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ArbitrLikvidat [17]

2 years ago

4

Carmen is having her basement refinished. She notices that the flooring was installed hastily and is uneven. Who should she talk

to about the problem?

2 answers:

Mamont248 [21]2 years ago

6 0

A contractor that works on housing projects

lianna [129]2 years ago

4 0

Answer:

the answer is foreman (lmk if it's wrong)

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Refrigerant-134a enters a diffuser steadily as saturated vapor at 600 kPa with a velocity of 160 m/s, and it leaves at 700 kPa a

Zina [86]

Answer:

a) V_2 = 82.1 m/s

b) m = 0.298 Kg/s

Explanation:

from A-11 to A-13 we have the following data

P_1 = 600 kpa

V_1 = 0.033925 m^3/kg

h_1 = 262.52 kJ/kg

P_2 = 700 kpa

V_2 = 0.0313 m^3/kg

T_2 = 40°C = 313K

h_2 = 278.66 kJ/kg

Now, from the conversation of mass,

A_2*V_2/u_2 = A_1*V_1/u_1

V_2 = A_1/A_2*u_2/u_1*V_1

V_2 = A_1/1.8*A_1 * 0.0313 /0.033925*160

V_2 = 82.1 m/s

now from the energy balance equation

E_in = E_out

Q_in + m(h_1 + V_1^2/2) = m(h_2 + V_2^2/2)

m = 0.298 Kg/s

4 0

2 years ago

Read 2 more answers

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt

= $\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}$ watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = $\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}$

15999.86 $=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}$

$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

4 0

2 years ago

A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and th

Inessa05 [86]

Answer:

11 hours approximately

Explanation:

We are to calculate mean cell residence time mcrt

= Mass of solid in reactor/mass of solid wasted in a day

Q = Qe + We

Q = 2.5

Qw = 0.5

Qe = 2.5 - 0.5

= 2 MGD

10⁶/svi

= 10⁶/125

= 8000

X = 3500

Xe = 20mg/

1MGD = 0.1337million

Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337

= 262500000/[4000000000+40000000} x 0.1337

= 262500000/574800000

= 0.45668 days

= 0.45668 x 24 hours

= 10.9603 hours

Approximately 11 hours

3 0

2 years ago

Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the

defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

$P = \frac{weight}{base}$

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

$weight = \delta * v * g$

Meanwhile, the volume of a column is the area of the base multiplied by the height:

$V = base * h$

Replacing:

$P = \frac{\delta * base * h * g}{base}$

The base cancels out, so:

$P = \delta * h * g$

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

$\Delta P = \rho * g * (h1 - h2)$

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

$\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa$

4 0

2 years ago

A three-phase motor rated 25 hp, 480 V, operates with a power factor of 0.74 lagging and supplies the rated load. The motor effi

coldgirl [10]

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

= 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

3 0

2 years ago