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3 years ago

8

Two containers initially contain dry air at 40 °C. Water is added to both containers so that the relative humidity is 30% in con

tainer #1, and 80% in container #2. The wet bulb temperature in container #1 is _________ the wet bulb temperature in #2.
a. less than
b. greater than
c. the same as

1 answer:

enot [183]3 years ago

8 0

Answer:

A) Wet bulb temperature of #1 is less than that of #2

Explanation:

This can be gotten from pinpointing the states of the two containers on a psychometric chart.

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g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What

earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ( $v$ ), measured in meters per second, is determined by the following expression:

$v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}$ (1)

Where:

$\dot V$ - Flow rate, measured in cubic meters per second.

$D$ - Diameter, measured in meters.

If we know that $\dot V = 0.01\,\frac{m^{3}}{s}$ and $D = 0.05\,m$ , then the flow velocity is:

$v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}$

$v \approx 5.093\,\frac{m}{s}$

The density and dinamic viscosity of the glycerin at 20 ºC are $\rho = 1260\,\frac{kg}{m^{3}}$ and $\mu = 1.5\,\frac{kg}{m\cdot s}$ , then the Reynolds number ( $Re$ ), dimensionless, which is used to define the flow regime of the fluid, is used:

$Re = \frac{\rho\cdot v \cdot D}{\mu}$ (2)

If we know that $\rho = 1260\,\frac{kg}{m^{3}}$ , $\mu = 1.519\,\frac{kg}{m\cdot s}$ , $v \approx 5.093\,\frac{m}{s}$ and $D = 0.05\,m$ , then the Reynolds number is:

$Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }$

$Re = 211.230$

A pipeline is in turbulent flow when $Re > 4000$ , otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ( $f$ ), dimensionless, is determined by the following expression:

$f = \frac{64}{Re}$

If we get that $Re = 211.230$ , then the friction factor is:

$f = \frac{64}{211.230}$

$f = 0.303$

The friction factor is 0.303.

4 0

2 years ago

You are a designer of a new processor. You have to choose between two possible implementations (called M1 and M2) of the same ar

Kaylis [27]

Answer:

A ) CPI : M1 = 2.4 , M2 = 2.65

B ) MIPS : M1 = 1083, M2 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

Explanation:

A) The CPI for each machine

CPI = ( Total number of execution cycles ) / ( instruction counter executed )

For Machine 1 ( M1 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2.5 , number of times C was executed = 2.5, Number of times D was executed = 1. and this was based on the frequency given above

hence CPI for M1 =[ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] / 10

CPI for M1 = 2.4

For Machine 2 ( M2 )

we have to make some assumptions : number of instructions = 10

number of times A was executed = 4 , Number of times B was executed = 2. number of times C was executed = 1.5, Number of times D was executed = 2.5 times. and this was based on the frequency given above

Hence CPI for M1 = [ ( 2 * 4 ) + ( 2 * 2 ) + ( 3 * 1.5 ) + ( 4 * 2.5 ) ] / 10

CPI for M2 = 2.65

B ) Calculate the native MIPS ratings for M1 and M2

MIPS = ( instruction counts ) / ( Execution time * 10^6 )

For M1

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.6 Ghz

first we calculate the total execution time which is equal to :

= [ ( 1 * 4 ) + ( 3 * 2.5 ) + ( 3 * 2.5 ) + ( 5 * 1 ) ] * 0.3846 * 10 ^-9

= 9.2304 * 10 ^-9 secs

therefore the MIPS for M1

= 10 / ( 9.2304 * 10^-9 ) * 10^6 = 1083

For M2

Assumptions : number of instructions executed = 10

each clock cycle = 0.3846 * 10^-9. frequency = 2.8 Ghz

first we calculate the total execution time which is equal to :

= [ (2*4) + (2*2) + (3 * 1.5 ) + ( 4 * 2.5 ) ] * 0.3846 * 10^-9 = 9.4631 * 10^-9 secs

therefor the MIPS for M2

= 10 / ( 9.4631*10^-9) * 10^6 = 1056

C ) The machine that has a better performance based on MIPS is M1 and this is by 27 million number of instructions per sec

8 0

3 years ago

A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 5

goblinko [34]

Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

7 0

3 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

If an object has the same number of positive and negative charges, its electrical charge is

N76 [4]

When an object has the same number of positive and negative charges, its electrical charge will become neutral.

What is an electric charge?

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

Now when the two equal magnitude charges with opposite natures come together they become neutral.

To know more about charges follow

brainly.com/question/24391667

#SPJ4

7 0

2 years ago