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2 days ago

Consider film condensation on a vertical plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?

Engineering

1 answer:

pantera1 [220]2 days ago

8 0

Answer:

In film condensation on a vertical plate, the heat flux at the upper section will be greater since the film thickness is less at this location, which translates to reduced thermal resistance.

Explanation:

https://www.docsity.com/pt/cengel-solution-heat-and-mass-transfer-2th-ed-heat-chap10-034/4868218/

https://arc.aiaa.org/doi/pdf/10.2514/1.43136

https://arxiv.org/ftp/arxiv/papers/1402/1402.5018.pdf[[TAG_22]]

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A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t

mote1985 [204]

Answer:

The calculated result is 11.7 ft

Explanation:

You can apply the combined gas law, which incorporates Boyle's law, Charles's law, and Gay-Lussac's Law, because hydrogen demonstrates ideal gas behavior under these specific conditions.

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

where the subscripts indicate "p" for pressure, "V" for volume, and "T" for temperature (in Kelvin) at varying moments. Let's denote $t_1$ as the balloon at 150,000 ft so

$p_1 = 0.14 \ lb/in^2$

$V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3$

and $T_1 = -67^\circ F = 218.15\ K$ .

Then $t_2$ represents the point at which the balloon is on the ground.

$p_2 = 14.7 \ lb/in^2$ and $T_2 = 68^\circ F = 293.15\ K$ .

Based on the first equation

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}$ , we find

$V_2 = 6701.07 ft^3$ and consequently the radius turns out to be

$R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft$ .

5 0

5 days 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 [279]

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

16 days ago

1. Mark ‘N’ if a wrong type of units is used. Mark ‘Y’ otherwise. (1 point each)

pantera1 [220]

Response:

bsbsdbsd

Clarification:

7 0

23 days ago

Six years ago, an 80-kW diesel electric set cost $160,000. The cost index for this class of equipment six years ago was 187 and

grin007 [219]

Answer:

total expense for the new boiler = $229706.825

total expense for new boiler = $127512

Explanation:

provided information

initial power p1 = 80 kW

price C = $160000

cost index CI 1 = 187

cost index CI 2= 194

capacity factor f = 0.6

subsequent power p2 = 120 kW

present cost = $18000

to determine

total expense and cost for 40 kW

solution

we evaluate CN cost for the new boiler and CO cost for the existing boiler

where x represents the capacity of the new boiler

first we compute the current cost of the old boiler which is

current cost CO = C × $\frac{CI 1 }{CI 2 }$ .............1

substituting the value here

current cost = 160000 × $\frac{194 }{187 }$

updated current cost = $165989.304

and

employing power sizing strategy for 124 kW

CN/CO = $(\frac{p2}{p1} )^{f}$ ...............2

insert value and calculate CN

CN/CO = $(\frac{p2}{p1} )^{f}$

CN / 165989.304 = $(\frac{120}{80} )^{0.6}$

CN = 211706.825

therefore the new expense = $211706.825

hence

total expense for the new boiler amounts to

total expense = new expense + current cost

total exp = 211706.825 + 18000

boiler expense total = $229706.825

and

concerning a 40 kW unit, the calculated new cost will be

applying equation 2

CN/CO = $(\frac{p2}{p1} )^{f}$

CN / 165989.304 = $(\frac{40}{80} )^{0.6}$

CN = 109512

thus the new cost is $109512

therefore

total expense for the new boiler is

total exp = new cost + current cost

total expense = 109512 + 18000

total cost for the new boiler = $127512

7 0

13 days ago

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

Viktor [230]

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.

3 0

16 days ago