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3 years ago

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Two long copper rods of diameter D 10 mm are soldered together end to end, with solder having a melting point of 650 C. The rods

are in air at 25 C with a convection coefficient of 10 W/m2 K. What is the minimum power input needed to effect the soldering?

2 answers:

nadezda [96]3 years ago

5 0

Answer:

Qf 194.2 W

Explanation:

Given data:

Diameter of copper rod D is 0.01 ,

Temperature at junction is Tb = 650 + 273 = 923 K

Temperature of air is 15+273 = 288 K

$Tmean = \frac{923 + 288}{2} = 605.5 K$

For temperature 605.5 degree celcius thermal conductivity is K = 379 W/m K

Heat transfer is calculated as

$qf = \sqrt{hPkAc(Tb -T\infty)}$

$qf = \sqrt{h\pi D k\frac{\pi}{4} D^2 (Tb -T\infty)}$

$qf = \sqrt{25 \pi \times 0.01\times 379 \times \frac{\pi}{4} \times 0.01^2(920 -288)}$

qf = 97.1 W

Hence, the rate of heat is

$Qf = 2qf = 2\times 97.1 = 194.2 W$

saw5 [17]3 years ago

3 0

Answer:

125pi W/m

Explanation:

This is the formula for heat transfer by convection

Q=hA(T2-T1)

Q=Power

h=convection coefficient

A=area

T2= temperature of melting point

T1=temperature of air

Area of tow rods

A=2*pi*D=2*pi*0.01=pi/50

Q=10*(´pi/50)(650-25)

Q=125pi W/m

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3 years ago

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$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

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a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

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$\dot W_{out} = 18276.307\,kW$

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