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MatroZZZ [7]
3 years ago
13

A stationary boat in the ocean is experiencing waves from a storm. the waves move at 56 km/h and have a wavelength of 160 m, bot

h typical values. the boat is at the crest of a wave. how much time elapses until the boat is first at the trough of a wave?
Physics
1 answer:
Lera25 [3.4K]3 years ago
5 0
The wavelength \lambda of the wave is 160 m, and this is the distance between two consecutive crests. The boat is located at a crest of the wave, this means that the first trough is located 80 meters from the boat (because the distance between a crest and a trough is half the wavelength).

The speed of the wave is 
v=56 km/h = 15.6 m/s
so the time the boat takes to reach the first trough is
t= \frac{S}{v} = \frac{80 m}{15.6 m/s}=5.1 s
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Answer:

Explanation:

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To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

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Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

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Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

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3 years ago
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