A thermocouple is immersed in a liquid contained in a cylinder of inside diameter of 3 mm and an outside diameter of 3.5 mm. The cylinder material is copper with a specific heat of 383 J/kg^◦C and density 8900 kg/m^3. The heat transfer coefficient on the outer and inner surfaces, respectively, of t

Question

A thermocouple is immersed in a liquid contained in a cylinder of inside diameter of 3 mm and an outside diameter of 3.5 mm. The cylinder material is copper with a specific heat of 383 J/kg°C and density [latex]8900 kg/m^{3} [/latex]. The heat transfer coefficient on the outer and inner surfaces, respectively, of the cylinder are 15 and [latex]45 W/m^{2}°C[/latex]. The density and specific heat of the liquid are known to be [latex]1040 kg/m^{3} and 3650 J/kg°C[/latex], respectively. The effective length of the cylinder is 10 mm. Obtain the step response of the thermocouple treating it as a second-order system. Make a plot of this response using suitable non-dimensional coordinates.

Accepted Answer

Step 1 We follow the notation used in the text and specify the given data.

Outside heat transfer coefficients: [latex]h_{1} = 15 W/m^{2} °C[/latex]
Inside heat transfer coefficients: [latex]h_{2} = 45 W/m^{2} °C[/latex]
Outer diameter of cylinder: [latex]D_{1} = 0.0035 m[/latex]
Inner diameter of cylinder: [latex]D_{2} = 0.003 m[/latex]
Effective cylinder length: [latex]L = 0.01 m[/latex]
Density of cylinder material: [latex]ρ_{1} = 8900 kg/m^{3}[/latex]
Specific heat of cylinder material: [latex]c_{1} = 383 J/kg°C[/latex]
Density of liquid: [latex]ρ_{2} = 1040 kg/m^{3}[/latex]
Specific heat of liquid: [latex]c_{2} = 3650 J/kg°C[/latex]

Heat transfer areas are calculated as

[latex]S_{1} = πD_{1}L = π × 0.0035 × 0.01 = 1.1 × 10^{-4} m^{2}[/latex]

[latex]S_{2} = πD_{1}L = π × 0.003 × 0.01 = 9.425 × 10^{-5} m^{2}[/latex]

Volume of cylinder material is

[latex]V_{1} =\frac{\pi (D^{2}_{1}-D^{2}_{2})L}{4} = \frac{\pi imes (0.0035^{2}-0.003^{2} ) imes 0.01}{4} =2.553 imes 10^{-8} m^{3}[/latex]

Volume of the liquid is calculated as

[latex]V_{2} =\frac{\pi D^{2}_{2}L}{4} = \frac{\pi imes 0.003^{2} imes 0.01}{4} =7.069 imes 10^{-8} m^{3}[/latex]

Step 2 Characteristic times:
Characteristic time [latex]τ_{1}[/latex] is calculated as

[latex] au _{1}=\frac{ ho _{1}V_{1}c_{1}}{h_{1}S_{1}} =\frac{8900 imes 2.553 imes 10^{-8} imes 383}{15 imes 1.1 imes 10^{-4} }=52.75s[/latex]

Characteristic time [latex]τ_{2}[/latex] is

[latex] au _{2}=\frac{ ho _{2}V_{2}c_{2}}{h_{2}S_{2}} =\frac{1040 imes 7.069 imes 10^{-8} imes 3650}{45 imes 9.425 imes 10^{-5} }=63.27s[/latex]

Characteristic time [latex]τ_{3}[/latex] is

[latex] au _{3}=\frac{ ho _{1}V_{1}c_{1}}{h_{2}S_{2}} =\frac{8900 imes 2.553 imes 10^{-8} imes 383}{45 imes 9.425 imes 10^{-5} } =20.52s[/latex]

With these, the two parameters that govern the problem are

[latex] au _{p}= au _{1} au _{2}=52.75 imes 63.27=3337.5s^{2}[/latex]

[latex] au _{s}=\left[\frac{1}{ au _{1}}+\frac{1}{ au _{2}} +\frac{1}{ au _{3}} ight] ^{-1} = \left[\frac{1}{52.75}+\frac{1}{63.27} +\frac{1}{20.52} ight] ^{-1} =11.98s[/latex]

Step 3 Solution:
Using Eq. 3.32, we obtain the exponents [latex]m_{1}[/latex] and [latex]m_{2}[/latex].

