Measurement of forced and free convection heat transfer coefficients - Lab Report Example

FORCED AND FREE CONVECTION HEAT TRANSFER The of the The of the School The and State where it is locatedThe DateResultsForced and Free Convection Heat Transfer1) Relevant Physical PropertiesFor the disc d = 75 mm (0.075 m), L = 6 mm (0.006 m)For aluminum, Density (ℓ) = 2702 kg/m3Specific Heat, Cp = 903 J/kgKi. Area, A = (Pi)d2/4 = ((22/7)*0.0752)/4 m2 = 0.004418 m2ii. Mass, m = ℓ A L = (2702*0.004418*0.006) = 0.

0716 Kg iii. Length scale for the disc, x, = exposed surface area/perimeter = (Pi)*(d) = (22/7)*75 = 235.62 mm = 0.236 miv. Jet pipe diameter, D = 34 mmv. Distance from jet exit to plate = 68 mm2) Estimate Biot Number, Bi = hL/kL = volume/exposed and uninsulated surface area = 6 mm, k = 185 W/mKFrom Bi = hL/k,Bi = 0.006 h/185 = 3*105 h (i.e., less than unity for all physically possible values of h.)From Loge Θ = - (h A /m Cp) t,Therefore, h = - ((Loge Θ)/t)* m Cp)/ATaking an instance at t = 10, from the table below, therefore:i.

Loge Θ = Loge 0.95 = - 0.05ii. M = 0.0716 Kgiii. Cp = 1006 J/kgKiv. A = 0.004418 m2h = - ((- 0.05)/10)* 0.0716*1006)/0.004418h = 81.53Therefore, Bi = 3*105 *81.53 = 244590003) Choice of Value of nEstimate GrPr, GrPr = (gβ△Tx3/V2) (η Cp/k) where the characteristic length x, is taken as the value of surface area /perimeter and △T is the surface to air temperature difference at the beginning of the free convection experiment. It is convention to evaluate all properties at the mean film temperature, T film that is-T film = (Ts, i + T∞, i) / 2.

The relevant physical properties are, for T film in Kelvin1. Cp = 1006 J/kgK2. T film = (Ts, i + T∞, i) / 2 = (54.9 + 24.3)/2 = (39.6 + 273) = 312.6 K3. k* = 0.0034 + 7.58 *10-5 T film = 0.0034 + 7.58 *10-5 * 312.6 W/mK = 0.0271 W/mK4. η* = 1.46 * 10-6 (T film)1.5 / (110 + T film) = 1.46 * 10-6 (312.6)1.5/(110 + 312.6) kg/ms = 1.91*10-5 Kg/ms5. ℓ = P /R T film = 105 / 287 * T film = 105/(287 * 312.6) = 1.11 Kg/m36. V = η/ℓ = (1.91*10-5)/1.

11 = 1.721 * 10-5 m2/s7. β = 1/ T film = 1 / 312.6= 0.0032 K-18. △T = (Ts, i - T∞, i) = (54.9 – 24.3) = 30.6 KBut, GrPr = GrPr = (gβ△Tx3/V2) (η Cp/k)Therefore, GrPr = ((9.8*0.0032*30.6*0.2363*10-6)/ (1.721 * 10-5)2) (1.91*10-5*1006/0.0271) = 30. 1949, which is less than the transition value of 107, so n = 1/4.4.) Forced Convection Measurements (Logarithmic Graph)From the Equation for Conservation of Energy on a solid, (m Cp d (Ts -T∞)/ dt = - h A (Ts - T∞))……………………………………………………………………. (i)Where:I.

A = Area II. M = Mass III. Cp = Specific Heat,Loge Θ = - (h A /m Cp) t……………………………………………….. (ii), and Θ = (Ts - T∞)/ (Ts, i - T∞, i)…………………………………….. (iii), and, i, denotes conditions at initial time, t.From Equation (iii), we can get the values of Θ and fill the table below as shown: Time, t,( SECONDS)Jet Temp., T∞ oCDisc Temp. Ts oCΘ0561.524.11.001061.724.90.981561.625.10.982061.626.30.943061.727.70.915061.730.00.857061.732.20.7910061.635.10.7113061.937.80.6416062.339.90.6019062.441.90.5522062.643.60.5125062.745.00.4730062.647.10.4136062.349.10.3542062.650.50.3248062.851.80.2954062.752.60.2760062.953.30.

