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4 years ago

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Snoopy is inflated for the Macy’s Thanksgiving parade with 50,000 L of He at 25 oC and 1.2 atm. What is the percent decrease in

Snoopy’s volume if the temperature drops to 4 oC before the parade? Assume constant pressure.

1 answer:

Dvinal [7]4 years ago

5 0

Answer:

D.V.= - 1.34 %

Explanation:

Given that

V₁=50,000 L

T₁=25⁰C = 273 + 25 K =298 K

The final temperature ,T₂= 25 - 4 ⁰C = 21 ⁰C

T₂ = 21 + 273 K= 294 K

We know that for constant pressure process

$V_2=\dfrac{T_2}{T_1}\times V_1$

$V_2=\dfrac{294}{298}\times 50000\ L$

V₂=49328.85 L

The percent decrease in the volume is given as

$D.V.=\dfrac{49328.85-50000}{50000}\times 100$

D.V.= - 1.34 %

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Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per

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Answer:

pipe is old one with increased roughness

Explanation:

discharge is given as

$V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}$

V = 7.07 m/s

from bernou;ii's theorem we have

$\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l$

as we know pipe is horizontal and with constant velocity so we have

$\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}$

$P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma$

$135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81$

solving for friction factor f

f = 0.0324

fro galvanized iron pipe we have $\epsilon = 0.15 mm$

$\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025$

reynold number is

$Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}$

Re = 378750

from moody chart

$For Re = 378750 and \frac{\epsilon}{d} = 0.0025$

$f_{new} = 0.025$

therefore new friction factor is less than old friction factoer hence pipe is not new one

now for Re = 378750 and f = 0.0324

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we have $\frac{\epsilon}{d} =0.006$

$\epsilon = 0.006 \times 60$

$\epsilon = 0.36 mm$

thus pipe is old one with increased roughness

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3 years ago

The rate of energy transfer by work is called power. a)-True b)-False

Pie

Answer:

Yes the statement is true.

Explanation:

Power is defined as the rate at which energy is transferred by an object on account of work done.

Mathematically

$Power=\frac{dE}{dt}$

An object that does work loses it's energy while an object on which work is done gains energy.

Power is often dependent on the type of energy transfer thus we have Electrical Power, Mechanical Power depending on the type of energy involved in the system.

Concept of power is important since it gives us a measure of how fast energy can be derived to given to a system.

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4 years ago

In primary processing their are 3 different steps, what are the steps?

Nana76 [90]

Answer:

Primary processing involves cutting, cleaning, packaging, storage and refrigeration of raw foods to ensure that they are not spoilt before they reach the consumer.

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3 years ago

Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi

VladimirAG [237]

Answer:

Source 2.

Explanation:

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$\eta_{th} = 1 - \frac{T_{L}}{T_{H}}$

Where:

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$T_{H}$ - Temperature of the hot reservoir, in K.

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$\eta_{th} = 0.0798 \,(7.98\,\%)$

The power produced by each device is presented below:

Source 1

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Source 2

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$\dot W = 146.3\,kW$

The source 2 produces the largest amount of power.

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3 years ago

Consider a cylinder of height h, diameter d, and wall thickness t pressurized to an internal pressure P_0 (gauge pressure, relat

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Explanation:

Note: For equations refer the attached document!

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Eq 1

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Eq 2

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Eq 3

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Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

Eq 5

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Eq 6

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Eq 7

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Eq 8

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Eq 9

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Eq9

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Eq 10

part b

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