Question #79207

The heat transfer from a 2 m diameter sphere to a 25 °C air stream over a time interval of one
hour is 3000 kJ. Estimate the surface temperature of the sphere if the heat transfer coefficient is
10 W/m2 K
1

Expert's answer

2018-07-23T07:29:55-0400

Question #79207

The heat transfer from a 2 m diameter sphere to a 25 °C air stream over a time interval of one hour is 3000 kJ. Estimate the surface temperature of the sphere if the heat transfer coefficient is 10 W/m²K

Answer:

The convective heat transfer (Q=3000 kJ=3000000 JQ = 3000\ \mathrm{kJ} = 3000000\ \mathrm{J}) is given by:


Q=hA(TwTf)τ,Q = h A \left(T _ {w} - T _ {f}\right) \tau ,


where h=10 W/m2Kh = 10\ \mathrm{W/m^2K} is the heat transfer coefficient,


A=πd2A = \pi d ^ {2}


is the surface area of the sphere,

d=2 md = 2\ \mathrm{m} is the diameter of the sphere,

TwT_{w} is the surface temperature,

Tf=25CT_{f} = 25^{\circ}\mathrm{C} is the temperature of the air,

τ=1 hr=3600 s\tau = 1\ \mathrm{hr} = 3600\ \mathrm{s} is the period of time.

So, we can derive TwT_{w} from (1) as follows:


Tw=Tf+QhAτ,T _ {w} = T _ {f} + \frac {Q}{h A \tau},


Substitute into (2) and (3):


A=π22=12.566 m2,A = \pi \cdot 2 ^ {2} = 12.566\ \mathrm{m ^ {2}},Tw=25+30000001012.5663600=31.6C.T _ {w} = 25 + \frac {3000000}{10 \cdot 12.566 \cdot 3600} = 31.6^{\circ}\mathrm{C}.

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