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konstantin123 [22]
3 years ago
8

Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i

s (a) Higher for larger diameter pipes (b) The same for all pipes independent of the diameter (c) Lower for larger diameter pipe
Engineering
1 answer:
leonid [27]3 years ago
5 0

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

You might be interested in
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

3 0
2 years ago
A person is planning a bungee jump from a 40 meter high bridge. Under the bridge is a river with crocodiles, so the person does
Nonamiya [84]

Answer:

a. l = 19.7m, b. 18.55m, c. Impact Force = 3889.84 N

Explanation:

The total energy of the system when the person is at top of the bridge is

Potential energy = mgh, Kinetic energy = 0

The total energy of the  system when the person reaches just above the surface

Potential energy = 0, Kinetic energy = 0, Spring energy = ½ K X2, where k is the spring constant and X is the deflection

Applying conservation of energy

mgh = 0 + 0 + ½ K X²

80 x 9.81 x 40 = ½ (3600/l) X²

31392 = ½ (3600/l) X²

We can also conclude that

l+ X + 1.75 = 40

l + X = 38.25

a. <u>Substitute the value of x from above into the energy conversion expression</u>

31392 = ½ (3600/l)(38.25 - l)²

31392 x 2/3600 = (38.25 + l² – 2l(38.25))/l

17.44l = l2 – 76.5l + 38.25²

l² – 76.5l – 17.44l +1463.0625 = 0

Solving for l we get

L = 19.7

Hence, length of the rope is 19.7m

b. <u>The deflection is calculated by using the relation between l and X</u>

L + X = 38.25

X = 38.25 – 19.7 = 18.55m

c. <u>The impact force is calculated using the impact force formula which relates the impact force with the deflection</u>

F = KX

F = (3600/l) . X

F = (3600/19.7) . (18.55) = 3889.84 N

Thus, the impact force is 3889.84 N

3 0
3 years ago
What is the De Broglie wavelength of an electron under 150 V acceleration?
yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

4 0
3 years ago
Read 2 more answers
A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th
prohojiy [21]

Answer:

Q=486.49 KJ/kg

Explanation:

Given that

V= 0.2 m³

At initial condition

P= 2 MPa

T=320 °C

Final condition

P= 2 MPa

T=540°C

From steam table

At P= 2 MPa and T=320 °C

h₁=3070.15 KJ/kg

At P= 2 MPa and T=540°C

h₂=3556.64  KJ/kg

So the heat transfer ,Q=h₂ - h₁

Q= 3556.64 - 3070.15  KJ/kg

Q=486.49 KJ/kg

7 0
3 years ago
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