Consider a constant temperature bath that is in the form of a cylindrical vessel of 0.1m diameter and 0.25m height. The vessel contains water whose density and specific heat may be taken as 1000 kg/m^3 and 4200 J/kg◦C respectively. The vessel is fairly well insulated and hence has an overall heat t

Question

Consider a constant temperature bath that is in the form of a cylindrical vessel of 0.1m diameter and 0.25m height. The vessel contains water whose density and specific heat may be taken as [latex]1000 kg/m^{3}[/latex] and 4200 J/kg°C respectively. The vessel is fairly well insulated and hence has an overall heat transfer coefficient of [latex]5W/m^{2}°C[/latex] and interacts with an environment at 25 °C. The vessel is equipped with a small electric heater that provides 200W when it is “on”. The bath has an on-off controller that is adjusted to switch on when the temperature reduces to 53 °C and switches off when the temperature goes up to 57 °C. The bath is expected to have a set point temperature of 55 °C. Perform a simulation of the system.

Accepted Answer

Step 1 The required background for the simulation has been presented above. The given data is written down first.

System related data:

Diameter of cylindrical vessel: [latex]D = 0.1 m[/latex]
Height of cylindrical vessel: [latex]H = 0.25 m[/latex]
Density of water: [latex]ρ = 1000 kg/m^{3}[/latex]
Specific heat of water: [latex]c = 4200 J/kg°C[/latex]
Heater power: [latex]Q = 200 W[/latex]

Environment related data:

Ambient temperature: [latex]T_{∞} = 25 °C[/latex]
Heat transfer coefficient: [latex]h = 5 W/m°C[/latex]

Controller related data:

Set point temperature: [latex]T_{sp} = 55 °C[/latex]
Heater “on” temperature: [latex]T_{ lp} = 53 °C[/latex]
Heater “off” temperature: [latex]T_{hp} = 57 °C[/latex]

The volume of the bath is calculated as

[latex]V=\frac{\pi D^{2}H }{4}=\frac{\pi imes 0.1^{2} imes 0.25 }{4} =0.00196m^{3}[/latex]

The surface area of the bath interacting with the environment is

[latex]S=\pi DH + 2 \frac{\pi D^{2} }{4}=\pi imes 0.1 imes 0.25+2 imes \frac{\pi imes 0.1^{2} }{4} =0.0942 m^{2}[/latex]

The time constant of the bath treated as a first-order system is

[latex] au =\frac{ ho Vc}{hS}=\frac{1000 imes 0.00196 imes 4200}{5 imes 0.0942} =17500s[/latex]

The internal heat generation rate is calculated assuming that the heater output is uniformly released throughout the volume of the bath.

[latex]G=\frac{Q}{V} =\frac{200}{0.00196} =101859W/m^{3}[/latex]

Step 2 The simulation is done in two parts. In the first part, the bath starts from cold and the temperature is brought up to the operating temperature. For this part, the initial temperature is 25 °C and the heater is switched on at t = 0. The final temperature attained is 57 °C when the controller will switch off the heater. The bath temperature follows the first of relation 3.19.

[latex]Cold start : heta =\frac{G au }{ ho c} \left[1-e^{-\frac{t}{ au } } ight][/latex]

[latex]Subsequent operation : heta =(T_{lp}-T_{\infty } )e^{-\frac{t-t_{0} }{ au }}+\frac{G au }{ ho c} \left[1-e^{-\frac{t-t_{0}}{ au } } ight][/latex] (3.19)

[latex]T=25+\frac{101859 imes 17500}{1000 imes 4200}\left(1-e^{\frac{-t}{17500} } ight) =25+424.41 imes \left(1-e^{\frac{-t}{17500} } ight)[/latex]

Temperature reaches [latex]T_{hp}[/latex] at t = 1372 s at which the heater is switched off by the controller. The cooling of the bath follows the Eq. 3.21 with the origin for time at 1372 s.

[latex] heta =(T_{hp}-T_{\infty })e^{-\frac{t-t_{hp}}{ au } }[/latex] (3.21)

[latex]T=25+32e^{-\frac{t-1372}{14500 } }[/latex]

The temperature reaches a value of 53°C at t = 3714 s. Thus the very first on-off cycle from cold start is of duration equal to 3714 s.

Step 3 Second and subsequent on-off cycles are now considered. During the on period the temperature variation with time is given by the second of Eq. 3.19 with [latex]t_{0}= 3714 s[/latex] . Hence we have

[latex]T=25+(53-25)e^{-\frac{t-t_{0} }{17500} }+\frac{101859 imes 17500}{1000 imes 4200} \left[1-e^{-\frac{t-t_{0}}{17500 } } ight][/latex]

[latex]=25+28e^{-\frac{t-3714 }{17500} }+424.4 \left[1-e^{-\frac{t-3714}{17500 } } ight][/latex]

Calculations show that the temperature reaches the upper limit of 57 °C at [latex]t_{hp}[/latex] = 3891.5 s. Subsequent cooling process is governed by Eq. 3.21 which translates to

[latex] heta =(T_{hp}-T_{\infty })e^{-\frac{t-t_{hp}}{ au } }[/latex]

or

[latex]T=25+(57-25)e^{-\frac{(t-3891.5)}{17500 } }=25+32e^{-\frac{(t-3891.5)}{17500 } }[/latex]

Calculations show that the temperature reduces to 53 °C at [latex]t_{lp}[/latex] = 6161.5 s. Thus one on-off cycle takes 6161.5 − 3714 = 2447.5 s. The temperature variation of the bath during one on-off cycle is shown in Fig. 3.9. The duration of an on-off cycle is decided by the choice of the temperatures [latex]T_{lp}[/latex] and [latex]T_{hp}[/latex]. If these are closer to the set point, the cycle time reduces. The on time is also determined by the heater dissipation. Larger the Q smaller the “on” time. If the range [latex]T_{lp}[/latex], [latex]T_{hp}[/latex] is chosen very small the on-off periods will reduce and in the limit the switch may go in to rapid flutter!

Question 3.5: Consider a constant temperature bath that is in the form of ...

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