An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min

The Reynolds number is a dimensionless parameter used in fluid mechanics to predict flow patterns in different fluids. It can be calculated using the formula pVD/u.

The Reynolds number, represented by Re, is a dimensionless number used to predict the flow behavior of fluids. It is determined by the ratio of the inertial forces to the viscous forces in a fluid.

The equation for Reynolds number is Re = pVD/u, where p is the density of the fluid, V is the velocity of the fluid, D is the diameter of the pipe, and u is the dynamic viscosity of the fluid.

Given the flow velocity of 2 m/s and pipe diameter of 5 inches (0.127 m), we need to find the dynamic viscosity of ethyl alcohol and density to calculate the Reynolds number. At 20°C, the dynamic viscosity of ethyl alcohol is 1.2 x 10⁻³ Pa·s and the density is 789 kg/m³.

Therefore, the Reynolds number for ethyl alcohol flowing at 2 m/s through a 5-inch diameter pipe is:

Re = (789 kg/m³) x (2 m/s) x (0.127 m) / (1.2 x 10⁻³Pa·s) = 1.05 x 10⁶  

So, the Reynolds number for this flow is relatively high, indicating that the flow will be turbulent.

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Answer:

1417.05 kw

Explanation:

From the question,

The Mass flow rate = 3kg

The Saturated temperature = 275⁰c

The Thermal efficiency = 30%

The enthalpy of vaporization is 1574.5kj/kg

Rate of heat transfer = 3x1574.5 = 4723.5 kW

Thermal efficiency of engine =

0.30 x 4723.5

= 1417.05 kW

This answer is the net power output of this engine, in kW.

Thank you!

This question is incomplete, the complete question is;

A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N.

Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.

answer in mm please

Answer:

the minimum required wire radius is 5.3166 mm

Explanation:

Given that;

Load F = 11100N

N = 3

∝y = 1500 MPa

∝workmg = ∝y / N = 1500 / 3 = 500 MPa

now stress of Wire:

∝w = F/A

500 × 10⁶ = 11100 / A

A = 22.2 × 10⁻⁶ m²

so

(π/4)d² = A

(π/4)d² = 22.2 × 10⁻⁶

d² = 2.8265 × 10⁻⁵

d = 5.3165 7 × 10⁻³ m³

now we convert to mm(millimeters)

d = 5.3166 mm

Therefore the minimum required wire radius is 5.3166 mm

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\(\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}\)

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\(\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}\)

\(P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa\)

\(\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}\)

\(\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right\)

\(5^{0.33} = \left \dfrac{T_1}{T_2} \right\)

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×\(c_v\)×(T₂ - T₁)

\(c_v\) = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×\(c_v\)×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = \(U_g = h_g -pV_g\)

\(h_g\) = 2777.12 kJ/kg

\(V_g\) = 0.194349 m³/kg

p = 1000 kPa

\(U_{g1}\) = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, \(U_{g1}\) + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

Answer: ⚠this is not my answer it from hamzaahmeds⚠

Water: h = 35.53 W/m².k

Engine oil: h = 18.84 W/m².k

Mercury: h = 1.19 W/m².k

Explanation:

Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.

Therefore,

mCΔT = hAΔT

mC = hA

h = mC/A

where,

h = convection coefficient

m = mass flow rate =  0.01 kg/s

C = specific heat capacity of fluid

A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²

FOR WATER:

C = 4186 J/Kg.k

Therefore,

h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)

h = 35.53 W/m².k

FOR ENGINE OIL:

C = 2220 J/Kg.k

Therefore,

h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)

h = 18.84 W/m².k

FOR LIQUID MERCURY:

C = 140 J/Kg.k

Therefore,

h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)

h = 1.19 W/m².k

Python program to accept string inputs from the user and display them in ascending order (alphabetical order) until the user enters zzz or has entered 15 strings.``

`pythondef main():

