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Grace [21]
3 years ago
6

Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kep

t at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Engineering
1 answer:
Talja [164]3 years ago
7 0

Answer:

95.914\ \text{GJ}

\$272.78

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

\Delta T = Change in temperature = 80.6-50=30.6^{\circ}\text{F}

Cost of 1 GJ of energy = $2.844

Heat required is given by

Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}

Amount of heat required to heat the water is 95.914\ \text{GJ}.

Cost of heating the water is

95.914\times 2.844=\$272.78

Cost of heating the water to the required temperature is \$272.78.

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Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

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Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

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A = 6.28 x 10^(-4) m²

We are given that;

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Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

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8 0
4 years ago
A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal
nirvana33 [79]

Answer:

a)  the temperature to which the pin must be cooled for assembly is T_2 = -101.89^ \ ^0}C

b) the radial pressure at room temperature after assembly is P_f = 62.8 \ MPa

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Explanation:

Coefficient of thermal expansion \alpha = 12.3*10^{-6} \  ^0 C

Yield strength \sigma_y = 400 MPa

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Room Temperature T_1 = 20°C

outer diameter of the collar D_o = 95 \ mm

inner diameter of the collarD_i = 60 \ mm

pin diameter D_p = 60.03 \ mm

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

(D_i - c )-D_p = \alpha * D_p(T_2-T_1)

(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}(T_{2}-20^{0}C)

-0.09 = 7.38369*10^{-4}T_2  \ \ - \ \ 0.01476738

-0.09 +  0.01476738 = 7.38369*10^{-4}T_2

−0.07523262 =7.38369*10^{-4}T_2

T_2 = \frac{-0.07523262}{7.38369*10^{-4}}

T_2 = -101.89^ \ ^0}C

b)

To determine the radial pressure at room temperature after assembly ;we have:

P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\  P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2}  \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa

c)  the safety factor of the resulting assembly is calculated as:

safety factor =  \frac{Yield \ strength }{walking \ stress}

safety factor =  \frac{400}{62.8}

safety factor = 6.4

Thus, the  safety factor in the resulting assembly = 6.4

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