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Olin [163]
3 years ago
8

Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the diameter from 10 cm to 2 cm. Calculate the water’s vel

ocity leaving the nozzle and the flow rate.

Engineering
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

See the pictures attached

Explanation:

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Answer:

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Explanation:

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6 0
2 years ago
Does the music contribute to the performance of its production?
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Answer:

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6 0
2 years ago
Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–
Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

C_p = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]

Substitute all parameters in the equation

Δs = 5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]

Δs = 5 kg × 0.14629 kJ/kg.K

    = 0.73145 kJ/K

b)

Δs = m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

           T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = 5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]

    = 5 kg (0.13795 kJ/kg.K)

    = 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0
3 years ago
A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons
antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given  :

PV^{1.3}=C

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}

where r_{c} is compression ratio = \frac{v_{c}+v_{s}}{v_{c}}

           r_{c} = 1+\frac{v_{s}}{v_{c}}

where v_{c} is compression volume

           v_{s} is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, v_{30}=v_{c}+0.7v_{s}

and at 70% compression, 30% of the swept volume remains

    ∴    v_{70}=v_{c}+0.3v_{s}  

We know,

\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}

\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}

\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\

1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}

v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}

0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}

0.2218v_{s} = 0.594v_{c}

v_{c}=0.3734 v_{s}

∴   r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}

Therefore, compression ratio is r_{c} = 3.678

Now efficiency, \eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}

 \eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}

 \eta =0.32342 , this is the ideal efficiency

Therefore actual efficiency, \eta_{act} =0.5\times \eta _{ideal}

           \eta_{act} =0.5\times 0.32342

           \eta_{act} =0.1617

Therefore total power required = 1 kW x 3600 J

                                                    = 3600 kJ

∴ we know efficiency, \eta=\frac{W_{net}}{Q_{supply}}

Q_{supply}=\frac{W_{net}}{\eta _{act}}

Q_{supply}=\frac{3600}{0.1617}

Q_{supply}=22261.78 kJ

Therefore fuel required = \frac{22261.78}{45000}

                                        = 0.4947 kg/hr      

5 0
3 years ago
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