# Define an empty list for storing the strings lst = []

# Loop until the user enters zzz or 15 strings have been entered while len(lst) < 15: s = input("Enter a string (or 'zzz' to quit): ") if s == "zzz": break lst.append(s)

# Sort the list in ascending order lst.sort()

# Display the sorted list for string in lst: print(string)if __name__ == "__main__": main()```

This program defines an empty list called lst, and then uses a loop to accept string inputs from the user until they enter "zzz" or have entered 15 strings. The input() function is used to accept user input for each string.Once the user has entered a string, the program adds it to the list lst. The program then sorts the list in ascending order using the sort() function. Finally, the program loops through the list and displays each string in ascending order using the print() function.

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Answer:

Explanation:

i took the test and screenshot it

Answer:

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

Explanation:

As,

Categorical syllogisms follow an "If A is part of C, then B is part of C" logic.

Conditional syllogisms follow an "If A is true, then B is true" pattern of logic.

So,

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

The statement predicts that half the land will return to pre-construction conditions, while the other half of the land will not because half the land is underwater, meaning that the environmental impact will be lessened. The correct option is b.

Pre-construction is when hazards connected with the project are carefully evaluated and planned to be minimized.

This includes assessing the construction site, determining the need for permits and inspections, and addressing any other unique circumstances that may arise before or during construction.

Therefore, the correct option is B.

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Answer:

According to search result [1], the driving environment that has been proven to have fewer collisions is the expressway.

Explanation:

The exclusive OR (XOR) gate has a special property that makes it useful for building adder-subtractor circuits using full adders. The XOR gate output is high only when its inputs are different; otherwise, the output is low. This property allows us to use XOR gates to perform the subtraction operation in the adder-subtractor circuit.

In the adder-subtractor circuit, the XOR gate is used to determine the sign of the numbers being added or subtracted. If the two input numbers have the same sign (both positive or both negative), then the output of the XOR gate is low, indicating that an addition operation should be performed. However, if the two input numbers have opposite signs (one positive and one negative), then the output of the XOR gate is high, indicating that a subtraction operation should be performed.

By using full adders and XOR gates in combination, we can perform both addition and subtraction operations within the same circuit. The full adders are used to perform the addition operation, while the XOR gates are used to perform the subtraction operation by changing the sign of one of the input numbers.

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Micropulse arc welding is a well-known tool for speeding up simple bench operations during the welding process like ring size, making connections near thermally sensitive gemstones, and mending porosity castings.

The practice performs effectively on gold, silver, copper, and stainless steel whether joining tiny jewelry materials or bigger pieces for miniature metal sculptures.

The welding arc temperatures typically range from 6000°C to 8000°C, which translates to about 10000F to 15000F.

This increased range of thermal heat is high enough to turn Gold and silver rings into a molten state. A molten state is a phase at which a solid element is being melted and turns into a liquid state under the influence of a high temperature.

Therefore, from the above explanation, we can conclude that it is possible and true for gold and silver rings to receive an arc and turn molten.

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Answer:  System Consists Of 1 Kg Of CO2 (Cp = 46.4 J Moll K:') Gas Initially At 1 Bar And 300K. The System Undergo

Explanation:

Answer:

  A accountant

  D teacher

  B cook

  C beautician

Explanation:

C is on the right of the accountant, so D (opposite C) is the teacher. The accountant is to the right of the teacher, so cannot be B, and must be A. Then B (opposite A) is the cook, and C (left of B) is the beautician.

__

Clockwise from A:

  A accountant

  D teacher

  B cook

  C beautician

Answer:

it should be accuracy

Explanation:

Answer:

option C

Explanation:

Answer:

with turbo or nos

Explanation:

Answer:s

Explanation:s

It's the law. Under New York law, you can get a citation for riding your bike while listening to music using headphones or earbuds in both ears. It is lawful, however, to listen to music with only one earbud keeping the other ear free.

Sensory deprivation. One of the main ways that people alert each other to danger on the road is through sound. Whether it's honking, yelling or ringing a bicycle bell, you're less likely to hear any of these warnings if you're listening to music putting you at higher risk of an accident.

