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blsea [12.9K]
2 years ago
9

A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and gauge pressure of 300 kPa. The gas is h

eated, and the gauge pressure at the final state is 367 kPa. Determine the final temperature, in °C. The local atmospheric pressure is 1 atm.
Engineering
1 answer:
Sergio [31]2 years ago
3 0

Answer:

the final temperature is 77.1 °C

Explanation:

Given the data in the question;

Initial temperature; T₁ = 27°C = ( 27 + 273)K = 300 K

Initial absolute pressure P₁ = 300 kPa = ( 300 + 101.325 )kPa = 401.325 kPa

Final absolute pressure P₂ = 367 kPa = ( 367 + 101.325 )kPa = 468.325 kPa

Now, to calculate the final temperature, we use the ideal gas equation;

P₁V/T₁ = P₂V/T₂

but it is mentioned that the rigid tank is closed,

so the volume is the same both before and after.

Change in volume = 0

hence;

P₁/T₁ = P₂/T₂

we substitute

401.325 kPa / 300 K = 468.325 kPa / T₂

T₂ × 401.325 kPa  = 300 K × 468.325 kPa

T₂ = [ 300 K × 468.325 kPa ] / 401.325 kPa

T₂ = 140497.5 K / 401.325

T₂ =  350.08 K

T₂ = ( 350.08 - 273 ) °C

T₂ = 77.1 °C

Therefore, the final temperature is 77.1 °C

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Answer:

(a) 94.25 m/min

(b) 219911.5 mm^{3}/min

(c) 1 min

(d)  10.996 kW and 15.027 kW

(e) 7 kN and 9.57 kN

Explanation:

The maximum cutting speed, v_c is observed at the outer diameter hence

v_c=N\piD_o where N is the rate of rotation and D_o is the outer diameter. Substituting 75mm for D_o and 400 rpm for N we obtain

v_c=\frac {400\times \pi\times 75}{1000}=94.24778 m/min\approx 94.25 m/min

(b)

The material removal rate, MRR is given by

MMR=\pi\times D_{avg}\times d\times f \times N where D_{avg} is the average between the inner and outer diameters, d is the depth of cut, f is the feed which is given by \frac {V}{N} where V is the axial velocity

In this case, the average diameter is \frac {75+65}{2}=70mm

The feed f is f=\frac {200}{400}=0.5  

The depth of cut is d=\frac {75-65}{2}=5mm

Therefore, MRR=\pi\times 70mm\times 5\times 0.5 \times 400 \approx 219911.5 mm^{3}/min

(c)

Time of cut is given by

T=\frac {L}{fN} where L is the length of rod. Substituting L for 200mm, f as seen in part b is 0.5 and 400 for N we obtain

T=\frac {200}{0.5*400}=1 min

(d)

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Power, P1=\frac {3\times 219911.5}{60}=10995.57 W\approx 10.996 kW

Power, P2=\frac {4.1\times 219911.5}{60}=15027.28 W\approx 15.027 kW

Therefore, power ranges between 10.996 kW and 15.027 kW

(e)

Cutting force is given by

F_c=\frac {P}{v_c} where P is power and v_c is already calculated in part a

First, v_c is converted to m/s hence v_c=\frac {400\times \pi\times 75}{1000\times 60}= 1.570796\approx 1.57 m/s

From the power range in part d,

F_{c1}=\frac {10.996 kW}{1.57}= 7.003822\approx 7 kN

F_{c2}=\frac {15.027 kW}{1.57}= 9.571338\approx 9.57 kN

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Answer:

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As per the question:

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