All categories

2 months ago

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.

Engineering

1 answer:

grin007 [323]2 months ago

3 0

Answer:

a. The rate of heat removal will increase

b. The rate of heat removal will decrease

Explanation:

Provided that

One end of the rod is attached to a furnace and the rod is sufficiently long, it can be treated as an infinite fin.

The heat transfer in a fin is represented as:

$Q_{fin}=\sqrt{hPKA}\ \Delta T$

We recognize that the area is

$A=\dfrac{\pi}{4}d^2$

When the diameter triples, it results in:

$A_f=\dfrac{\pi}{4}{\left (3d \right )}^2$

$A_f=9A$

$Q'_{fin}=\sqrt{9hPKA}\ \Delta T$

$Q'_{fin}=3\sqrt{hPKA}\ \Delta T$

$Q'_{fin}=3Q$

The new heat transfer will therefore increase by a factor of 3.

When substituting the copper rod with an aluminum rod:

It is known that the thermal conductivity (K) of Aluminum is lower compared to that of Copper, which indicates a reduction in heat transfer.

You might be interested in

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

Mrrafil [318]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

Final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

Substituting all values yields

0.0130 = $\frac{\pi }{4} *2^2*dl$

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0

2 months ago

Effects of biological hazards are widespread. Select the answer options which describe potential effects of coming into contact

pantera1 [306]

Answer:

- Allergic Responses

- Events Posing Life Threats

Explanation:

Biological hazards can originate from a variety of sources such as bacteria, viruses, insects, plants, birds, animals, and humans. These can lead to numerous health issues, which may range from skin allergies and irritations to infections (like tuberculosis or AIDS) and even cancer.

7 0

2 months ago

A liquid food with 12% total solids is being heated by steam injection using steam at a pressure of 232.1 kPa (Fig. E3.3). The p

iogann1982 [368]

Answer:

$m_{s}=20kg/min$

$H_{s}=1914kJ/kg$

Explanation:

A liquid food containing 12% total solids is heated via steam injection at a pressure of 232.1 kPa (see Fig. E3.3). The product starts at a temperature of 50°C and has a flow rate of 100 kg/min, being elevated to a temperature of 120°C. The specific heat of the product varies with its composition as follows:

$c_{p}=c_{pw}(mass fraction H_{2}0)+c_{ps}(mass fraction solid)$ and the

specific heat of the product at 12% total solids is 3.936 kJ/(kg°C). The goal is to calculate the quantity and minimum quality of steam required to ensure that the leaving product has 10% total solids.

Given

Product total solids in ( $X_{A}$ ) = 0.12

Product mass flow rate ( $m_{A}$ ) = 100 kg/min

Product total solids out ( $X_{B}$ ) = 0.1

Product temperature in ( $T_{A}$ ) = 50°C

Product temperature out ( $T_{B}$ ) = 120°C

Steam pressure = 232.1 kPa at ( $T_{S}$ ) = 125°C

Product specific heat in ( $C_{PA}$ ) = 3.936 kJ/(kg°C)

The mass equation is:

$m_{A}X_{A}=m_{B}X_{B}$

$100(0.12)=m_{B}(0.1)\\m_{B}=\frac{100(0.12)}{0.1} =120$

Also $m_{a}+m_{s}=m_{b}\\$

Therefore: $100}+m_{s}=120\\\\m_{s}=120-100=20$

The energy balance equation is:

$m_{A}C_{PA}(T_{A}-0)+m_{s}H_{s}=m_{B}C_{PB}(T_{B}-0)$

$3.936 = (4.178)(0.88) +C_{PS}(0.12)\\C_{PS}=\frac{3.936-3.677}{0.12} =2.161$

$C_{PB}= 4.232*0.9+0.1C_{PS}= 4.232*0.9+0.1*2.161=4.025$ kJ/(kg°C)

By substituting values into the energy equation:

$100(3.936)(50-0)+20H_{s}=120(4.025)}(120-0)$

$19680+20H_{s}=57960\\20H_{s}=57960-19680 \\20H_{s}=38280\\H_{s}=\frac{38280}{20} =1914$

$H_{s}=1914kJ/kg$

From the properties of saturated steam at 232.1 kPa,

$H_{c}$ = 524.99 kJ/kg

$H_{v}$ = 2713.5 kJ/kg

% quality = $\frac{1914-524.99}{2713.5-524.99} =63.5%$

Any steam quality above 63.5% will result in higher total solids in the heated product.

3 0

2 months ago

Which of the following types of protective equipment protects workers who are passing by from stray sparks or metal while anothe

choli [298]

Answer:

Flame-resistant clothing and aprons

Explanation:

Workers involved with welding are generally mandated to wear flame-resistant clothing and aprons to shield them from various hazards, including heat, flames, burns, and potential radiation. In the context of welding, this gear protects individuals from flying sparks that can ignite and cause fires. Hence, such clothing helps to prevent accidents in these situations.

7 0

3 months ago

A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force

Viktor [391]

Response:

The solution to this question is 1273885.3 ∅

Clarification:

The first step is to ascertain the required hydraulic flow rate liquid based on the working pressure if a cylinder with a piston diameter of 100 mm is utilized.

Given that,

The distance = 50mm

The time t =10 seconds

The force F = 10kN

The piston diameter = 100mm

The pressure = F/A

10 * 10^3/Δ/Δ

P = 1273885.3503 pa

Subsequently

Power = work/time = Force * distance /time

= 10 * 1000 * 0.050/10

which amounts to =50 watt

Power =∅ΔP

50 = 1273885.3 ∅

5 0

1 month ago