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kolezko [41]
3 years ago
14

A preheater involves the use of condensing steam at 100o C on the inside of a bank of tubes to heat air that enters at I atm and

25o C. The air moves at 5 m/s in cross flow over the tubes. Each tube is 1 m long and has an outside diameter of 10 mm. The bank consists of 196 tubes in a square, aligned array for which ST = SL = 15 mm. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?
(b) Repeat the analysis of part (a), but assume that the tubes have a staggered arrangement with ST = 15 mm and SL = 10 mm.
Engineering
1 answer:
Anna35 [415]3 years ago
8 0

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
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You have the assignment of designing color codes for different parts. Three colors are used on each part, but a combination of t
Masja [62]

Answer:

7 available

Explanation:

Since 3 colors are available r = 3

Total combination = 35

nCr = 35 ---1

nCr = n!/(n-r)!r!---2

We put equation 1 and 2 together

n-1)(n-2)(n-3)!/n-3)! = 35x 3!

We cancel out (n-3)!

(n-1)(n-2) = 210

7x6x5 = 210

nC3 = 35

7C3 = 35

So If there are 35 combinations, 7 colors are available.

Thank you!

3 0
2 years ago
Propane burns at an equivalence ratio (ER) of 0.6, determine actual air-fuel ratio. If excess air is 5%, what will be the actual
dimaraw [331]

Answer:

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

Explanation:

Equivalence ratio = 0.6

Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR

combustion reaction of propane is

C_3H_8+ 5O_2 ----->3CO_2+4H_2O

From above reaction,  1 mole of propane, from the reaction, 5  moles of oxygen required,  

we know that air contains 21% O_2 and 79% N_2,

Therefore, moles of air based on stoichiometry = \frac{5}{0.21} = 23.81

Theoretical air to fuel ratio = \frac{23.81}{1} = 23.81

Given\frac{AFR}{SFR} = 0.6

Actual Air Fuel Ratio = 23.81\times 0.6 = 14.3

when 5% excess air is supplied, moles of air supplied/moles of fuel = 23.81\times 1.05 =25

6 0
3 years ago
¿Qué aspecto importante debemos conocer en el ámbito del
german
I would help but don’t speak that sorry
8 0
2 years ago
Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB Originally the cable is unstretc
AnnZ [28]

Answer:

e_{ab} =  4.18*10^(-3)

Explanation:

This question will be solved with the help of diagram (see attachment)

Given:

Δ∅ = 0.50 degrees (correction)

BC = 800 mm

AC = 600 mm

Solution:

We use coordinate system with point C as origin (0,0)

Hence,

Point A = (600,0)

Point B = (0,800)

The change in length or displacement can be calculated BB' :

BB' = BC * tan (Δ∅) = (800 mm) * tan (0.5) = 6.9815 mm

Hence, Point B' = (-6.9815,800) and we calculate distance A and B

AB = \sqrt{600^2 + 800^2} = 1000 mm

We calculate distance A and B'

AB' = \sqrt{(600 - (-6.9815))^2 + (0-800)^2} = 1004.2 mm

Normal Strain in AB is:

e_{ab} = \frac{AB' - AB}{AB} \\\\= \frac{1004.2 - 1000}{100}\\\\= 4.18 * 10^(-3)

The solution is :

e_{ab} =  4.18 * 10^(-3)

6 0
3 years ago
One kilogram of a contaminant is spilled at a point in a 4m deep reservoir and is instantaneously mixed over the entire depth. (
rosijanka [135]

Answer:

0.028 kg per Seconds; 8.4

Explanation:

N-S, 4 x 100 = 400 m^2, concentration 5/400= 0.0125 kg/s

E-W 4 x 100 =400 m^2 , => 10/400 =0.025 kg/s

(0.0125^2 + 0.025^2)^(1/2) =0.028 kg/s

in 5 minutes = 5 x 60 x 0.028 =8.4

8 0
3 years ago
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