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alexdok [17]
3 years ago
14

A thick steel slab ( 7800 kg/m3, 480 J/kg·K, 50 W/m·K) is initially at 300°C and is cooled by water jets impinging on one of its

surfaces. The temperature of the water is 25°C, and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach 50°C at a distance of 30 mm from the surface?

Engineering
2 answers:
AleksandrR [38]3 years ago
4 0

Answer: 67.392s

Explanation: detailed calculation is shown below

dlinn [17]3 years ago
3 0

Answer:

Time(t) = 2592.91 seconds

Explanation:

From the question, we have;

ρ = 7800 kg/m

c =480 J/kg⋅K

k = 50 W/m⋅K

Ti = 300°C

Ts = 25°C

x = 25 mm or in meters; = 0.025 m

From formula, we know that;

(T(x,t) - T(s))/(T(i) - T(s)) = erf(x/(2√(αt))

Where erf is error function

And α = k/(ρc)

So, solving we have;

(50 - 25)/(300 - 25) = erf(x/(2√(αt))

erf(x/(2√(αt)) = 0.0909

From the error function table which i attached, 0.0909 will give us approximately 0.0806 upon interpolation.

Thus, (x/(2√(αt)) = 0.0806

Let's make "t" the subject of the formula;

x² = 0.0806²(2²)(αt)

t = x²/0.026α

Let's find α

From earlier, we saw that;

α = k/(ρc)

Thus; α = 50/(7800 x 480) = 1.335 x 10^(-5) m²/s

Also, from the question, x = 30mm or 0.03m

So, t = 0.03²/(0.026 x 1.335 x 10^(-5)) = 2592.91 seconds

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At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t
nlexa [21]

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

Thus, the radius of a wind turbine is 691.1 ft

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

5 0
3 years ago
Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation
lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0
3 years ago
Multiple Choice
katrin [286]

Answer:

Zoning is like a hammer because it is used as a tool for urban planning.

8 0
3 years ago
What is the primary function of NCEES?
joja [24]
National Council of Examiners for engineering and surveying a nonprofit organization
7 0
3 years ago
A polyethylene rod exactly 10 inches long with a cross-sectional area of 0.04 in2 is used to suspend a weight of 358 lbs-f (poun
Nadya [2.5K]

Answer:

Final length of the rod = 13.90 in

Explanation:

Cross Sectional Area of the polythene rod, A = 0.04 in²

Original length of the polythene rod, l = 10 inches

Tensile modulus for the polymer, E = 25,000 psi

Viscosity, \eta = 1*10^{9} psi -sec

Weight = 358 lbs - f

time, t = 1 hr = 3600 sec

Stress is given by:

\sigma = \frac{Force}{Area} \\\sigma = \frac{358}{0.04} \\\sigma = 8950 psi

Based on Maxwell's equation, the strain is given by:

strain = \sigma ( \frac{1}{E} + \frac{t}{\eta} )\\Strain = 8950 ( \frac{1}{25000} + \frac{3600}{10^{9} } )\\Strain = 0.39022

Strain = Extension/(original Length)

0.39022 = Extension/10

Extension = 0.39022 * 10

Extension = 3.9022 in

Extension = Final length - Original length

3.9022 =  Final length - 10

Final length = 10 + 3.9022

Final length = 13.9022 in

Final length = 13.90 in

7 0
3 years ago
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