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Y_Kistochka [10]
3 years ago
8

Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate

of 300 x 10-6 kg/s. A constant heat flux is applied to the air from the tube surface. If the tube surface temperature at the exit is Ts,o=180oC, determine the heat rate entering the tube. Evaluate properties at T=400 K.

Engineering
1 answer:
Annette [7]3 years ago
6 0

Answer:

The heat transfer q = 18.32W

Explanation:

In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K

Please check attachment for complete solution and step by step explanation

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Answer:

attached below

Explanation:

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3 years ago
Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kep
Talja [164]

Answer:

95.914\ \text{GJ}

\$272.78

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

\Delta T = Change in temperature = 80.6-50=30.6^{\circ}\text{F}

Cost of 1 GJ of energy = $2.844

Heat required is given by

Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}

Amount of heat required to heat the water is 95.914\ \text{GJ}.

Cost of heating the water is

95.914\times 2.844=\$272.78

Cost of heating the water to the required temperature is \$272.78.

7 0
3 years ago
Im passed due someone help meeeeeee
vovangra [49]

Answer:

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8 0
3 years ago
Read 2 more answers
A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.
Monica [59]

Answer: (a). E = 3.1656×10³⁴ √k/m  

(b). f = 9.246 × 10¹² Hz

(c). Infrared region.

Explanation:

From Quantum Theory,

The energy of a proton is proportional to the frequency, from the equation;

E = hf

where E = energy in joules

h = planck's constant i.e. 6.626*10³⁴ Js

f = frequency

(a). from E = hf = 1 quanta

    f = ω/2π

where ω = √k/m

consider 3 quanta of energy is lost;

E = 3hf = 3h/2π × √k/m

E = (3×6.626×10³⁴ / 2π) × √k/m

E = 3.1656×10³⁴ √k/m    

(b). given from the question that K = 15 N/m

and mass M = 4 × 10⁻²⁶ kg

To get the frequency of the emitted photon,

Ephoton =hf = 3h/2π × √k/m (h cancels out)

f = 3h/2π × √k/m

f =  3h/2π × (√15 / 4 × 10⁻²⁶ )

f = 9.246 × 10¹² Hz

(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to  400 THz in the electromagnetic spectrum.

6 0
3 years ago
#5 Air undergoes an adiabatic compression in a piston-cylinder assembly from P1= 1 atm and Ti=70 oF to P2= 5 atm. Employing idea
otez555 [7]

Answer:

The work transfer per unit mass is approximately 149.89 kJ

The heat transfer for an adiabatic process = 0

Explanation:

The given information are;

P₁ = 1 atm

T₁ = 70°F = 294.2611 F

P₂ = 5 atm

γ = 1.5

Therefore, we have for adiabatic system under compression

T_{2} = T_{1}\cdot \left (\dfrac{P_{2}}{P_{1}}  \right )^{\dfrac{\gamma -1}{\gamma }}

Therefore, we have;

T_{2} = 294.2611 \times \left (\dfrac{5}{1}  \right )^{\dfrac{1.5 -1}{1.5 }} \approx 503.179 \ K

The p·dV work is given as follows;

p \cdot dV = m \cdot c_v \cdot (T_2 - T_1)

Therefore, we have;

Taking air as a diatomic gas, we have;

C_v = \dfrac{5\times R}{2} = \dfrac{5\times 8.314}{2} = 20.785 \ J/(mol \cdot K)

The molar mass of air = 28.97 g/mol

Therefore, we have

c_v = \dfrac{C_v}{Molar \ mass} = \dfrac{20.785}{28.97} \approx 0.7175 \ kJ/(kg \cdot K)

The work done per unit mass of gas is therefore;

p \cdot dV =W =   1 \times 0.7175 \times (503.179 - 294.2611) \approx 149.89 \ kJ

The work transfer per unit mass ≈ 149.89 kJ

The heat transfer for an adiabatic process = 0.

8 0
3 years ago
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