Heat transfre for a domestic radiator

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SUMMARY

The discussion focuses on calculating the total heat transfer from a domestic radiator to a room. The radiator dimensions are 2.5 m long and 0.6 m high, with water circulating at 90°C in a room at 14°C. The surface emissivity is 0.7, and the Stefan-Boltzmann constant is 5.67x10^-8. The calculated radiated heat transfer is 629.8W, and further calculations for convective heat transfer are necessary to complete the analysis.

PREREQUISITES
  • Understanding of heat transfer principles, specifically radiation and convection.
  • Familiarity with the Stefan-Boltzmann Law for radiative heat transfer.
  • Knowledge of Nusselt, Grashof, and Prandtl numbers in fluid dynamics.
  • Ability to perform calculations involving thermal conductivity and heat transfer coefficients.
NEXT STEPS
  • Calculate the convective heat transfer coefficient using appropriate formulas.
  • Explore the implications of varying surface emissivity on heat transfer rates.
  • Investigate the effects of different room temperatures on overall heat transfer.
  • Learn about advanced heat transfer modeling techniques for complex systems.
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Students studying thermodynamics, engineers involved in HVAC design, and anyone interested in optimizing heat transfer in domestic heating systems.

nickbone59
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Any suggestions anyone? Really struggling with this one !

Homework Statement


A domestic radiator is 2.5 m long and 0.6m high and is sited in a room whose temperature is 14C. Hot water is circulating through the radiator at a temperature of 90c. The radiator is convecting heat from both sides but only radiating from one side. Given that the surface emissivity is 0.7, calculate the total heat transfer from radiator to the room.



Homework Equations


Stefan-Boltzmann Constant = 5.67x10^-8

Nu = 0.59(Gr.Pr)^0.25 for laminar flow (GR.Pr)<10^9
Nu = 0.129(Gr.Pr)^0.33 for turbulent flow (GR.Pr)>10^9
where,
Grashof no. Gr= (\rho^2\beta(\theta1-\theta2)l^3)/\mu^3

Nusselt No. Nu= hl/k

Prandtl No. Pr=Cp\mu/k


The Attempt at a Solution



Q= \sigma\epsilonA(T1^4-T2^4)

= 5.67x10^-8 x 0.7 x 1.5 x (1.736x10^10-6.785x10^9)

= 629.8W
 
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Looks like you got the radiated heat correct. It would be preferable to include units with the numbers.

Now find the convective heat. One needs a formula for the convective heat transfer coefficient.
 

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