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galben [10]
3 years ago
14

In order to cool a cylindrical steel rod of diameter D, length L and uniform temperature T1, it is placed in a well-mixed water

bath at initial temperature T0 and volume V0. The heat transfer coefficient between the steel rod surface and the water is h1. Assume the specific heat, density and conductivity of the steel rod and water are known and constant. Write down heat transfer differential V1, P1, T1 equation and boundary and initial conditions for temperature distribution in both steel rod and water bath
Engineering
1 answer:
Jlenok [28]3 years ago
7 0

Answer:

attached below

Explanation:

Cylindrical steel rod : diameter ( D ) , length L, uniform temperature T

Initial temp of water : To, heat transfer coefficient between steel rod surface and water : h1

<u>The energy balance equation can be written as :</u>

Rate of convectional heat loss = Rate of decrease in internal energy with respect to time

cp = heat capacity of body, v = volume , р = density of body

attached below is the heat transfer differential equation and boundary and initial conditions

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Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
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Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

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\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

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v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

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h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

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The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

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