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Burka [1]
3 years ago
12

Match the following parts of a crane

Engineering
1 answer:
trasher [3.6K]3 years ago
5 0
It is auxillary sorry i couldn’t help it happens to the best of us
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Limestone scrubbing is used to remove SO2 in a flue gas desulfurization (FGD) system. Relevant reactions are given below. A lime
lord [1]

Answer:

hello some data related to your question is missing attached below is the missing data

answer : 63700 Ib/hr

Explanation:

Given data :

Limestone mix : consists of  94% CaCO3  and 6% inert material

Actual feed rate = 36,000 Ib/hr

SO2 in flue gas = 20,314 Ib/hr

FGD efficiency = 97%

resulting sludge contains 58% solids

<u>Calculate the Total sludge production rate </u>

First : determine  SO2 removed in sludge

 = 0.97 * 20314

 = 19704.58 Ib/hr

next : moles of SO2 removed

= 19704.58 / 64 Ib/ Ib mol

= 307.88 Ib mol / hr

also  moles of CaSO3 produced = 307.88 Ib mol / hr

mass of CaSO3 = 307.88 * 120 = 36946.09 Ib/hr

Therefore Total sludge production rate

= 36946.09 / 0.58

= 63700.15  Ib/hr

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7 0
2 years ago
The inspector should inspect insulation in unfinished spaces, including attics, _____ and foundation areas.
RoseWind [281]

Answer:

basements

Explanation:

6 0
2 years ago
A beam has been fixed to the floor by the pin at B and the roller at A as shown in figure 1 below.​
ahrayia [7]
What figure below???
3 0
3 years ago
How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?
nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \frac{N}{kg} · 1.50m

   = 35.316 \frac{N}{m}

   = 35.316 W

4 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
2 years ago
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