All categories

3 years ago

What are the effects of applying an additional layer of insulation to a cylindrical pipe or a spherical shell?

Engineering

1 answer:

Brums [2.3K]3 years ago

4 0

Answer:

When an additional layer of insulation is applied to a cylindrical pipe or a spherical shell, the insulation layer works to increase its conduction resistance but at the same time, lowers the convection resistance of the surface.

Explanation:

Generally, when more insulation is added to a wall, the resulting effect is that heat transfer decreases. With increasing thickness of the insulation, the heat transfer rate becomes lesser. This is because the thermal resistance of the wall becomes more with the added insulation, wherein the heat transfer area and convection resistance aren't affected.

However, a different scenario occurs when an additional layer of insulation is applied to a cylindrical pipe or spherical shell. The conduction resistance increases, but the surface convection resistance decreases since the outer surface area for convection also increases and does not remain constant.

Overall, the heat transfer from the cylindrical pipe or spherical shell may increase or decrease, which depends on the effect that dominates.

You might be interested in

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago

Explain why Chloe's design needs to be redone in the following scenario, and recommend the techniques she needs to include in he

Alenkinab [10]

Answer:

She believes he's weak and won't do what needs to be done to become a king.

Explanation:

She says ('I fear thy nature') and calls him 'too full o' th' milk of human kindness' which reflects that she feels his kindness makes him weak and may prevent him from proceeding the plan. Thus, she manipulates him to keep his kindness aside and do what she wishes him to do. She rather belittles him to get her purpose solved and

7 0

3 years ago

Read 2 more answers

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago

Assume a function requires 20 lines of machine code and will be called 10 times in the main program. You can choose to implement

Volgvan

Answer:

"Macro Instruction"

Explanation:

A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.

It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.

7 0

3 years ago

Though still is on the margins of the global agro-food system, organic farming practices are on the rise in the core. In growing

lisabon 2012 [21]

Answer: composted cow manure

Explanation:

A cow manure consists of high levels of ammonia, pathogens which can harm the plants thus cannot be used as fertilizer. The composted cow manure is a processed fertilizer which is prepared by eliminating the ammonia gas and pathogens. It is prepared by adding up the additional organic matter.

It is an excellent growing medium for the agricultural crops and garden plants as it is a nutrient rich fertilizer. It is mixed into the soil or can be used for top dressing of the agricultural field. It is a kind of organic manure.

4 0

3 years ago