26Gradient = (ln 0.26 – ln 0.27) / (600 – 540) = - 6. 29 * 10-4 S-1h = Gradient * m *Cp / A = (-6.29*10-4 *0.0716 * 1006)/0.004418 = -10.26 W/m2 K 5.) Forced Convection Measurements (Linear Graph)From the Equation for Conservation of Energy on a solid, (m Cp d (Ts -T∞)/ dt = - h A (Ts - T∞)) (Rolle, K. 2015, 12)……………………………………………………………………. (i)Where:IV. A = Area V. M = Mass VI. Cp = Specific Heat,Loge Θ = - (h A /m Cp) t……………………………………………….. (ii), and Θ = (Ts - T∞)/ (Ts, i - T∞, i)…………………………………….. (iii), and, i, denotes conditions at time t = 0.

From Equation (iii), we can get the values of Θ and fill the table below as shown: Time, t,( SECONDS)Air Temp., T∞ oCDisc Temp. Ts oCΘΘ-n( n=25)024.354.91.001.001025.654.70.953.612024.554.80.991.293023.354.01.001.004023.253.81.001.006023.053.30.991.297022.953.00.981.669022.952.50.972.1411022.552.00.962.7713022.851.50.944.7015022.850.90.928.0427022.848.30.83105.4533022.947.00.79362.5139022.945.90.751328.8345022.944.80.723687.1251022.943.80.6815391.3760022.942.30.63103869.0790023.038.20.5033554432.00120022.834.90.408881784197.00150022.932.50.315.

20*1012180022.930.60.251.126*1015210022.829.10.218.80*1016240022.928.00.171.73*1019Gradient = (8.04– 4.70) / (150 – 130) = 0.167 S-1h = Gradient * m *Cp / A = (0.167*10-4 *0.0716 * 1006)/0.004418 = 0.272 W/m2 KUncertainty AnalysisFor the forced convection experiment, evaluate the uncertainty at t = 200 seconds:Ts = 50.1 ºC T∞ = 22.8 ºCTs, i = 54.9 ºC and T∞, i = 24.

3 ºC,Hence, Θ = 0.90, and from Equation (7)UΘ = {[U (Ts) / (Ts, i- T∞, i)] 2 +[U (T∞) / (Ts, i - T∞, i)] 2 +[U (Ts, i - T∞, i) *(Ts - T∞) / (Ts, i - T∞, i) 2]2}1/2UΘ = (- or +1/ (54.9-24.3)2) + (- or +1/ ((54.9-24.3)2)) + (0)*(50.1-22.8)/ ((54.9-24.3)2)2)0.5= +/- 0.00213Substituting into Equation (8):Uh = UΘ *m * Cp/Θ * △t* A = +/- 0.00213*0.0716*1006/0.90*27.3*0.004418 = +/- 0.021 W/m2K.For the forced convection experiment, evaluate the uncertainty at t = 1000 seconds:Ts = 36.

5 ºC T∞ = 22.9 ºCTs, i = 54.9 ºC and T∞, i = 24.3 ºC,Hence, Θ = 0.48, and from Equation (7)UΘ = {[U (Ts) / (Ts, i- T∞, i)] 2 +[U (T∞) / (Ts, i - T∞, i)] 2 +[U (Ts, i - T∞, i) *(Ts - T∞) / (Ts, i - T∞, i) 2]2}1/2UΘ = (- or +1/ (54.9-24.3)2) + (- or +1/ ((54.9-24.3)2)) + (0)*(36.5- 22.9)/ ((54.9-24.3)2)2)0.5= +/- 0.00213Substituting into Equation (8):Uh = UΘ *m * Cp/Θ * △t* A = +/- 0.00213*0.0716*1006/0.48*13.6*0.

004418 = +/- 0.0192 W/m2K.Discussion1. The existence of a vacuum between two surfaces often enhances the rates of Energy transfer between the said surfaces.Heat transfer by radiation =A* ε × σ × T4 =0.4418* 0.95* 5.67*10-8* 307.514 = 212.7989 W/m2K2. Radiative heat transfer was employed in the generation of heat used in varying the temperatures in this experiment. 3. The free convection coefficient of the whole heat transfer is 38.4 from the result. The anticipated coefficient should, however, range between 0.88 and 4.41 thus the actual value is much bigger than the expected value.4. If the heated surface of the plate is made to face downwards, the film outside the container would make the convection coefficient to originate from the outside values of the convectional film coefficients.

ReferencesRolle, K., 2015. Heat and Mass Transfer, Cambridge: Cengage Learning.