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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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Answer:

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Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please
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Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

<u>Input</u>

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

<u>Processing</u>

Assign variable to the input number (x)

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Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

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PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

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