Question #140238
A solid cylinder of radius r(1)=2.5cm length L(1)=5.0cm and temperature 40°C is suspended in an environment of temperature 60°C .The thermal radiation transfer rate for cylinder is 1.0 Watt.If the cylinder is stretched until its radius becomes r(2)=0.50 cm the thermal radiation transfer rate is changed to
Ans 3.35W
1
Expert's answer
2020-11-02T07:13:55-0500

a)P=e×A×(T24T14)=e×(2×pi×r12+2×pi×r1×L1)×(T24T14)=125.309Wa) P = e\times A\times (T2^4 - T1^4) = e\times (2\times pi\times r1^2+2\times pi\times r1\times L1) \times (T2^4 - T1^4)=125.309 W


B)- as volume is constant

r12×L1=r22×L2r1^2\times L1 = r2^2 \times L2

L2=1.25mL2 = 1.25m

P2=e×(2×pi×r2×L2)×(T24T14)=459.46P2 = e\times (2\times pi\times r2\times L2) \times (T2^4 - T1^4) = 459.46WW


P2P1=459.46125.309\frac{P2}{P1} = \frac{459.46}{125.309} = 3.67





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