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Dmitrij [34]
3 years ago
10

Assume that a surface-grinding operation is being carried out under the following conditions: D = 250 mm, d 0.1 mm, ys 0.5 m/s a

nd V = 50 m/s. These conditions are then changed to the following: D-150 mm, d 0.1 mm, v 0.3 m/s and V-25 m/s, what is the difference in the temperature rise from the initial condition? Assume that C does not change between the two cases. Please give your answer in percent form. For example, "the temperature rise is 50% lower in the second case"
Engineering
1 answer:
Misha Larkins [42]3 years ago
5 0

Answer:

modify temperature is lower than by the 0.196 %

Explanation:

given data

D = 250 mm

d = 0.1 mm

v = 0.5 m/s

V = 50 m/s

D = 150 mm

d = 0.1 mm

v = 0.3 m/s

V = 25 m/s

solution

first we get here initial coating condition for temperature change ΔT is

ΔT = A \times D^{1/4} \times d^{3/4}\times \frac{V}{v}^{1/2}    ...............1

put here value for both condition

ΔT = A \times 250^{1/4} \times 0.1^{3/4}\times \frac{50}{0.5}^{1/2}  

ΔT = 7.07 A      ......................2

and

ΔT = A \times 150^{1/4} \times 0.1^{3/4}\times \frac{25}{0.3}^{1/2}  

ΔT = 5.68 A       .......................3

so here percentage change is

percentage change = \frac{5.68 A-7.07 A }{7.07 A }  

percentage change  = - 0.196

so that modify temperature is lower than by the 0.196 %

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A mass of air occupying a volume of 0.15m^3 at 3.5 bar and 150 °C is allowed [13] to expand isentropically to 1.05 bar. Its enth
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Answer:

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Explanation:

V0 = 0.15

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Now we calculate the parameters at point 1

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The ideal gas equation:

P0 * v0 / T0 = P1 * v1 / T1

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Then the second transformation. It is at constant pressure.

P2 = P1 = 105 kPa

The enthalpy is raised in 52 kJ

Cv * T1 + P1*v1 = Cv * T2 + P2*v2 + Δh

And the idal gas equation is:

P1 * v1 / T1 = P2 * v2 / T2

T2 = T1 * P2 * v2 / (P1 * v1)

Replacing:

Cv * T1 + P1*v1 + Δh = Cv * T1 * P2 * v2 / (P1 * v1) + P2*v2

Cv * T1 + P1*v1 + Δh = v2 * (Cv * T1 * P2 / (P1 * v1) + P2)

v2 = (Cv * T1 + P1*v1 + Δh) / (Cv * T1 * P2 / (P1 * v1) + P2)

The Cv of air is 0.7 kJ/kg

v2 = (0.7 * 300 + 105*0.16 + 52) / (0.7 * 300 * 105 / (105 * 0.16) + 105) = 0.2 m^3/kg

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The work is:

L2 = P2*V2 - P1*V1

L2 = 105 * 0.45 - 105 * 0.36 = 9.45 kJ

The total work is

L = L1 + L2

L = -14.7 + 9.45 = -5.25 kJ

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