Question:
Water flows through a pipe of diameter 0.02 m. The Reynolds number of the flow is 1000. The pipe is heated from outside with a uniform heat flux. The flow and heat transfer in the pipe are steady and fully developed. The thermal conductivity of water is 0.66 W/(m-K). The convective heat transfer coefficient, in W/(m$^2$-K), is ................. (round off to 2 decimal places).
Water flows through a pipe of diameter 0.02 m. The Reynolds number of the flow is 1000. The pipe is heated from outside with a uniform heat flux. The flow and heat transfer in the pipe are steady and fully developed. The thermal conductivity of water is 0.66 W/(m-K). The convective heat transfer coefficient, in W/(m$^2$-K), is ................. (round off to 2 decimal places).
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The convective heat transfer coefficient is related to the Nusselt number, which is influenced by the Reynolds number and Prandtl number for fully developed flow.
Updated On: Sep 4, 2025
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Solution and Explanation
Given:
Diameter of pipe, \( D = 0.02 \, \text{m} \)
Reynolds number, \( Re = 1000 \)
Thermal conductivity of water, \( k = 0.66 \, \text{W/m-K} \)
For fully developed laminar flow in a pipe, the convective heat transfer coefficient is given by the following empirical relation: \[ Nu = 0.023 Re^{0.8} Pr^{0.3} \] where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, and \(Pr\) is the Prandtl number.
For water, at moderate temperature: \[ Pr = 7.0 \] The Nusselt number can be calculated as: \[ Nu = 0.023 \times (1000)^{0.8} \times (7.0)^{0.3} \] \[ Nu = 0.023 \times 158.49 \times 2.229 = 7.933 \] The convective heat transfer coefficient \(h\) is related to the Nusselt number as: \[ h = \frac{Nu \times k}{D} \] Substitute the values: \[ h = \frac{7.933 \times 0.66}{0.02} = 261.4 \, \text{W/m}^2\text{-K} \] Thus, the convective heat transfer coefficient is 261.4 W/m$^2$-K.
Diameter of pipe, \( D = 0.02 \, \text{m} \)
Reynolds number, \( Re = 1000 \)
Thermal conductivity of water, \( k = 0.66 \, \text{W/m-K} \)
For fully developed laminar flow in a pipe, the convective heat transfer coefficient is given by the following empirical relation: \[ Nu = 0.023 Re^{0.8} Pr^{0.3} \] where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number, and \(Pr\) is the Prandtl number.
For water, at moderate temperature: \[ Pr = 7.0 \] The Nusselt number can be calculated as: \[ Nu = 0.023 \times (1000)^{0.8} \times (7.0)^{0.3} \] \[ Nu = 0.023 \times 158.49 \times 2.229 = 7.933 \] The convective heat transfer coefficient \(h\) is related to the Nusselt number as: \[ h = \frac{Nu \times k}{D} \] Substitute the values: \[ h = \frac{7.933 \times 0.66}{0.02} = 261.4 \, \text{W/m}^2\text{-K} \] Thus, the convective heat transfer coefficient is 261.4 W/m$^2$-K.
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