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9 days ago

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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

tially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg K and 4.4 kJ/kg K, respectively, determine the final equilibrium temperature after quenching, in K.

1 answer:

mote1985 [204]9 days ago

8 0

Answer:

provided below

Explanation:

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A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o

Daniel [215]

Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

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1 day ago

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A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plat

Mrrafil [253]

Answer:

t = 5.27 years

Explanation:

Firstly, the corrosion penetration rate is defined by the formula;

CPR = (KW)/(ρAt)

Where;

K = constant based on exposed area A.

W - mass lost over time

t- duration

ρ - density

A - area exposed

From the problem, we have;

W = 7.6kg or 7.6 x 10^(6) mg

CPR = 4 mm/yr

ρ = 4.5 g/cm³

Area = 800 cm²

K is a constant valued at 87.6cm

Rearranging the CPR formula to isolate t, we derive;

t = KW/(ρA(CPR))

t = (87.6 x 7.6 x 10^(6))/(4.5 x 800 x 4) = 46233.3 hours

The duration in question needs to be expressed in years.

Thus, converting hours to years;

There are 8760 hours in a year.

Therefore;

t = 46233.3/8760 = 5.27 years.

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4 days ago

3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a

Kisachek [217]

Answer:

Ps=19.62N

Explanation:

A thorough explanation of the answer can be found in the attached files.

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29 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.

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16 days ago

The uniform dresser has a weight of 90 lb and rests on a tile floor for which the coefficient of static friction is 0.25. If the

Kisachek [217]

Answer:

a) $F = 736.065\,lbf$ , b) $\mu_{k} = 0.15$

Explanation:

a) The uniform dresser can be modeled using specific equilibrium equations:

$\Sigma F_{x} = F - \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

Following some algebraic manipulations, the formulated equation is derived:

$F = \mu_{k}\cdot m \cdot g$

$F = (0.25)\cdot (90\,lbm)\cdot (32.714\,\frac{ft}{s^{2}} )$

$F = 22.5\,lbf$

b) Similarly, the man can be represented by a set of equilibrium equations:

$\Sigma F_{x} = -F + \mu_{k}\cdot N = 0$

$\Sigma F_{y} = N-m\cdot g=0$

After some algebraic changes, the expression for the coefficient of static friction comes out as:

$\mu_{k} = \frac{F}{m\cdot g}$

$\mu_{k} = \frac{22.5\,lbf}{150\,lbf}$

$\mu_{k} = 0.15$

3 0

7 days ago