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barxatty [35]
3 years ago
6

You’ve experienced convection cooling if you’ve ever extended your hand out the window of a moving vehicle or into a flowing wat

er stream. With the surface of your hand at a temperature of 30°C, consider the two following conditions.
Condition A: a vehicle speed of 35 km/h in air at -5°C with a convection coefficient of 40 W/m2·K.
Condition B: a velocity of 0.2 m/s in a water stream at 10°C with a convection coefficient of 800 W/m2·K.

What is the convection heat flux for these two condition?
Engineering
1 answer:
Anna35 [415]3 years ago
7 0

Answer:

Condition A

Heat flux is 1400 W/M^2

Condition B

Heat flux is 12800 w/m^2

Explanation:

Given that:

T_s is given as  30 degree celcius

condition A

Air temperature =  - 5 degree c

convection coefficient h = 40 w/m^2. k

heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2

condition A

water temperature  = 10 degree c

convection coefficient = 800 w/m^2.k

heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2

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Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca
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Answer:

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

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Explanation:

<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

Capacitance of the second capacitor C_{2}= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

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C_{s}=\frac{Q}{V}

Then the charge stored in the series combination is:  

Q=C_{s} V

Energy stored in the series combination is:  

U_{c}=\frac{1}{2}  V^{2} C_{s}

<u>Solve the Problem:  </u>

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

<u>b)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

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<u>Solve the Problem: </u><em>  </em>

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

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see attached file below

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