Answer:
Look at the load capacity chart in the cab
Explanation:
A crane can be defined as a large, tall metallic machine or equipment that is designed with a long horizontal arm (jib) used for the lifting and movement of very heavy objects through the air. They're usually designed to be operated by a human operator, who typically uses a remote controller and a beam to control the direction of movement of the crane.
Due to the fact that cranes are used for lifting and moving very heavy objects, they are powered by an internal combustion engine and electric motors.
Furthermore, all cranes have the maximum capacity of load they're able to lift at a particular point in time. In order to determine the rated or gross capacity of a crane and maintain safe operation, it is important to check its load capacity chart. The actual load a crane can lift is its net capacity and it must not be exceeded at any time, so as to avoid structural failure or overturning of the crane.
Hence, the best way to find the load capacity of a crane is to look at the load capacity chart in the cab.
The load can include rigging components, hooks, blocks, and other lifting equipment that is considered part of the load. The maximum load capacity of the crane is about 18 metric tons.
The load carried has made the crane in such as way that it can carry the load to certain levels. Hence they also provide the chart which says about the types and quantities of load.Hence the option C look at the chart is correct.
Learn more about the way to find the load capacity of a crane.
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How are engine bearings lubricated?
Answer:
Engine bearings are lubricated by motor oils constantly supplied in sufficient amounts to the bearings surfaces. Lubricated friction is characterized by the presence of a thin film of the pressurized lubricant
Explanation:
What value of filter capacitor is required to produce 1% ripple factor for a full wave rectifier having load resistance of 1.5kohm? Assume rectifier produces peak output of 18v
Answer:
Explanation:
A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked materials to be dried. The base is maintained at a temperature of 370K, while the dome of the dryer is maintained at 1000 K. If both surfaces behave as blackbody, determine the drying rate per unit length experienced by the wet materials.
Answer:
0.0371 kg/s.m
Explanation:
From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)
Assuming the base surface of both ends of the cylinder is denoted by:
[tex]A_1 \ and \ A_2[/tex]
Thus, using the summation rule, the view factor [tex]F_{11[/tex] and [tex]F_{12[/tex] is as follows:
[tex]F_{11}+F_{12}=1[/tex]
Let assume the surface (1) is flat, the [tex]F_{11} = 0[/tex]
Now:
[tex]0+F_{12}=1[/tex]
[tex]F_{12}=1[/tex]
However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder [tex]A_2[/tex] to the flat base surface [tex]A_1[/tex]; we have:
[tex]A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}[/tex]
Suppose, we replace DL for [tex]A_1[/tex] and
[tex]A_2[/tex] = [tex]\dfrac{\pi D}{2}[/tex]
Then:
[tex]F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64[/tex]
Now, we need to employ the use of energy balance formula to the dryer.
i.e.
[tex]Q_{21} = Q_{evaporation}[/tex]
But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:
[tex]Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)[/tex]
where;
[tex]\sigma = Stefan \ Boltzmann's \ constant[/tex]
[tex]T_1 = base \ temperature[/tex]
[tex]T_2 = temperature \ of \ the \ dome[/tex]
∴
[tex]Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m[/tex]
Recall the energy balance formula;
[tex]Q_{21} = Q_{evaporation}[/tex]
where;
[tex]Q_{evaporation} = mh_{fg}[/tex]
here;
[tex]h_{fg}[/tex] = enthalpy of vaporization
m = the water mass flow rate
∴
[tex]83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}[/tex]
The drying rate per unit length is 0.037 kg/S.m
Given data;
Base temperature (T1) = 370KTemperature of the dome (T2) = 1000KF[tex]_1_2[/tex] = 1.5mD = 1.5mBoltzmann's constant (δ) = [tex]5.67 * 10 ^-^8 W/m^2.K^4[/tex]From the attached diagram, the surface 1 is flat, it is a view factor, f[tex]_1_1[/tex] = 0.
Applying summation rule and solving the view factor from the base surface A[tex]_1[/tex] to the cylindrical dome A[tex]_2[/tex].
[tex]f_1_1+f_1_2=1[/tex]
Put F[tex]_1_2=0[/tex]
[tex]0+f_1_2=1[/tex]
This makes [tex]f_1_2=1[/tex]
Applying reciprocal rule and solving the view factor from the cylindrical dome A[tex]_2[/tex] to the base surface A[tex]_1[/tex].
[tex]A_2F_2_1=A_1F_1_2\\F_2_1=(\frac{A_1}{A_2})F_1_2[/tex]
Where A is the area of the surface.
Substitute DL for A[tex]_1[/tex] and [tex]\frac{\pi D}{2}[/tex] for A[tex]_2[/tex]
[tex]F_2_1 = \frac{DL}{(\frac{\pi D}{2}})L *1 = \frac{2 }{\pi } =2/3.14 = 0.64[/tex]
Using the energy balance equation to the dryer,
[tex]Q_2_1=Q_e_v_a_p[/tex]
Let's calculate the radiation heat exchange between the dome and the base surface per unit length by using the equation below
[tex]Q_2_1=F_2_1A_2[/tex]δ[tex](T_2^4-T_1^4)[/tex]
[tex]Q_2_1= F_2_1 * \frac{\pi D}{2}[/tex]δ[tex](T_2^4-T_1^4)[/tex]
substitute the respective values into the equation
[tex]Q_2_1=0.64*(\frac{\pi }{2}*1.5)*5.67*10^-^8*(1000^4-370^4)\\Q_2_1=8.3899.15W/m[/tex]
Mass flow rateLet's calculate the mass flow rate of water using the amount of heat required for drying up.
[tex]Q_2_1=Q_e_v_a_p\\Q_e_v_a_p=mh_f_g\\[/tex]
where [tex]h_f_g= 2257*10^3J/kg[/tex]
and this is the enthalpy of vaporization and mass flow rate of water.
[tex]83899.15=m*2257*10^3\\m=0.037kg/S.m[/tex]
The drying rate per unit length is 0.037kg/S.m
Learn more about mass flow rate here
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For the siphon shown in Figure, determine the flowrate out of the siphon and the absolute
pressure at the crest of the siphon assuming that there is no losses through the pipe.
2.00 m
Del
50-mm-diameter
Oil (s.g. =0.82)
5.00 m
Suppose a population of rabbits is introduced to an environment that has hot summers and extremely cold winters. Every winter, many rabbits die because of the cold and the lack of food. The rabbits have a range of ear sizes. Small Ears Medium Ears Large Ears Rabbits lose a lot of their body heat through their ears. Every winter, more of the large-eared rabbits die than the others. Every summer, more of the short-eared rabbits die than the others. At first, there are an equal number of rabbits with each size of ears. After several years, what will the frequency of ear sizes in the population be like?
Answer:
medium sized ears
Explanation:
