A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?
Answers
Answer:
The geometry is treated as a two surface enclosure because the two surfaces have the same properties.
Let's take the base surface to be surface 1, while the side surfaces are surface 2.
Let's take the heat transfer expression:
Where,
= base temperature
= surface 2 temperature = 500K
= emissivity of surface 1 = 0.8
= emissivity of surface 2 = 0.5
= Area
= shape factor
Substituting figures in the equation, we have:
The base temperature is 523.038 k
Related Questions
Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.113V. Assuming that errors are due to random processes, how many of the readings are expected to be greater than 2.70V?
Consider a mixture of hydrocarbons that consists of 60 percent (by volume) methane, 30 percent ethane, and 10 percent propane. After passing through a separator, the mole fraction of the propane is reduced to 1 percent. the mixture pressure before and after the separation is 100 kPa. Determine the change in the partial pressures of all the constituents in the mixture.
Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.
Answers
Answer:
Concentration = 10.33 kg/m³
Explanation:
We are given;
Mass of solids; 10,000 kg
Volume; V = 440,000 L = 440 m³
Rate at which water is pumped out = 40,000 liter/h
Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³
Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;
200m³/440m³ = 5/11
Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;
Thus,
Amount left; = 10000 x (5/11) = 4545 kg
The concentration would be calculated by:
Concentration = amount left/initial volume
Thus,
Concentration = 4545/440 = 10.3 kg/m³
Answers
The most upstream process with issues would be a good location to start exploring the cause of the variance.
Chain processes
A manufacturing technique would be a specific procedure for generating a commodity.
Throughout manufacturing, a six sigma process has been utilized just to generate a product throughout which 99.99966 percent among all possibilities to produce certain aspects of a part seem to be likely toward being defect-free.
Thus the response above is correct.
Find out more information about chain processes here:
Answer: The furthest upstream process that has problems.
A process in manufacturing is a particular method used for producing a product.
A six sigma process is used in processing to produce a product that is 99.99966% of all opportunities to produce some feature of a part are statistically expected to be free of defects.
According to the rules of the six sigma process, when there's a defect, the best thing to do is investigate the furthest upstream process that has problems.
Answers
Answer:
A. 450 btu/h
Explanation:
We solve this problem by using this formula:
Q = U x TD x area
U = U value of used material
TD = Temperature difference = 60°
Q = heat loss
Area = 3x5 = 15
We first find U
R = 1/u
2 = 1/U
U = 1/2 = 0.5
Then when we put these values into the formula above, we would have:
Q = 0.5 x 15 x 60
Q = 450Btu/h
Therefore 450btu/h is the answer
Answers
Answer:
The change in entropy is found to be 0.85244 KJ/k
Explanation:
In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.
P1/T1 = P2/T2
T2/T1 = P2/P1
T2/T1 = 180 KPa/120KPa
T2/T1 = 1.5
Now, the change in entropy is given as:
ΔS = m(s2 - s1)
where,
s2 = Cv ln(T2/T1)
s1 = R ln(V2/V1)
ΔS = change in entropy
m = mass of CO2 = 3.2 kg
Therefore,
ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]
Since, V1 = V2, therefore,
ΔS = mCv ln(T2/T1)
Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K
Therefore,
ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)
ΔS = 0.85244 KJ/k
Answers
Answer:
B) gate-source junction is reverse-biased
Explanation:
FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow" in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".
In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".
Answers
Answer:
The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Explanation:
The distance that the truck starts slowing down = 80 ft from the stop sign
Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.
u = initial velocity of the truck = 40 mph = 58.667 ft/s
v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)
x = horizontal distance covered during the deceleration
a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration
v² = u² + 2ax
0² = 58.667² + 2(-12)(x)
24x = 3441.816889
x = 143.41 ft
143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.
Corrected Question:
A truck traveling at 40 mph is approaching a stop sign. At time t₀ and at a distance of 80 ft, the truck begins to slow down by decelerating at 12 ft/s2, will the truck be able to stop in time?
Answer:
The truck will not be able to stop in time.
Explanation:
==> First lets convert all variables to SI units
1 mph = 0.45m/s
40mph = 40 miles per hour = 40 x 0.45 m/s
40mph = 18m/s
1 ft = 0.3048m
80 ft = 80 x 0.3048m
80 ft = 24.38m
Also;
12ft/s² = 12 x 0.3048m/s²
12ft/s² = 3.66m/s²
==> Now, consider one of the equations of motion as follows;
v² = u² + 2as -----------------(i)
Where;
v = final velocity of motion
u = initial velocity of motion
a= acceleration/deceleration of motion
s = distance covered during motion
Using this equation, lets calculate the distance, s, covered during the acceleration;
We know that;
v = 0 [since the truck comes to a stop]
u = 40mph = 18m/s
a = -12ft/s² = -3.66m/s² [the negative sign shows that the truck decelerates]
Substitute these values into equation (i) as follows;
0² = 18² + 2 (-3.66)s
0 = 324 - 7.32s
7.32s = 324
s =
s = 44.26m
The distance from where the truck starts decelerating to where it eventually stops is 44.26m which is past the stop sign (which is at 80ft = 24.38m). This means that the truck stops, 44.26m - 24.38m = 19.88m, after the stop sign. Therefore, the truck will not be able to stop in time.