Decreased attention. The more stimuli you have to contend with, the more difficult it is to dedicate your full attention to one particular task. If you get caught up in a song or story you're listening to while cycling, it can increase your distractibility which could have disastrous consequences.

The majority of cycling-related accidents involve a cyclist and a motor vehicle. Not surprisingly, the injuries a cyclist suffers in such collisions tend to be disproportionately serious and often fatal. Therefore, if you're cycling in traffic, foregoing the headphones can prove to be a life-saving decision.

The principal stresses are 9, 6, and -3, the maximum in-plane shear stress is 1.5, and the average normal stress is 4.

To determine the principal stresses, we need to find the eigenvalues of the stress tensor. The stress tensor for this problem can be represented by the following matrix:
[9  0  0]
[0  6  0]
[0  0 -3]
To find the eigenvalues, we need to solve the equation det(A-λI) = 0, where A is the stress tensor, I is the identity matrix, and λ is the eigenvalue. Solving this equation gives us the following eigenvalues:
λ1 = 9
λ2 = 6
λ3 = -3
The principal stresses are simply the eigenvalues, so the principal stresses for this problem are:
σ1 = 9
σ2 = 6
σ3 = -3
To find the maximum in-plane shear stress, we can use the formula:
τmax = (σ1 - σ2) / 2
Plugging in the values for σ1 and σ2 gives us: τmax = (9 - 6) / 2 = 1.5
Finally, to find the average normal stress, we can use the formula:
σavg = (σ1 + σ2 + σ3) / 3

Plugging in the values for σ1, σ2, and σ3 gives us:

σavg = (9 + 6 - 3) / 3 = 4

Therefore, the principal stresses are 9, 6, and -3, the maximum in-plane shear stress is 1.5, and the average normal stress is 4.

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Out of the given options, the incorrect expression information to perform the same is: df.rename (mapper lambda x: x.upper(), axis = 1).

above expression contains syntax error as the `mapper` is not defined correctly. The correct way to define a `mapper` function in pandas is by using the `lambda` keyword followed by the variable that stores the column names. Hence, the correct syntax to define `mapper` in the above code should be `mapper = lambda x: x.upper()`. Therefore, the corrected version of this expression would be: df.rename(mapper=lambda x: x.upper(), axis=1)Alternatively, we can perform the same task using other given expressions as well which are: df.rename(mapper=lambda x: x.upper(), axis=1, inplace=True)df = df.rename(mapper=lambda x: x.upper(), axis='columns')We can also perform the above task using the `columns` attribute of a DataFrame as shown below:df.columns = [col.upper() for col in df.columns]Therefore, we can say that all of the given options, except for df.rename(mapper lambda x: x.upper(), axis = 1) are correct to change the original DataFrame df in a way that all the column names are cast to upper case.

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Answer:

For most uses you'll want your water heated to 120 F(49 C) In this example you'd need a demand water heater that produces a temperature rise and it will take about 2 hours

To achieve 100% statement coverage, additional tests are needed to cover different combinations of drink preferences (milk and sugar). For 100% decision coverage, tests should cover both the selection of drink type and the decisions related to adding milk and sugar.

a. Control Flow Diagram:

Start

|

V

Choose Drink Type (Hot or Cold)

|

V

IF Hot Drink

|   |

|   V

|   Ask for Milk Preference

|   |

|   V

|   IF Milk Required

|   |   |

|   |   V

|   |   Add Milk

|   |   |

|   |   V

|   |   Ask for Sugar Preference

|   |   |

|   |   V

|   |   IF Sugar Required

|   |   |   |

|   |   |   V

|   |   |   Add Sugar

|   |   |   |

|   |   |   V

|   |   V

|   V

|   Dispense Hot Drink

|

V

ELSE (Cold Drink)

   |

   V

   Dispense Cold Drink

|

V

End

b. Given the tests:

Test 1: Cold drink

Test 2: Hot drink with milk and sugar

Statement Coverage achieved: The statement coverage achieved would be 10 out of 15 statements (66.7%).

