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3 years ago

Define volume flow rate of air flowing in a duct of area A with average velocity V.

Engineering

1 answer:

KATRIN_1 [288]3 years ago

6 0

Explanation:

Step1

Volume flow rate is the rate of change of volume of fluid that is flowing in the duct of pipe per unit time. It is measure in m³/s or l/s. Volume flow rate is very important parameter in fluid analysis.

Step2

For the given duct, the volume flow rate is the product of average velocity to the cross section area of duct.

Expression for volume flow rate is given as follows:

Q=AV

Here, Q is the flow rate, A is area of the duct and V is the average velocity of flowing fluid.

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How can feeding plant crops to animals be considered an efficient use of those crops?

stira [4]

Answer:

Most of Our Grain is Being Consumed by Animals, Not Humans. How the Planting of Crops Used to Feed Livestock is Contributing to Habitat Destruction.

Explanation:

The major feed grains are corn, sorghum, barley, and oats.

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3 years ago

The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.

Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL $_{max}$ / D = 57.915

we substitute

(4×0.005×L $_{max}$ ) / 4 = 57.915

0.005L $_{max}$ = 57.915

L $_{max}$ = 57.915 / 0.005

L $_{max}$ = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0

2 years ago

Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure

iris [78.8K]

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3

P5/P6 = P7/P8 = √9 =3

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg

Calculate the net work done

Wnet = (Wturb)out - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

4 0

3 years ago

A compound sliding miter saw can be used to make

Sunny_sXe [5.5K]

Answer:

Explanation:

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2 years ago

A driver takes 3.5 s to react to a complex situation while traveling at a speed of 60 mi/h. How far does the vehicle travel befo

victus00 [196]

Answer:

x = 93.8 m.

Explanation:

During the entire the reaction time interval, the vehicle continues moving at the same speed that it was moving, i.e., 60 mi/hr.

In order to calculate the distance in meters, travelled at that speed, it is advisable first to convert the 60 mi/hr to m/seg, as follows:

$60 mi/hr = 60*\frac{1hr}{3,600s}*\frac{1,605m}{1mi} = 26.8 m/s$

Applying the definition of average velocity, we can solve for Δx, as follows:

Δx = 26.8 m/s* 3.5 s = 93.8 m

7 0

3 years ago