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masya89 [10]
3 years ago
7

S1.1 The Acre-Foot. Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as

the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 minutes on a town of area 26 km2. What volume of water, in acre-feet, fell on the town? S1.2 The Thunderstorm. What mass of water fell on the town in S1.1The Acre-Foot during the thunderstorm? One cubic meter of water has a mass of 103kg.
Engineering
1 answer:
Mars2501 [29]3 years ago
4 0

Answer:

1072 acre foot

1331424000 kg

Explanation:

1 feet has 12 inches, so 2 in is 0.167 feet.

1 km^2 has 1 million m^2.

1 acre is 4074 m^2.

So, 1 km is 247 acres.

Then 26 km^2 is 6422 acres.

So, the volume of water is

6422 * 0.167 = 1072 acre-foot

Since one cubic meter of water has 1000 kg

One inch is 25.4 mm = 0.0254 m

One feet is 12 * 0.0254 = 0.3048 m

An acre-feet has a volume of

4074*0.3048 = 1242 m^3

And that is a mass of water of

1242 * 1000 = 1242000 kg/acre-feet

Therefore the mass of rainwater in the town is of

1072 * 1242000 = 1331424000 kg = 1331424 tons

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Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-
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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
Sliva [168]

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}

   = \frac{144}{8}

   = 18 teeth/inch

The diametral is 18 teeth/inch

3 0
2 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
2 years ago
John wants to construct a device using quartz crystal, Which device can he construct?
tatiyna

Answer: Option D, piezoelectric pressure guage

Explanation: Quartz crystal possess a very useful quality in science as they can generate small charges when pressure is applied to them or when they are hit. This property can be harnessed to construct a piezoelectric pressure gauge which would be used to measure and indicate changes in pressure, the quartz crystal releases little voltage each time there is an applied pressure . This device would be able to sense changes in pressure as there would voltage proportional to the applied pressure.

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3 years ago
Read 2 more answers
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
jekas [21]

Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

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Generally the equation for Power is mathematically given by

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Since

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Where

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Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

 V'=5m/s

5 0
3 years ago
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