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1 month ago

6

Three return steam lines in a chemical processing plant enter a collection tank operating at steady state at 1 bar. Steam enters

inlet 1 with flow rate of 0.8 kg/s and a quality of 0.9. Steam enters inlet 2 with flow rate of 2 kg/s at 200 C. Steam enters inlet 3 with flow rate of 1.2 kg/s at 95 C. Steam exits the tank at 1 bar. The rate of heat transfer from the collection tank is 40 kW. Neglecting kinetic and potential engery effects, determine for the steam exiting the tank
(a) the mass flow rate, in kg/s
(b) the temperature, in degrees C

1 answer:

alex41 [359]1 month ago

5 0

Response:

a) 4 kg/s

b) 99.61 °C

Rationale:

Refer to the pictures provided.

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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.

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3 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.

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2 months ago

A 227 pound compressor is supported by four legs that contact the floor of a machine shop. At the bottom of each leg there is a

choli [298]

Answer:

1.312 in

Explanation:

The details provided in the question are:

The weight of the compressor, W is 227 pounds.

It has 4 legs.

The maximum permissible pressure is 42 psi.

Let F represent the force exerted by each leg.

Thus,

W = 4F,

or

227 pounds = 4F,

implying that:

F = 56.75 pounds.

Furthermore,

Force = Pressure × Area,

therefore:

56.75 pounds = 42 psi × πr² [ r signifies the radius of one leg]

Consequently, we find:

r² = 0.4301,

and thus:

r = 0.656;

resulting in a diameter equal to 2r = 2 × 0.656,

which equals 1.312 in.

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3 months ago

6.15. In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20

Viktor [391]

Explanation:

At a temperature of $33^{\circ} C$ and relative humidity of 86%, the humidity ratio stands at 0.0223 with a specific volume of 14.289.

At a temperature of $33^{\circ} C$ and relative humidity of 40%, the humidity ratio is 0.0066 while the specific volume is 13.535.

To determine the mass of air, the following formula can be used:

$\begin{aligned}m _{1} &=\frac{ v }{ v }(1- w ) \\&=\frac{1 \times 10^{5}}{13.535}(1-0.0066) \\&=7339.49 lb / min \\v _{ a } &=\frac{ m _{1} v }{(1- w )} \\v _{ a } &=\frac{7339.49 \times 14.289}{(1-0.0223)} \\v _{ a } &=107266.0 ft ^{3} / min\end{aligned}$

Now, we will calculate the volume

$\begin{aligned}m _{ w } &=\frac{ v _{ a }}{ v _{ a }} w _{ a }-\frac{ v _{ i }}{ v _{ i }} w _{ i } \\&=\frac{107266.0}{14.289} \times 0.0223-\frac{100000}{13.535} \times 0.0066 \\&=118.64 lb / min\end{aligned}$

The duration required to fill the cistern can be determined with the equation:

$Time \(=\frac{\text { cistern volume }}{\text { removal water perminute volume }}\)$

By substituting the values into the preceding formula, we find:

$\(\frac{\left(15 \times 10^{3} L\right) \times\left(0.0353147 ft ^{3} / L \right)}{(118.641 b / min ) \times\left(\frac{1}{62.41 lb / ft ^{3}}\right)}\)\\\(=279.09\) minutes\\\(=4.65\) hours.$

Thus, the hours necessary to fill the cistern amount to 4.65 hours.

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2 months ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Mrrafil [318]

The explanation and answer for this query can be found in the attached document below.

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2 months ago