Steam at $120^{o} C$ is continuously passed through a $50 c m$ long rubber tube of inner and outer radii $1.0 c m$ and $1.2 c m$ . Find the transfer of heat?

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Solution

Given data,
$T_{s t e a m} = 120^{o} C$
$L = 50 c m = 0.5 m$
$r_{i} = 0.01 c m$
$r_{0} = 0.012 m$
$T_{0} = 30^{o} C$
$K_{1} = 0.15 J / m . s^{o} c$ or $w / m c$

$Q_{c y l i n d e r} = \frac{2 π L K (T_{s t e a m} - T_{0})}{l n (r_{0} / r_{i})}$

$\frac{2 π \times 0.5 \times 0.15 (T_{s t e a m} - T_{0})}{l n (\frac{0.012}{0.01})}$
$= 232.63 ≅ 233 J / s e c$ or $W$

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Steam at 120oC is continuously passed through a 50cm long rubber tube of inner and outer radii 1.0cm and 1.2cm. Find the transfer of heat?

Given data,Tsteam=120oCL=50cm=0.5mri=0.01cmr0=0.012mT0=30oCK1=0.15J/m.soc or w/mcQcylinder=2πLK(Tsteam−T0)ln(r0/ri)2π×0.5×0.15(Tsteam−T0)ln(0.0120.01)=232.63≅233J/sec or W

Steam at $120^{o} C$ is continuously passed through a $50 c m$ long rubber tube of inner and outer radii $1.0 c m$ and $1.2 c m$ . Find the transfer of heat?