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Mars2501 [29]
2 years ago
5

A 0.91 m diameter corrugated metal pipe culvert (n = 0.024) has a length of 90 m and a slope of 0.0067. The entrance has a squar

e edge in a headwall. At the design discharge of 1.2 m3 /s, the tailwater is 0.45 m above the outlet invert. Determine the head on the culvert at the design discharge. Repeat the calculation for head if the culvert is concrete.

Engineering
1 answer:
NikAS [45]2 years ago
3 0

Answer:

HW=1.71m

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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The author states that chemical engineering is one of the most difficult and complex aspects of engineering. Why do you think th
marishachu [46]

Answer:

hope this helps

Explanation:

answers:

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8 0
3 years ago
Read 2 more answers
What was the role of the rotors in the enigma machine?.
Lisa [10]

Answer:

Implements a reordering of the letters of the alphabet.

Explanation: GIVE ME 5 STARS AND a HEART!!!                                      Those contacts are wired across the rotor so that each contact on the left connects to the contact on the right in some scrambled arrangement. Each rotor, therefore, implements a reordering of the letters of the alphabet, which mathematicians call a permutation.

3 0
2 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
2 years ago
Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.
levacccp [35]

Answer:

\dot W_{out} = 3863.98\,kW

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

h_{in} = 3353.1\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mixture)

h_{out} = 890.1\,\frac{kJ}{kg}

The power produced by the turbine is:

\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})

\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )

\dot W_{out} = 3863.98\,kW

8 0
3 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
3 years ago
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