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Lemur
3 hours ago
5

Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that

the result of the function is returned in register %eax. You will find it helpful to examine the assembly code before, during, and after the loop to form a consistent mapping between the registers and the program variables.
Engineering
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The Machine Shop has received an order to turn three alloy steel cylinders. Starting diameter = 250 mm and length = 625 mm. Feed
alex41 [359]

Response:

The cutting speed is calculated at 365.71 m/min

Clarification:

Given parameters include

diameter D = 250 mm

length L = 625 mm

Feed f = 0.30 mm/rev

cut depth = 2.5 mm

n = 0.25

C = 700

To find

the cutting speed that ensures the tool life coincides with the cutting time for the three parts

The formula for cutting time is given as

Tc =

....................1

where D refers to diameter, L refers to length and f refers to feed while V represents speed \frac{\pi DL}{1000*f*V}

Thus, we derive

Tc = \frac{\pi (250)*625}{1000*0.3*V}

Tc = \frac{1636.25}{V}

Given the tool life is expressed as

T = 3 × Tc............................2

where T denotes tool life and Tc is the cutting duration

Calculating tool life by substituting values into equation 2 yields

T = 3 × \frac{1636.25}{V}

According to the Taylor tool formula, cutting speed is expressed as

VT^{0.25} = 700

 × V × 8.37 = 700

This yields V = 365.71

Thus, the cutting speed calculates to 365.71 m/min

5 0
2 months 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 [356]

Answer:

Ps=19.62N

Explanation:

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

5 0
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
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