5.0 RESULT

HEATERBRASS

SAMPLE REGIONCOOLER

x (mm) 0 10 20 30 40 50 60 70 80x (m) 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

Test # Q̇ (W)T1

(℃)T2

(℃)T3

(℃)T4

(℃)T5

(℃) T6(℃) T7(℃) T8(℃) T9(℃)

A 10 49.5 43.2 48.3 35.3 34.2 34.1 32.3 31.2 30.9B 20 83.2 80.9 79.3 45.9 42.5 42.1 33.3 31.6 31.2C 30 117.5 115.6 111.4 59.5 54.3 52.5 35.5 33.2 32.6

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090

50

100

150

200

250

300

350

400

450 Graph of Temperature Versus Length

10 Watt20 Watt30 Watt

Length (m)

Tem

pera

ture

(K)

Graph 5.1: Graph of temperature versus length for entire length

Sample Calculation

Converting the measured temperature to degrees Kelvin by using following formula:

T (K )=T (℃ )+273.15

T (K )=49.5+273.15

T (K )=322.65K

Data tabulation after converting measured temperature to degrees Kelvin.

HEATERBRASS

SAMPLE REGIONCOOLER

x (mm) 0 10 20 30 40 50 60 70 80x (m) 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

Test # Q̇ (W) T1(K ) T2(K ) T3(K ) T4(K ) T5(K ) T6(K ) T7(K ) T8(K ) T9(K )A 10 322.65 316.35 321.45 308.45 307.35 307.25 305.45 304.35 304.05B 20 356.35 354.05 352.45 319.05 315.65 315.25 306.45 304.75 304.35C 30 390.65 388.75 384.55 332.65 327.45 325.65 308.65 306.35 305.75

Cross-sectional area (A) of the circular cylinder is calculated by using the following equation:

A=π4×d2

A=π4×0.0252

A=4.91×10−4m2

Thermal conductivity of the brass at T4, T5, and T6 can be determined by using the following equation:

k= Q̇A∆ x∆T

units[ Wm. K ]

Noting that the gradient, m, can be obtained from Graph 5.2 and noted approximately as

m=∆T∆ x

∴ k= Q̇A∆ x∆T

=Q̇A

1m

0.025 0.03 0.035 0.04 0.045 0.05 0.055290

295

300

305

310

315

320

325

330

335

f(x) = − 350 x + 342.583333333333

f(x) = − 190.000000000001 x + 324.25

f(x) = − 59.9999999999994 x + 310.083333333333

Graph Of Temperature Versus Length

10 WattLinear (10 Watt)20 WattLinear (20 Watt)30 WAttLinear (30 WAtt)

Length (m)

Tem

pera

ture

(K)

Thermal conductivity, k, for 10 Watt input power.

k= Q̇A∆ x∆T

= Q̇A

1m

By comparing the straight line equation, y=mx+c with the straight line equation obtained from the Graph 5.2 for 10 Watt input power, y=−60x+310.0, m10W=60.

k= 10

4.91×10−4

160

k=339.44Wm.k

Thermal conductivity, k, for 20 Watt input power.

k= Q̇A∆ x∆T

= Q̇A

1m

By comparing the straight line equation, y=mx+c with the straight line equation obtained from the Graph 5.2 for 10 Watt input power, y=−190x+324.2, m20W=190.

k= 20

4.91×10−4

1190

k=214.39Wm.k

Graph 5.2: Graph of temperature versus length for T4, T5, and T6.

Thermal conductivity, k, for 20 Watt input power.

k= Q̇A∆ x∆T

= Q̇A

1m

By comparing the straight line equation, y=mx+c with the straight line equation obtained from the Graph 5.2 for 30 Watt input power, y=−350x+342.5, m20W=350.

k= 30

4.91×10−4

1350

k=174.57Wm.k