[latex]m_{1} =-\frac{\frac{1}{ au _{s}}+\sqrt{\frac{1}{ au _{s}^{2} }-\frac{4}{ au _{p}}} }{2} , m_{2} =-\frac{\frac{1}{ au _{s}}-\sqrt{\frac{1}{ au _{s}^{2} }-\frac{4}{ au _{p}}} }{2}[/latex] (3.32)

[latex]m_{1} =-\frac{\frac{1}{11.98}+\sqrt{\frac{1}{11.98^{2} }-\frac{4}{3338}} }{2}=-0.0797s^{-1}[/latex]

[latex]m_{2} =-\frac{\frac{1}{11.98}-\sqrt{\frac{1}{11.98^{2} }-\frac{4}{3338}} }{2}=-0.00376s^{-1}[/latex]

The constants of integration are obtained now. Using Eq. 3.36

[latex]A=-\frac{ heta _{0} }{2} imes \frac{\frac{1}{ au _{s}}-\sqrt{\frac{1}{ au _{s}^{2} }-\frac{4}{ au _{p}}} }{\sqrt{\frac{1}{ au _{s}^{2} }-\frac{4}{ au _{p}}} }[/latex] (3.36)

[latex]A=-\frac{ heta _{0} }{2} \frac{\frac{1}{11.98}-\sqrt{\frac{1}{11.98^{2} }-\frac{4}{3338}} }{\sqrt{\frac{1}{11.98^{2} }-\frac{4}{3338}} }=1.0495 heta _{0} , and B= heta _{0}-A=-0.0495 heta _{0}[/latex]

Introduce the following non-dimensional variables

[latex]\phi =\frac{ heta }{ heta _{0}} , au =\frac{t}{ au _{2}}[/latex]

to obtain the response as

[latex]\phi =1.0495e^{-0.0797t} -0.0495e^{-0.00376t}[/latex]

Step 4 Plot of response of second-order system: The response of the secondorder system shown in Fig. 3.11 uses the above non-dimensional variables along the two axes.

Comments

The second-order response is also compared with the first-order response with time constant equal to [latex] au _{2}[/latex]. The second-order system response is sluggish as compared to the first-order system. There is a time delay due to the wall that has to respond before the liquid starts responding to the input.

The two roots of the characteristic equations may be interpreted as two reciprocal time constants. One of them is labeled [latex] au _{h}[/latex] and the other is labeled [latex] au _{l}[/latex]. These may be obtained as

[latex] au _{h} =\frac{1}{0.00376} =266.2s[/latex]

[latex] au _{l} =\frac{1}{0.0797} =12.54s[/latex]

It is interesting to note that,when [latex]h_{2} \gg h_{1}[/latex], the smaller of the two time constants tends to 0 and the larger of the two time constants tends to ∞. The response of the system becomes a first-order response with the bigger time constant. To demonstrate this, we redo Example 3.6 with [latex]h_{1} = 5W/m^{2}°C[/latex] and [latex]h_{2} = 100 W/m^{2}°C[/latex]. The coefficients now become [latex]A = −0.0102[/latex] and [latex]B = 1.0102[/latex]. Also the two time constants become [latex]τ_{l} = 6.75 s[/latex] and [latex]τ_{h} = 668 s[/latex]. The temperature response becomes

[latex]\phi =1.0102e^{-\frac{t}{668} } -0.0495e^{\frac{t}{6.75} }[/latex]

Since the second term on the right-hand side is very small compared to the first term, the response is basically that of a first-order one with a time constant of 668 s.

Question 3.6: A thermocouple is immersed in a liquid contained in a cylind...

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