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elena55 [62]
3 years ago
8

Classify the terms as related to a thermal system or mechanical system.

Engineering
1 answer:
amid [387]3 years ago
6 0

The definition of the terms are given below:

Explanation:

Conduction

  • Conduction causes the heat to transfer from an object to another by the direct contact between them.

Motion

  • The action of moving of any object or substance is called motion.

Distance

  • The measurement of, how far two objects are located.

Temperature

  • It is the degree of hotness and coldness of body.

Absolute Zero

  • It is the lowest temperature possible where the particles have no motion. Absolute Zero is considered as 0 kelvin.

Speed

  • Rate of change of distance is called speed. It determines at what rate the object is changing its position.
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Kipish [7]

Answer:

tHE answer is b

Explanation:

7 0
3 years ago
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2 Air enters the compressor of a cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compres
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5 0
2 years ago
A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th
Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ    

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588  = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

 cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 =  2552.64 kJ    

b)  \Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)

Where:

x_A and x_B are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, x_A = 248 /(248 + 42.8) = 0.83

x_B = 42.8/(248 + 42.8) = 0.1472

∴ \Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472) =  1066.0279 J/K

5 0
3 years ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
White raven [17]

Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr
navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So  atmospheric pressure is equal to  755 mm Hg.

We know that P= ρ g h

Density of Hg=13600 \frac{kg}{m^3}

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30   KPa

So

Absolute pressure=70.72 KPa

8 0
3 years ago
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