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1 month ago

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On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of mercury. What would the barometer reading

be when you carry it up to the observation deck 500 ft above the base of the monument?

1 answer:

Mrrafil [253]1 month ago

8 0

Answer:

The barometer will read 29.43 in

Explanation:

Using the pressure variation equation,

p2 - p1 = -yair * H

= 7.65 * $10^{-2} \frac{lb}{ft^{3} } * 500 ft\\$

= 38.5 $\frac{lb}{ft^{2} }$

According to how pressure relates to the height of the mercury column,

p = yHg * h --> with yHg and h being the barometer reading

yHg $(\frac{29.97}{12} ft)$ - yHg * h1 = 38.5 $\frac{lb}{ft^{2} }$

h1 = ( $\frac{29.97}{12} ft$ ) - $\frac{38.5 \frac{lb}{ft^{2} } }{847 \frac{lb}{ft^{3} } }$

$[(\frac{29.97}{12} ft) - 0.0455 ft] - 12 \frac{in}{ft} \\\\h1 = 29.43 in$

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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.

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16 days ago

Your driver license will be _____ if you race another driver on a public road, commit a felony using a motor vehicle, or are fou

iogann1982 [279]

Hello there,

In this scenario, the driver's license has been both confiscated and suspended.

Therefore, the answer is: A)

Achievements.

6 0

1 month ago

Read 2 more answers

In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini

mote1985 [204]

Answer:

provided below

Explanation:

8 0

9 days ago

The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Cel

iogann1982 [279]

Answer:

The heat transfer rate into the wall is $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

The change in stored energy in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

The convection coefficient is h = 4.26 W/m².K

Explanation:

Considering the problem:

The temperature profile across the wall is expressed as:

$T(x) = ax+bx+cx^2$

where:

T = temperature in °C

and a, b, & c are constants.

Substituting a = 200° C, b = -200° C/m, and c = 30° C/m² results in:

$T(x) = 200x-200x+30x^2$

This follows the application of Fourier's Law of heat conduction.

$q_x = -k \dfrac{dT}{dx}$

where the heat input rate $q_{in} = q_k$ ; Then x= 0

<pThus:

$q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}$

$q_{in}= -1 (-200+60x)_{x=0}$

$\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

Consequently, the heat transfer rate into the wall measures $\mathbf{q__{in}} \mathbf{ = 200 W/m^2}$

The heat output rate is:

$q_{out} = q_{x=L}$ ; where x = 0.3

$q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}$

Replacing T with $200x-200x+30x^2$ and k with 1 W/m.K

$q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}$

$q_{out} = -1 (-200+60x)_{x=0.3}$

$q_{out} = 200-60*0.3$

$\mathbf{q_{out} =182 \ W/m^2}$

Thus, the heat output rate is $\mathbf{q_{out} =182 \ W/m^2}$

Applying the energy balance to find the change in energy (internal energy) stored in the wall.

$\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\$

$\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

Thus, the energetic change rate stored in the wall is $\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }$

We know that in a steady state, the heat reaching the end of the plate must be convected to the surrounding fluid.

Thus:

$q_{x=L} = q_{convected}$

$q_{x=L} = h(T(L) - T _ \infty)$

where;

h represents the convective heat transfer coefficient.

Therefore:

$Replacing \ 182 W/m^2 \ for \ q_{x=L}, (200-200x +30x \ for \ T(x) \, 0.3 m \ for \ x \ and \ 100^0 C for \ T$ We find:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient equals h = 4.26 W/m².K

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1 month ago

A long, horizontal, pressurized hot water pipe of 15cm diameter passes through a room where the air temperature is 24degree C. T

Daniel [215]

The heat transfer rate to the surrounding air per meter of pipe length is quantified as 521.99 W/m. Given the negligible radiation losses from the pipe, convection remains the sole method of heat transfer. The rate of heat transfer via convection can be defined as such, using the specified heat transfer coefficient of 10.45 for air and calculating the surface area of the pipe.

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7 days ago