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A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is

patriot [66]

Answer:4050 W

Explanation:

Given

Heat transfer Coefficient(h)= $20 W/m^2-K$

Air temperature =75 F

surface area(A)= $7.5 m^2$

Temperature of hot tube is 102 F

We know heat transfer due to convection is given by

$Q=hA\left ( \Delta T\right )$

$Q=20\times 7.5\left ( 102-75\right )=4050 W$

7 0

3 years ago

Air is compressed by a 30-kW compressor from P1 to P2. The air temperature is maintained constant at 25°C during this process as

adell [148]

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

4 0

2 years ago

What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?

Contact [7]

Answer: 100% (double)

Explanation:

The question tells us two important things:

Mass remains constant
Volume remains constant

(We can think in a gas enclosed in a closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1
State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.

6 0

3 years ago

You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

Katen [24]

Answer:

The required wall thickness is $1.506 \times 10^{-3}$ m

Explanation:

Given:

Fluid density $\rho = 1200$ $\frac{kg}{m^{3} }$

Diameter of tank $d = 8$ m

Length of tank $l = 4$ m

F.S = 4

For A-36 steel yield stress $\sigma = 250$ MPa,

Allowable stress $\sigma _{allow} = \frac{\sigma}{F.S}$

$\sigma _{allow} = \frac{250}{4} = 62.5$ MPa

Pressure force is given by,

$P = \rho gh$

$P = 1200 \times 9.8 \times 4$

$P = 47088$ Pa

Now for a vertical pipe,

$\sigma _{allow} = \frac{Pd}{4t}$

Where $t =$ required thickness

$t = \frac{Pd}{4 \sigma _{allow} }$

$t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }$

$t = 1.506 \times 10^{-3}$ m

Therefore, the required wall thickness is $1.506 \times 10^{-3}$ m

8 0

3 years ago

Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each

Elina [12.6K]

The initial void ratio is the parameter which is used to show the structural foundations for each specimen of sand so that the method and speed of compression would be measured.

Relative density is the mass per unit volume of each specimen of sand which is measured and it has to do with the relative ratio of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept