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Lana71 [14]
2 years ago
6

A 2"" Sch 40 stainless steel (k = 14.9 W/m-K) pipe is to be used as the interior pipe of a double pipe heat exchanger. The expec

ted heat transfer coefficients of the hot and cold fluids are 1800 W/m2-K and 1600 W/m2-K respectively. Determine the total resistance to heat flow per unit length (i.e. R  L) with and without the resistance of the pipe wall. What is the percentage difference? (18.2%)
Engineering
1 answer:
Scrat [10]2 years ago
8 0

Answer:

the percentage difference is E = = 55.7% respect to the real value ( with resistance of the wall)

Explanation:

the resistance to heat flow R is

R=  ( 1/h₁ + ln (re/ri)/(2π*k*L) + 1/h₂)

where

h₁= heat transfer coefficient for the hot fluid = 1600 W/m²K

h₂= heat transfer coefficient for the cold fluid = 1800 W/m²K

re/ri = ratio between external radius and internal radius of the pipe = 2.375 /2.067  for 2"" Sch 40 stainless steel

k= thermal conductivity = k = 14.9 W/m-K

L = length of the pipe = 1 ( unit length)

therefore replacing values

R₁=  1/h₁ + ln (re/ri)/(2π*k*L) + 1/h₂ = 1/1600 W/m²K + ln(2.375 /2.067)/(2π*14.9 W/m-K*1) + 1/1800 W/m²K = 2.664 *10⁻³m²K/W

when the resistance of the pipe wall is neglected then R would be

R₂=  1/h₁ + 1/h₂ =  1/1600 W/m²K 1+ 1/1800 W/m²K =1.180 *10⁻³ m²K/W

the percentage difference between the total resistance with and without the pipe wall resistance per unit length  is

E = 1-(R₂/L) /( R₁/L) = 1- R₂/R₁=1- (1.180 *10⁻³/2.664 *10⁻³) = 0.557= 55.7%

E = = 55.7%

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3. If nothing can ever be at absolute zero, why does the concept exist?
Tanzania [10]

The absolute zero in temperature refers to the minimal possible temperature. It is the temperature at which the molecules of a system stop moving, so it is a really useful reference point.

<h3>Why absolute zero can't be reached?</h3>

It would mean that we need to remove all the energy from a system, but to do this we need to interact with the system in some way, and by interacting with it we give it "some" energy.

Actually, from a quantum mechanical point of view, the absolute zero has a residual energy (so it is not actually zero) and it is called the "zero point". This happens because it must meet <u>Heisenberg's uncertainty principle</u>.

So yes, the absolute zero can't be reached, but there are really good approximations (At the moment there is a difference of about 150 nanokelvins between the absolute zero and the smallest temperature reached). Also, there are a lot of investigations near the absolute zero, like people that try to reach it or people that just need to work with really low temperatures, like in type I superconductors.

So, concluding, why does the concept exist?

  • Because it is a reference point.
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If you want to learn more about the absolute zero, you can read:

brainly.com/question/3795971

3 0
2 years ago
Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr
omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

                                   W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                         c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

                  c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

                             q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

                             q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

                            q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                          Q_in = Δ u_2,3

                                         q_2,3 = u_3 - u_2

                                         q_2,3 = c_v*(T_H - T_2)  

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

                                         q_2,3 = c_v*(T_H - T_L*r^(n-1) )    

                                         q_2,3 = 0.718*(1173-288*8(1.3-1) )

                                        q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

                                     q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

                           c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n +  q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

                  c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

                             q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

                             q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

                            q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

                                          Q_out = Δ u_4,1

                                         q_4,1 = u_4 - u_1

                                         q_4,1 = c_v*(T_4 - T_L)  

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

                                         q_4,1 = c_v*(T_H*r^(1-n) - T_L )    

                                         q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

                                        q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

                                         q_net,in = q_3,4+q_2,3

                                         q_net,in = 129.8+456

                                         q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

                                         q_net,out = q_4,1+q_1,2

                                         q_net,out = 244+60

                                         q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

                                         n = 1 - q_net,out / q_net,in

                                         n = 1 - 304/585.8

                                         n = 0.481

6 0
3 years ago
A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa
Fed [463]

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
4 0
3 years ago
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A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,
disa [49]

Answer: (a) 36.18mm

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3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

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load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

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δ = \frac{PL}{AE}  

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