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Arte-miy333 [17]
3 years ago
13

Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The cha

nnel is having a bed slope of 1 in 2000. Take Chezy's coefficient, C=55
Engineering
1 answer:
Elza [17]3 years ago
5 0

Answer:

V = 1.5062 m/s

Explanation:

look to the photos

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A 200-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is t
Neko [114]

Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

total strain of step process = 0.40 + 0.29 = 0.69 units

SO TRUE STRAIN IS ADDITIVE.

4 0
2 years ago
A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al
marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0
3 years ago
The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in
amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be Q\frac{gallons}{minute}

Now replacing the gallon and the minute by the above values we get

Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}

Thus Q'=0.631Q\frac{liters}{second}

Now since we know that 1 liter = 0.0353ft^{3}

Using this in above relation we get

Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0
3 years ago
How is the energy harnessed and converted into useful energy?
garri49 [273]

Answer:

1. How energy is harnessed?  

Another way to tap solar energy is by collecting the sun's heat. Solar thermal power plants use heat from the sun to create steam, which can then be used to make electricity. On a smaller scale, solar panels that harness thermal energy can be used for heating water in homes, other buildings, and swimming pools.

2. How is solar energy converted into useful energy?

Solar panels convert the sun's light into usable solar energy using N-type and P-type semiconductor material. When sunlight is absorbed by these materials, the solar energy knocks electrons loose from their atoms, allowing the electrons to flow through the material to produce electricity.

Explanation:

hope it helps, please mark as brainliest

6 0
3 years ago
Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu
Sonja [21]

Answer:

a

The rate of work developed is \frac{\r W}{\r m}= 300kJ/kg

b

The rate of entropy produced within the turbine is   \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

Explanation:

     From  the question we are told

          The rate at which heat is transferred is \frac{\r Q}{\r m } = -  30KJ/kg

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is c_p = 1.1KJ/kg \cdot K

          The inlet temperature is  T_1 = 970K

          The outlet temperature is T_2 = 670K

           The pressure at the inlet of the turbine is p_1 = 400 kPa

          The pressure at the exist of the turbine is p_2 = 100kPa

           The temperature at outer surface is T_s = 315K

         The individual gas constant of air  R with a constant value R = 0.287kJ/kg \cdot K

The general equation for the turbine operating at steady state is \

               \r Q - \r W + \r m (h_1 - h_2) = 0

h is the enthalpy of the turbine and it is mathematically represented as          

        h = c_p T

The above equation becomes

             \r Q - \r W + \r m c_p(T_1 - T_2) = 0

              \frac{\r W}{\r m}  = \frac{\r Q}{\r m} + c_p (T_1 -T_2)

Where \r Q is the heat transfer from the turbine

           \r W is the work output from the turbine

            \r m is the mass flow rate of air

             \frac{\r W}{\r m} is the rate of work developed

Substituting values

              \frac{\r W}{\r m} =  (-30)+1.1(970-670)

                   \frac{\r W}{\r m}= 300kJ/kg

The general balance  equation for an entropy rate is represented mathematically as

                       \frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma  = 0

          =>          \frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)

    generally (s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]

substituting for (s_1 -s_2)

                      \frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} +  c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]

                      Where \frac{\sigma}{\r m} is the rate of entropy produced within the turbine

 substituting values

                \frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]

                    \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

           

 

                   

   

5 0
3 years ago
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