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s344n2d4d5 [400]
3 years ago
14

The atmosphere within a room is at 70 °F dry-bulb temperature, 50 percent degree of saturation, and 14.696 psia pressure. The in

side surface tempera- ture of the windows is 40 °F. Will moisture con- dense out of the air upon the window glass?

Engineering
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

Given that the temperature of the window is below the dew point it will condensate.

Explanation:

A psychrometric chart (like the one attached) will give you the information needed. This chart is for 14.696 psia.

On the bottom horizontal axes you have the dry-bulb temperature, in this case 70°F, going up from this point you can reach the 50% relative humidity curve (red point on chart), going horizontally from this point to the 100% relative humidity you get the dew point temperature (the point at which moisture will condensate) (blue point on chart). In this case the dew point is 50°C. Given that the temperature of the window is below the dew point it will condensate.

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Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
4 years ago
What are the standard procedures involved in the fixing/securing of cables?​
Sladkaya [172]

<u>Cable should be pre-cut and hung suspended for 48 hours to develop its most natural set and lay prior to installation.</u>

<u>Cable should be installed with, not against, its natural set. ... </u>

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4 0
2 years ago
In modern construction, studs are usually cut to exact length at the mill. They are designated by the letters
ehidna [41]

Answer:

P.E.T

Explanation:

Stud is a short form of stud wall which are typically vertical framing members in a wall of a building. Wall studs are available in standard lengths but since the height of ceiling vary, sometimes one has to cut the stud walls on site to fit. Therefore, there are pre-cut studs which are more efficient  in implementation. These are usually referred as Precision-End Trimmed studs, (P.E.T). Therefore, stud walls which are cut to exact length at the mill for modern construction are designated by letters P.E.T

3 0
4 years ago
A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an
konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

P_4=50\ kPa

h_4=0.7(2304.7)+340.5

    =1953.83 \ KJ/Kg

and,

P_3=15 \ mPa

h_3=hg

    =2610.8 \ KJ/Kg

s_3=sg

    =5.3108 \ KJ/Kgh

At 45,

⇒ x_{45} = \frac{5.3108-1.0912}{6.5019}

          =0.66

At P_4=50 \ Kpa,

h_f=340.54

or,

V_f=0.001030 \ m^3/Kg

then,

⇒ h_2=340.54+0.001030(15\times 10^{3}-50)

        =355.94 \ kJ/kg

hence,

The isentropic efficiency of turbine will be:

⇒ n_T=\frac{h_3-h_4}{h_3-h_{45}}

         =\frac{2610.8-1953.83}{2610.8-1836.26}

         =84.818 (%)

The thermal efficiency of cycle will be:

⇒ n_C=\frac{W_T-W_P}{2_{in}}

         =\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}

         =28.45 (%)  

4 0
3 years ago
A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction.
Jlenok [28]

Solution :

Finding the cohesion of the soil(c) using the relation:

$c = \frac{q_u}{2}$

Here, $q_u$ is the unconfined compression strength of the soil;

$c = \frac{800}{2}$

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction $0^\circ$

    $N_c = 5.14$

   $N_q = 1.0$

   $N_r = 0$

Therefore,

$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$

     =  2386 psf

∴ Allowable bearing capacity $q_{a} = \frac{Q_{allow}}{A}$

                                                     $=\frac{30}{B^2}$

∴ $q_a = \frac{q_{ult}}{F.O.S}$

  $\frac{30}{B^2} = \frac{2386}{3}$

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 $=0.04 \ ft^2$

7 0
3 years ago
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