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just olya [345]
2 years ago
14

100 kg of refrigerant-134a at 200 kPa iscontained in a piston-cylinder device whose volume is 12.322 m3. The piston is now moved

until the volume is one-half its original size. This is done such that thepressure of the refrigerant-134a does not change. Determine (a) the final temperatureand (b) the change in the specific internal energy.
Engineering
1 answer:
natali 33 [55]2 years ago
5 0

Answer:

(a) Final temperature is 151.2 K

(b) Change in the specific internal energy is -30.798 kJ/kg

Explanation:

(a) P1 = P2 = 200 kPa

V1 = 12.322 m^3

V2 = 1/2 × V1 = 1/2 × 12.322 = 6.161 m^3

mass of refrigerant-134a = 100 kg

MW of refrigerant-134a (C2H2F4) = (12×2) + (2×1) + (19×4) = 24 + 2 +76 = 102 g/mol

number of moles of refrigerant-134a (n) = mass/MW = 100/102 = 0.98 kgmol

R = 8.314 kJ/kgmol.K

Final temperature (T2) = P2V2/nR = 200×6.16/0.98×8.314 = 151.2 K

(b) Change in specific internal energy = change in internal energy (∆U)/mass (m)

∆U = Cv(T2 - T1)

Cv = 20.785 kJ/kgmol.K

T1 = P1V1/nR = 200×12.322/0.98×8.314 = 302.4 K

∆U = 20.785(151.2 - 302.4) = 20.785 × -151.2 = -3142.7 kJ/kgmol × 0.98 kgmol = -3079.8 kJ

Change in specific internal energy = -3079.8 kJ/100 kg = -30.798 kJ/kg

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Phoenix [80]

Answer: C) Sump pit

Explanation:

Wastewater from the house is allowed to flow out of places such as the bathroom or the kitchen sink through a floor drain.

This floor drain is connected to a sump pit which is typically located at the lowest point in the house (lowest point in basement) so that water can flow into it easier.

When the wastewater is released from the floor drain, it will flow into the sump pit. This water is in turn removed from the pit by a sump pump.

8 0
2 years ago
What are the BENEFITS and RISKS of using automobiles?
Alona [7]

Answer:

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Explanation:

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5 0
2 years ago
A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg w
zysi [14]

Answer:

HB \approx 200.484

Explanation:

The Brinell Hardness is obtained from the following formula:

HB = \frac{2\cdot P}{\pi \cdot D^{2}}\cdot \left(\frac{1}{1 - \sqrt{1-\frac{d^{2}}{D^{2}} } } \right )

Where P is the load measured in kilograms-force, D is the diameter of indenter measured in millimeters and d is the diameter of indentation measured in millimeters.

HB=\frac{2\cdot (1000\,kgf)}{\pi\cdot (10\,mm)^{2}}\cdot \left[ \frac{1}{1-\sqrt{1-\frac{(2.50\,mm)^{2}}{(10\,mm)^{2}} } }  \right ]

HB \approx 200.484

4 0
2 years ago
A two-stroke CI. engine delivers 5000 kWwhile using 1000 kW to overcome friction losses. It consumes 2300 kg of fuel per hour at
Lady_Fox [76]

Answer:

(a) Indicating power(IP)=6000 KW

(b)   \eta_{mech}=0.833

(c) Consumption of air per hour =46000 kg/hr

(d) \eta_{BPth}=0.1865

Explanation:

Break power(BP) =5000 KW

Friction power(FP)=1000 KW

Consumption of fuel per hour=2300 kg/hr

CV=42000 KJ/kg

We know that

Indicating power(IP)=Break power(BP)+Friction power(FP)

⇒IP=5000+1000 KW

  IP=6000 KW

(a)

Indicating power(IP)=6000 KW

(b)

Mechanical efficiency  \eta_{mech}=\dfrac{BP}{IP}

 \eta_{mech}=\dfrac{5000}{6000}      

  \eta_{mech}=0.833

(c)

Air fuel ratio=\dfrac{mass \ of \ air}{mass \ of \ fuel}

consumption of air per hour=20\times2300 kg/hr

So consumption of air per hour =46000 kg/hr

(d)

Break thermal efficiency  \eta_{BPth}=\dfrac{IP}{\dot{m_f}\times CV}

\dot{m_f}=\dfrac{2300}{3600}  

                      =0.638 kg/s

\eta_{BPth}=\dfrac{5000}{{0.638}\times 42000}

\eta_{BPth}=0.1865

 

5 0
3 years ago
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}

  r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Now by putting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

  Ф=18.03°

So the shear angle is 18.03°.

4 0
3 years ago
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