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zysi
2 months ago
7

The tightening rule of thumb for threaded stud shocks is

Engineering
1 answer:
mote1985 [299]2 months ago
7 0

Answer:
Righty tighty lefty loosey
Explanation:
You're welcome
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A cylinder in space is of uniform temperature and dissipates 100 Watts. The cylinder diameter is 3" and its height is 12". Assum
Kisachek [356]

Answer:

Temperature T = 394.38 K

Explanation:

The full solution and detailed explanation regarding the above question and its specified conditions can be found below in the accompanying document. I trust my explanation will assist you in grasping this particular topic.

7 0
2 months ago
A shopaholic has to buy a pair of jeans , a pair of shoes l,a skirt and a top with budgeted dollar.Given the quantity of each pr
choli [298]

Answer:

you may be struggling to pinpoint the separation between your inquiry and my perspective

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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.

3 0
2 months 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 [299]

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
2 months ago
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