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anastassius [24]
3 years ago
5

Consider a 1.2-m-high and 2-m-wide glass window with a thickness of 6 mm, thermal conductivity k = 0.78 W/m·K, and emissivity ε

= 0.9. The room and the walls that face the window are maintained at 25°C, and the average temperature of the inner surface of the window is measured to be 5°C. If the temperature of the outdoors is −5°C, determine (a) the convection heat transfer coefficient on the inner surface of the window, (b) the rate of total heat transfer through the window, and (c) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case?

Engineering
1 answer:
babunello [35]3 years ago
3 0

Answer:

Explanation:

The solutions to this question can be seen in the following screenshots taken from the solution manual.

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As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can
andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

2) 1.744 x 10 raised to power 9 firings

3) 2.18 x 10 raised to power 8  intake strokes

Explanation:

The step by step explanation is as shown in the attachment

8 0
3 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir
Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

                                                  = $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

            $=2 \pi (0.23)^2$

            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

     = 7043.42 N

3 0
3 years ago
A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min

4 0
3 years ago
Consider this example of a recurrence relation. A police officer needs to patrol a gated community. He would like to enter the g
SashulF [63]

Answer:

the police officer cruise each streets precisely once and he enters and exit with the same gate.

Explanation:

NB: kindly check below for the attached picture.

The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.

The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.

Here in this question, the door is the vertex and the road is the edge.

The information needed to detemine a Euler circuit and a Hamilton circuit is;

"the police officer cruise each streets precisely once and he enters and exit with the same gate."

Check attachment for each type of circuit and the differences.

7 0
3 years ago
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