg-K Table
ν_{f} = 505 × 10^{−9} m^{2}/s Table
k_{f} = 0.648 W/m-K Table
Pr = 3.22 Table
For \left\langle u_{f} \right\rangle = 0.08 m/s
\dot{M}_{f}=\frac{\pi D^{2}}{4}\rho _{f}\left\langle u_{f}\right\rangle
=\frac{\pi× (10^{−2})^{2} (m)^{2}}{4}× 986.8(kg/m^{3} ) × 0.08(m/s) = 6.201 × 10^{−3} kg/s
For \left\langle u_{f} \right\rangle = 0.5 m/s
\dot{M}_{f}=\frac{\pi× (10^{−2})^{2}}{4}× 986.4 × 0.5 = 3.875 × 10^{−2} kg/s
(b) The Nusselt number \left\langle Nu\right\rangle _{D} depends on the Reynolds number Re_{D}. For circular tubes, the correlations are listed in Table and Table. The Reynolds number is given by Re_{D}=\frac{\left\langle u_{f}\right\rangle D}{v_{f}} ,Pr=\frac{v_{f}}{\alpha _{f}} , i.e.,
Re_{D}=\frac{\left\langle u_{f}\right\rangle D}{v_{f}}
For \left\langle u_{f} \right\rangle = 0.08 m/s we have
Re_{D}=\frac{0.08(m/s) × 10^{−2} (m)}{5.05 × 10^{−7} (m^{2}/s)}
= 1,584 < Re_{D,t} = 2,300 laminar flow regime.
For \left\langle u_{f} \right\rangle = 0.5 m/s
Re_{D}=\frac{0.5(m/s) × 10^{−2} (m)}{5.05 × 10^{−7}(m^{2}/s)}
= 9,901 > Re_{D,t} = 2,300 turbulent flow.
For \left\langle u_{f} \right\rangle = 0.08 m/s, and under uniform temperature Ts and fully developed conditions, we have from Table
Re_{D} = 1,584 < Re_{D,t} = 2,300, \left\langle Nu\right\rangle _{D} = 3.66 Table
For \left\langle u_{f} \right\rangle = 0.5 m/s, and under fully developed conditions, we have from Table
Re_{D} = 9,901 > Re_{D,t} = 2,300, \left\langle Nu\right\rangle _{D} = 0.023Re^{4/5}_{D} Pr^{n} Table
Here we have T_{s} > \left\langle T_{f} \right\rangle and from Table , n = 0.4. Then
\left\langle Nu\right\rangle _{D} = 0.023 × (9,901)^{4/5} (3.22)^{0.4} = 57.73.
Note the larger Nusselt number associated with turbulent flows.
(c) The effectiveness for the heat exchange between a bounded fluid and its
constant-temperature bounding surface is given by \epsilon _{h,e}\equiv \frac{\left\langle T_{f}\right\rangle _{L}-\left\langle T_{f}\right\rangle _{0}}{T_{s}-\left\langle T_{f}\right\rangle _{0}} =\frac{\left\langle T_{f}\right\rangle _{0}-\left\langle T_{f}\right\rangle _{L}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} =1-e^{-NTU} as
\epsilon _{h,e} =\frac{\left\langle T_{f}\right\rangle _{0}-\left\langle T_{f}\right\rangle _{L}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} =1-e^{-NTU}
Here we are given \epsilon _{he} = 0.5, or
0.5 = 1 − e^{−NTU}
e^{−NTU }= 0.5
NTU = − ln 0.5 = 0.6931.
From NTU\equiv \frac{R_{u,f}}{\left\langle R_{ku}\right\rangle _{D}} =\frac{1}{\left\langle R_{ku}\right\rangle_{D} (\dot{M}c_{p})_{f}}=\frac{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}/D}{(\dot{M}c_{p})_{f}} , the definition of NTU is
NTU=\frac{1}{(\dot{M}c_{p})_{f}\left\langle R_{ku}\right\rangle_{D} }=\frac{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}}{D(\dot{M}c_{p})_{f}}
where
A_{ku} =πDL.
Then
NTU=\frac{\pi Lk_{f}\left\langle Nu\right\rangle _{D}}{\dot{M}_{f}c_{p.f}}
Solving for L, we have
L=\frac{NTU\dot{M}_{f}c_{p,f}}{\pi k_{f}\left\langle Nu\right\rangle _{D}} required length.
For \left\langle u_{f} \right\rangle = 0.08 m/s
L=\frac{0.6931 × 4,183(J/kg-K) × 6.201 × 10^{−3} (kg/s)}{π × 0.648(W/m-K) × 3.66} = 2.413 m
For \left\langle u_{f} \right\rangle = 0.5 m/s
L=\frac{0.6931 × 4,183 × 3.875 × 10^{−2}}{π × 0.648 × 57.73} = 0.9560 m
Note the shorter length required for the turbulent flow.
