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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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A gear train has two gears. The driver gear has 8 teeth and a diametral pitch of 6 teeth/inch. the follower gear has 24 teeth. W
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Answer:

18 teeth/inch

Explanation:

Given that: i. driver gear has 8 teeth and diametral pitch of 6 teeth/inch.

                  ii. follower gear has 24 teeth.

Let the followers diametral pitch be represented by x.

Then,

8 teeth ⇒  6 teeth/inch

24 teeth ⇒ x teeth/inch

So that;

x = \frac{24 x 6}{8}

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3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
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Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

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It is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

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