5000 = CONSTANT, H = 1000 W/m2K, ṁ = 0.01 kg/s, T1 = 20°C, d = 50 mm = 0.05 m, L = 3 m, cp = 4.18 kJ/kgK = 4.18 × 103 K/kgKCalculation:\(q_w'' × A = mC_p{\rm{\Delta }}T\)5000 × π × 0.05 × 3 = 0.01 × 4.18 × 103 × (T0 - 20)T0 – 20 = 56.3 = T0 = 76.3°CHeat flux between any two sections is sameQ = hA(Tp – T0)But \(\frac{Q}{A} = q_w'' = 5000 = \left( {{T_p} - {T_0}} \right)\)5000 = 1000(Tp – 76.3)⇒ Tp = 76.3 + 5 = 81.3°C

"> 5000 = CONSTANT, H = 1000 W/m2K, ṁ = 0.01 kg/s, T1 = 20°C, d = 50 mm = 0.05 m, L = 3 m, cp = 4.18 kJ/kgK = 4.18 × 103 K/kgKCalculation:\(q_w'' × A = mC_p{\rm{\Delta }}T\)5000 × π × 0.05 × 3 = 0.01 × 4.18 × 103 × (T0 - 20)T0 – 20 = 56.3 = T0 = 76.3°CHeat flux between any two sections is sameQ = hA(Tp – T0)But \(\frac{Q}{A} = q_w'' = 5000 = \left( {{T_p} - {T_0}} \right)\)5000 = 1000(Tp – 76.3)⇒ Tp = 76.3 + 5 = 81.3°C

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If \(q_w'' = 5000\), and the convection heat transfer coefficient at the pipe outlet is 1000 W/m2K, the temperature in °C at the inner surface of the pipe at the outlet is

Heat Transfer Types Convection in Heat Transfer 1 year ago

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Concept:Heat transfer rate between the entire pipe is \(q_w'' × A = mC_p{\rm{\Delta }}T\)Given:q"w = 5000 = CONSTANT, H = 1000 W/m2K, ṁ = 0.01 kg/s, T1 = 20°C, d = 50 mm = 0.05 m, L = 3 m, cp = 4.18 kJ/kgK = 4.18 × 103 K/kgKCalculation:\(q_w'' × A = mC_p{\rm{\Delta }}T\)5000 × π × 0.05 × 3 = 0.01 × 4.18 × 103 × (T0 - 20)T0 – 20 = 56.3 = T0 = 76.3°CHeat flux between any two sections is sameQ = hA(Tp – T0)But \(\frac{Q}{A} = q_w'' = 5000 = \left( {{T_p} - {T_0}} \right)\)5000 = 1000(Tp – 76.3)⇒ Tp = 76.3 + 5 = 81.3°C

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Posted on 24 Oct 2024, this text provides information on Heat Transfer related to Types Convection in Heat Transfer. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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