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2 days ago

11

Suppose a steam locomotive is rated at 7500 horsepower. If its efficiency is 6%, how much wood must be burned in a 3-hour trip?

1 hp = 750 W. Wood has an energy density of 20.0 MJ/ kg.

1 answer:

choli [191]2 days ago

3 0

50,625 kg of wood. Power input can be calculated using power output divided by efficiency. The output is 7500 horsepower, which equates to 5625000 W. Given the efficiency of 6% (or 0.06), the input power is 93750000 W. Then, calculating the input energy over a 3-hour trip (10800 seconds), we find an energy input of 1.0125×10^12 J. Hence, the mass of wood required is the energy divided by its density of 20.0 MJ/kg, resulting in 50,625 kg.

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

choli [191]

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.

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1 month 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

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

Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and

choli [191]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data:

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

To determine final temperature in an Isentropic process

$T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}$ ----------- ( 1 )

T2 = 303 $( \frac{450}{90})^{(1.667- 1)/1.667}$ = 576.9K

The work performed in a piston-cylinder device is calculated using the subsequent formula

$w_{in} = c_{v} ( T2 - T1 )$ ------- ( 2 )

where: cv = 3.1156 kJ/kg.k for helium gas

T2 = 576.9K, T1 = 303 K

substituting values into equation 2

$w_{in}$ = 853.4 kJ/kg

the work done in a steady flow compressor is determined using this

$w_{in} = c_{p} ( T2 - T1 )$

where: cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

T2 = 576.9 k, T1 = 303 K

plugging values back into equation 3

$w_{in}$ = 1422.3 kJ / kg

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3 days ago

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:

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9 days ago