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svetoff [14.1K]
3 years ago
8

A heated plate is built into the wall of a large room. The room is maintained at 300 K. The exposed surface of the plate is main

tained at 400 K by a heater attached to the back of the plate. Assume the sides of the plate and back of the heater are perfectly insulated. The exposed surface of the plate (area 0.1 m^2 ) has an emissivity of 0.3 and a convection heat transfer coefficient of 6 W/m^2K with the room and air temperature of 300 K. Show resistance circuit for this problem and draw a picture to show relevant heat modes on system to receive full points. Calculate the following:
a. Find the net radiant heat flux from the plate surface, q1"=____
b. The required heater power, P=______
c. The radiosity of the plate surface, J1=
Engineering
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

B

Explanation:

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Provide two programming examples in which multithreading provides better performance than a single-threaded solution. Provide on
Kobotan [32]

Answer:

I dont kno

Explanation:

Im so sorry

5 0
2 years ago
A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted
Pavlova-9 [17]

Answer:

a) The minimum acceptable value is 387.5 HV using Vickers hardness test.

b) The minimum acceptable value is 39.4 HRC using Rockwell C hardness test.

Explanation:

To get the tensile strength of a material from its hardness, we multiply it by an empirical constant that depends on things like yield strength, work-hardening, Poisson's ratio and geometrical factors. The incidence of cold-work varies this relationship.

According to DIN 50150 (a conversion table for hardness), the constant for Vickers hardness is ≈ 3.2 (an empirical approximate):

\mbox{Tensile strength}=HV*3.2\\\\HV  = \frac{\mbox{Tensile strength}}{3.2} =\frac{1240}{3.2}=387.5

According to DIN 50150, the constant for Rockwell C hardness test is ≈31.5 around this values of tensile strength:

\mbox{Tensile strength}=HRC*31.5\\\\HRC  = \frac{\mbox{Tensile strength}}{31.5} =\frac{1240}{31.5}=39.4

8 0
3 years ago
Worth 20 points! Please help ASAP!
gulaghasi [49]

Answer:

I am in 6th grade, why are high school things popping up??

Explanation:

8 0
3 years ago
Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase
ludmilkaskok [199]

Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite (\% V_{Gr}) is determined by the following expression:

\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%

\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%

Where:

V_{Gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{Fe} - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}

V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}

Where:

m_{Gr}, m_{Fe} - Masses of the graphite and ferrite phases, measured in grams.

\rho_{Gr}, \rho_{Fe} - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%

\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

m_{Gr} = \frac{2.5}{100}\times (100\,g)

m_{Gr} = 2.5\,g

m_{Fe} = 100\,g - 2.5\,g

m_{Fe} = 97.5\,g

If m_{Gr} = 2.5\,g, m_{Fe} = 97.5\,g, \rho_{Fe} = 7.9\,\frac{g}{cm^{3}} and \rho_{Gr} = 2.3\,\frac{g}{cm^{3}}, the volume percent of graphite is:

\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%

\% V_{Gr} = 91.906\,\%

The volume percent of graphite is 91.906 per cent.

6 0
2 years ago
A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co
podryga [215]

Answer:

a) T_{H} = 1.967\,^{\circ}F, b) COP_{R} = 9.105, c) T_{H} = 115.934\,^{\circ}F, d) COP_{R} = 6.995, e) T_{H} = 25.129\,^{\circ}C

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}

10 = \frac{419.67\,R}{T_{H}-419.67\,R}

The temperature of the hot reservoir is:

10\cdot T_{H} - 4196.7 = 419.67

T_{H} = 461.637\,R

T_{H} = 1.967\,^{\circ}F

b) The coefficient of performance is:

COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}

COP_{R} = 9.105

c) The temperature of the hot reservoir can be determined with the help of the following relation:

\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}

\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}

5 = \frac{479.67\,R}{T_{H}-479.67\,R}

5\cdot T_{H} - 2398.35 = 479.67

T_{H} = 575.604\,R

T_{H} = 115.934\,^{\circ}F

d) The coefficient of performance is:

COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}

COP_{R} = 6.995

e) The temperature of the cold reservoir is:

8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}

8.9\cdot T_{H} - 2386.535 = 268.15

T_{H} = 298.279\,K

T_{H} = 25.129\,^{\circ}C

8 0
2 years ago
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