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2 years ago

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Consider a fan located in a 3 ft by 3 ft square duct. Velocities at various points at the outlet are measured, and the average f

low velocity is determined to be 22 ft/s. Taking the air density to 0.075 lbm/ft3, estimate the minimum electric power consumption of the fan motor.

1 answer:

natulia [17]2 years ago

5 0

Answer:

minimum electric power consumption of the fan motor is 0.1437 Btu/s

Explanation:

given data

area = 3 ft by 3 ft

air density = 0.075 lbm/ft³

to find out

minimum electric power consumption of the fan motor

solution

we know that energy balance equation that is express as

E in - E out = $\frac{dE \ system}{dt}$ ......................1

and at steady state $\frac{dE \ system}{dt}$ = 0

so we can say from equation 1

E in = E out

so

minimum power required is

E in = W = m $\frac{V^2}{2}$ = $\rho A V \frac{V^2}{2}$

put here value

E in = $\rho A V \frac{V^2}{2}$

E in = $0.075 *3*3* 22 \frac{22^2}{2}$

E in = 0.1437 Btu/s

minimum electric power consumption of the fan motor is 0.1437 Btu/s

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The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic

Leokris [45]

Answer:

m' = 4948.38 kg/s

Explanation:

For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion

GIVEN THAT

Power = 100 MW

rate of Potential energy = (m')*g*h

100*10^6 = (m')*9.81*206

m' = 4948.38 kg/s

3 0

2 years ago

Read 2 more answers

On a cold winter day, wind at 55 km/hr is blowing parallel to a 4-m high and 10-m long wall of a house. If the air outside is at

koban [17]

Answer:

16.21 kW

Explanation:

Solution

Given that,

The velocity of wind = 55 km/hr

The length of the wall L = 10m

The height of the wall w = 4m

The surface temperature at wall Ts = 12° C

Temperature of air T∞ = 5°C

Now,

The properties of the air at atm and average film temperature =( 12 + 5)/2 = 8.5°C, which is taken from the air table properties.

k= 0.02428 W/m°C

v= 1.413 *10 ^⁻5

Pr =0.7340

Now,

Recall Reynolds number when air flow parallel to 10 m side

[ 55 * 1000/3600) m/s (10 m)/1.413 *10^⁻5 m²/s

Rel =1.081 * 10⁷

This value is greater than Reynolds number.

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

Nu =1.336 * 10 ^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (1.336 * 10 ^4)

h = 32.43 W/m°C

The heat transfer area of surface,

As = 40 m²

= ( 4 m) (10 m)

As = 40 m²

The rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (32.43 W/m²°C) (40 m) (12 - 5)°C

=9081 W

Q = 9.08 kW

When the velocity is doubled,

let say V = 110km/hr

The Reynolds number is

Rel = VL/v

= [110 * 100/3600) m/s] (10 m)/ 1.413 *10^⁻5 m²/s

Rel = 2.163 * 10 ^7

This value is greater for critical Reynolds number

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

[0.037 ( 2.163 * 10 ^7)^0.08 - 871] (0.7340)^1/3

Nu =2.344 * 10^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (2.384 * 10 ^4)

h= 57.88 W/m²°C

The heat transfer area of surface,

As = wL

= ( 10 m) (4 m)

As = 40 m²

he rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (57.88 W/m²°C) (40 m²) (12 - 5)°C

= 16,207 W

= 16.21 kW

3 0

3 years ago

Which system provides an easier way for people to communicate with a computer than a graphical user interface (GUI)?

Fed [463]

Answer:

C.

Explanation:

Natural language processing or NLP can be defined as an artificial intelligence that aids computers to understand, interpret, and manipulate human language. NLP is subfiled of many disciplines, for instance, artifical intelligence, linguistics, computer science, etc. NLP helps computers to interact with humans in their own language.

So, the easier way through which people can communicate with computers apart from GUI is NLP. Thus option C is correct.

7 0

2 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago

A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0

3 years ago