Decision Coverage achieved: The decision coverage achieved would be 2 out of 3 decisions (66.7%).

c. Additional tests for 100% statement coverage:

Test 3: Hot drink without milk and sugar

Test 4: Hot drink with milk only

Test 5: Hot drink with sugar only

Test 6: Hot drink without milk and without sugar

Additional tests for 100% decision coverage:

Test 7: Cold drink

Test 8: Hot drink with milk and sugar

Test 9: Hot drink without milk and sugar

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Answer:

\(W_{net} = - 1223 kJ\)

Explanation:

State 1:

\(P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg\)

State 2:

\(P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg\)

State 3:

\(P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263\)

\(u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg\)

State 4:

\(P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg\)

Calculate the network done for the cycle

\(W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ\)

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

The measurement of the elevated rims of the rifle barrel represents the width of the land. Groove depth is a gauge for how deep grooves are. Each groove's pitch, angle, and diameter are all related.

Rifling, spiral imprints found inside handgun and rifle barrels, is a common term. Rifling is divided between lands, which are raised areas, and grooves, which are recessed portions.

According to the ATF, "the process for determining barrel length is to measure from the closed bolt to furthermost end of the barrel," and "the overall length of a firearm is the distance between the muzzle of the barrel and rearmost component of the weapon measured on a line parallel to axis."

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Answer: gas station employees

Explanation:

their not using the road

The employees and workers of a gas station and small aircraft pilots are not considered as road users. Road users are the people such as drivers and pedestrians who uses the road.

Road users include both the drivers as well as the pedestrians and cyclists. Their behavior is the area which has by far the biggest potential for improving the road safety. Road users means a motorist, passenger, public transportation operator, or user, truck driver, bicyclist, motorcyclist, or pedestrian, including a person with disabilities as well.

The important characteristics of the road user include intelligences, skills, experience, knowledge and literacy. Proper knowledge of the vehicle characteristics, the driving practices, rules of roads and traffic behavior are necessary for the purpose of safe traffic operations.

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The option that is not a derived units is known to be option  E) kg.

The units that are known to be used for any form of derived quantities are said to be called the derived units.

They are:

Note that this unit are derived because they are known to be derived because they have to be solved for using different ways. The others such as kg m-3 is one that need to be derived to arrive at it.

Therefore, based on the above, one can say that The option that is not a derived units is known to be option  E) kg because it is one that cannot be derived,

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The following are derived units EXCEPT

Options

A) kg m-3

B)N

C)Ns

D)m3

E)kg

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Solution :

Given :

d = diameter of the wire of the spring = 0.074 in

  = 0.188 cm

R = mean radius of coil \($=\frac{0.85}{2}$\)

                                      = 1.08 cm

G = modulus of rigidity of carbon steel wire = \($79 \ GN/m^2$\)

L = length of wire of spring = 2 inch

∴ L = 5.08 cm

k = spring constant

k = axial load/ axial deflection

 \($k= \frac{Gd^4}{64R^3}$\) ..................(1)

This is taken from the Torsion equation,

\($\frac{T}{I_P}=\frac{\tau_{max}}{d/2}=\frac{G \theta}{2 \pi R}$\)

\($\frac{WR}{\frac{\pi}{32} d^4} = \frac{\tau_{max}}{d/2} =\frac{G \theta}{2 \pi R}$\)

∴ From equation (1)

\($k =\frac{79 \times 10^9 \times (0.188)^4 \times 10^{-8}}{64 \times (1.08)^3 \times 10^{-6}}$\)

\($k = 11334 \ N/m $\)

So the axial compression is \($ \delta =0.618 \ inch$\) = 1.57 cm

Axial load required W= k x δ

                                   = 11334 x 1.57

                                   = 177.95 N

Answer:

"cool"

Explanation: