Engine oil is cooled by flowing through a tube at a rate of Mf = 0.05 kg/s with an inlet temperature 0 = 80^◦C. The surface of the tube is cooled and maintained at a constant temperature of Ts = 30^◦C (by, e.g., removing the heat from this surface to another fluid). This is rendered in Figure

Question

Engine oil is cooled by flowing through a tube at a rate of [latex]\dot{M}_{f} = 0.05 kg/s[/latex] with an inlet temperature [latex] \left\langle T_{f}\right\rangle _{0}= 80^{\circ }C[/latex]. The surface of the tube is cooled and maintained at a constant temperature of [latex]T_{s} = 30^{\circ }C[/latex] (by, e.g., removing the heat from this surface to another fluid). This is rendered in Figure (a).
The tube has a diameter D = 2 cm, and a length L = 20 m (this can be a coiled tube instead of a straight tube). The Nusselt number is [latex]\left\langle Nu\right\rangle _{D}= 3.66[/latex].

(a) Draw the thermal circuit diagram and show the expected fluid temperature distribution.
(b) Determine the number of transfer units NTU.

(c) Determine the effectiveness [latex]\epsilon _{he}[/latex].
(d) Determine the fluid exit temperature [latex] \left\langle T_{f}\right\rangle _{L}[/latex].
(e) Determine the average convection resistance [latex] \left\langle R_{u}\right\rangle _{L}[/latex].
(f) Determine the average convection heat transfer rate [latex] \left\langle Q_{u}\right\rangle _{L-0}[/latex].
Evaluate the oil thermophysical properties from Table and at T = 330 K.

Accepted Answer

(a) The thermal circuit diagram is shown in Figure (b). The anticipated decay of the fluid temperature is shown in Figure (c).
(b) The definition of the number of thermal units NTU is given by [latex]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}} [/latex], i.e.,

[latex]NTU=\frac{A_{ku}\left\langle Nu\right\rangle _{D}k_{f}/D}{(\dot{M}c_{p})_{f}} [/latex]

where
[latex]A_{ku} = πDL [/latex].
Then

[latex] NTU=\frac{\pi L\left\langle Nu\right\rangle _{D}k_{f}}{\dot{M}_{f}c_{p,f}} [/latex]

From Table C.23 and at T = 330 K, we have

[latex]c_{p,f} = 2,040 J/kg-K [/latex] Table
[latex]k_{f} = 0.14 W/m-K [/latex] Table .
Using these, we have

[latex] NTU=\frac{π × 20(m) × 3.66 × 0.14(W/m-K)}{0.05(kg/s) × 2,040(J/kg-K)} = 0.3156. [/latex]

(c) The heat exchanger effectiveness [latex]\epsilon _{he}[/latex] is defined by [latex]\epsilon _{he}\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}[/latex], i.e.,

[latex]\epsilon _{he}=1-e^{-NTU}= 1 − e^{−0.3156} = 0.2707[/latex].

(d) The fluid exit temperature [latex] \left\langle T_{f}\right\rangle _{L}[/latex] is found from the expression for[latex]\epsilon _{he}[/latex], i.e., [latex]\epsilon _{he}\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}[/latex],

[latex]\epsilon _{he}=\frac{\left\langle T_{f}\right\rangle _{0}-\left\langle T_{f}\right\rangle _{L}}{\left\langle T_{f}\right\rangle _{0}-T_{s}} = \frac{\left\langle T_{f}\right\rangle _{L}-\left\langle T_{f}\right\rangle _{0}}{T_{s}-\left\langle T_{f}\right\rangle _{0}}[/latex]

Solving for [latex]\left\langle T_{f}\right\rangle _{L}[/latex], we have

[latex]\left\langle T_{f}\right\rangle _{L}=\left\langle T_{f}\right\rangle _{0}-\epsilon _{he}(\left\langle T_{f}\right\rangle _{0}-T_{s})[/latex]

[latex] = 80(^{\circ } C) − 0.2707(80 − 30)(^{\circ } C) = 66.47^{\circ }C. [/latex]

(e) The average convection resistance [latex] \left\langle R_{u}\right\rangle _{L}[/latex] is given by [latex]\left\langle R_{u}\right\rangle _{L}=\frac{T_{s}-\left\langle T_{f}\right\rangle _{0}}{(\dot{M}c_{p})_{f}(\left\langle T_{f}\right\rangle _{L}-\left\langle T_{f}\right\rangle _{0})} =\frac{1}{(\dot{M}c_{p})_{f}\epsilon _{he}}=\frac{1}{(\dot{M}c_{p})_{f}(1-e^{-NTU})} [/latex], i.e.,

[latex]\left\langle R_{u}\right\rangle _{L}=\frac{1}{(\dot{M}c_{p})_{f}(1-e^{-NTU})} [/latex]

[latex] =\frac{1}{(0.05)(kg/s) × 2,040(J/kg-^{\circ }C)(1 − e^{−0.3156})} = 0.03622^{\circ }C/W. [/latex]

(f) The surface-averaged heat transfer rate [latex]\left\langle Q_{u}\right\rangle _{L-0}=\left\langle Q_{ku}\right\rangle _{L}[/latex] is given by

[latex]\left\langle Q_{u}\right\rangle _{L-0}= \frac{T_{s}-\left\langle T_{f}\right\rangle _{0}}{\left\langle R_{u}\right\rangle _{L}}[/latex]

[latex] = \frac{(30 − 80)(^{\circ }C)}{0.03622(^{\circ } C/W)} = −1.380 × 10^{3} W. [/latex]

Question 7.1: Engine oil is cooled by flowing through a tube at a rate of ...

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