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IceJOKER [234]
3 years ago
7

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

the air in and forces it through the resistors where it is heated. If the density of air is 1.20 kg/m3 at the inlet and 0.955 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the dryer.
Engineering
1 answer:
Inessa05 [86]3 years ago
6 0

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65

the percent increase in the velocity of air is 25.65%

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3 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

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Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

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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

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