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Aleonysh [2.5K]
3 years ago
12

3. Write down the total thermal resistance for a double-pipe heat exchanger. Show how to convert from total resistance to an ove

rall heat-transfer coefficient. Explain how flow regime may affect overall heat transfer rate.

Engineering
2 answers:
wlad13 [49]3 years ago
8 0

Answer: please check image for detailed calculation and explanation.

At the boundary between the pipes, there is discontinuity due to different regime; this gap can insulate heat flow causing a fluctuation in heat flow from the inner pipe to the outer pipe.

Nutka1998 [239]3 years ago
8 0

Answer:

\frac{1}{UA} =\frac{1}{A_{0}h_{0}  } +\frac{Rf_{0} }{A_{0} } +\frac{1}{2k\pi L} ln(\frac{di,0}{di,i} )+\frac{Rf_{i} }{A_{i} }

Explanation:

Global heat transfer determines the heat that is transferred from the inner tube to the outer tube. In the double tube heat exchanger, this global heat transfer coefficient is equal to:

\frac{1}{UA} =\frac{1}{A_{0}h_{0}  } +\frac{Rf_{0} }{A_{0} } +\frac{1}{2k\pi L} ln(\frac{di,0}{di,i} )+\frac{Rf_{i} }{A_{i} }

where subscript i refers to the inner surface and subscript 0 refers to the outer surface

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The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
3 years ago
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this e
liubo4ka [24]

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

​

Explanation:

thats all you said

7 0
3 years ago
Read 2 more answers
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure
belka [17]

The answer is c) atomic number

3 0
3 years ago
Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str
alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

         = 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

         = 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

                                                          = 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

                      ρ_avg= mass of unit cell/volume of unit cell

                      ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

                          λ^3 = 4.37

6 0
3 years ago
Consider the mixing of 0.8 kg/s of hot water at 348 K and 1 kg/s of cool water at 298 K that is generating warm water. Assume no
Drupady [299]

Answer:

T_warm = 47.22 C

Explanation:

Using energy balance for the system:

m_1*h_1 + m_2*h_2 = m_3*h3   ... Eq1

h_i = c_p. T_i   ... Eq 2

m_1 + m_2 = m_3   ... steady flow system (Eq 3)

Substitute Eq 2 and Eq3 in Eq1

m_3 = 0.8 + 1 = 1.8 kg/s

(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3

T_3 = 355.3 / (1.8*4.18) = 47.22 C

3 0
3 years ago
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