Heating element of a water heater is idealized as a 0.1 × 0.1m square vertical plate. The plate temperature is 27 °C while the water temperature is 15 °C. Determine the amount of heat transferred using the results of the integral method. Properties are to be taken at the mean of plate and ambient

Question

Heating element of a water heater is idealized as a 0.1 × 0.1m square vertical plate. The plate temperature is [latex]27^{\circ} C[/latex] while the water temperature is [latex]15^{\circ} C[/latex]. Determine the amount of heat transferred using the results of the integral method. Properties are to be taken at the mean of plate and ambient temperatures.

Accepted Answer

The given data, in the notation used in the above section is written down first.

Heater plate temperature: [latex]T_{w}=27^{\circ} C[/latex]

Ambient water temperature: [latex]T_{\infty}=15^{\circ} C[/latex]

Plate height: L = 0.1m

Plate width: W = 0.1m

Mean temperature for calculating water propertis is

[latex]T_{a}=\frac{T_{w}+T_{\infty}}{2}=\frac{27+15}{2}=21^{\circ} C[/latex]

Temperature difference for calculating the Grashof/Rayleigh number is

[latex]\Delta T=T_{w}-T_{\infty}=27-15=12^{\circ} C[/latex]

Properties of water at [latex]21^{\circ} C[/latex] are taken from from table of properties

[latex]k_{a}=0.597 W / m ^{\circ} C[/latex] , [latex]P r_{a}=7.0, \frac{g \beta_{a}}{v_{a}^{2}}=2.035 imes 10^{9} ext { (SI units) }[/latex]

The Rayleigh number based on plate height is calculated as

[latex]R a_{L}=\frac{g \beta_{a}}{v_{a}^{2}} imes L^{3} imes \Delta T imes P r_{a}=2.035 imes 10^{9} imes 0.1^{3} imes 12 imes 7=1.71 imes 10^{8}[/latex]

Flow is laminar since the Rayleigh number is less than [latex]10^{9}[/latex]. Eq.16.34 based on the integral solution gives

[latex]\overline{N u}_{L}=\frac{4}{3} N u_{L}=\frac{4}{3} \cdot 0.509\left[\frac{P r}{P r+0.952} ight]^{\frac{1}{4}} R a_{L}^{\frac{1}{4}}=0.677\left[\frac{P r}{P r+0.952} ight]^{\frac{1}{4}} R a_{L}^{\frac{1}{4}}[/latex] (16.34)

[latex]\overline{N u}_{L}=0.677\left[\frac{7}{7+0.952} ight]^{\frac{1}{4}}\left(1.71 imes 10^{8} ight)^{\frac{1}{4}}=75[/latex]

The average heat transfer coefficient may then be calculated as

[latex]\bar{h}_{L}=\frac{\overline{N u}_{L} k_{a}}{L}=\frac{75 imes 0.597}{0.1}=447.8 W / m ^{2 \circ} C[/latex]

Assuming that heat transfer takes place from both sides of the plate, the total heat transfer is given by

[latex]Q_{L}=2 L W \bar{h}_{L} \Delta T=2 imes 0.1 imes 0.1 imes 447.8 imes 12=107.5 W[/latex]

Question 16.3: Heating element of a water heater is idealized as a 0.1 × 0....

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