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3 years ago

8

Over 30 day period, a lake surface area is 1260 acres. The inflow is 36 cfs, thee outflow is 30 cfs. Seepage loss is 1.5 in. The

total precipitation is 4.0 in. Evaporation loss is 6.0 in. Determine the water level change (inch) for the lake during this period.

1 answer:

Elis [28]3 years ago

5 0

Answer:

-0.1 inches

Explanation:

The net inflow is ...

36 cfs -30 cfs = 6 cfs

The number of seconds in 30 days is ...

(3600 s/h)(24 h/da)(30 da) = 2,592,000 . . . . seconds/(30 days)

Then the volume of inflow is ...

(6 ft^3/s)(2,592,000 s) = 15,552,000 ft^3

The number of square feet in 1260 acres is ...

(1260 ac)(43560 ft^/ac) = 54,885,600 ft^2

So, the increase in depth due to the inflow is ...

(15,552,000 ft^3)/(54,885,600 ft^2) ≈ 0.283353 ft ≈ 3.4002 in

__

The net change in water level is then ...

inflow - seepage + precipitation - evaporation

3.4 in -1.5 in +4.0 in -6.0 in = -0.1 in

The water level change in the period is -0.1 inch.

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Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

$h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K$

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

$s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}$

$s_2 =0.51624 + x(7.82)$

$s_2 =0.51624 + 7.82x$

From the values obtained;

$s_1 =s_2= 6.52846 \ kJ/kg.K$

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222 = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

$h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 ) \\ \\ h_2 =2010.4084 \ kJ/kg$

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

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$0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}$

$0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}$

$0.8 * {1311.6116}= {3322.02-h_{2a}$

$1049.28928= {3322.02-h_{2a}$

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$w_p = h_4 -h_3$

From the process;

$h_3 = h_{f(6 kpa)} = 150.15 \ kJ/kg$

$10.0559628 = h_4 - 150.15$

$10.0559628+ 150.15 = h_4$

$160.2059628= h_4$

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3 years ago

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Vitek1552 [10]

Answer:

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to find out

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Now we put the value of heat (Q) and Temperature (T) in equation a

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so that

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ΔS = 0.25 J/K

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Hitman42 [59]

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rjkz [21]

Answer:

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using namespace std;

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double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

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double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

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{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

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{

double largest;

largest = NUM_MONTHS[0];

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while (rainFall[month] < 0)

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4 years ago

Okay bro let’s go man yes yes

r-ruslan [8.4K]

Sheeeeeesh bro same name ayoooo??

8 0

3 years ago

Read 2 more answers