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3 years ago

4 Points

Physics

2 answers:

Lyrx [107]3 years ago

5 0

Answer:

323 kg*m/s

Explanation:

lbvjy [14]3 years ago

4 0

Answer:

Impulse = 322.5[kg*m/s], the answer is D

Explanation:

This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.

If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.

$N*S=(kg*m/s^{2} )*s = kg*m/s$

Now replacing the values on the following equation that express the definition of impulse

$Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]$

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A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi

kati45 [8]

Explanation:

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2 years ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

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

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

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

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

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

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:

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