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PtichkaEL [24]
3 years ago
15

Consider steady one-dimensional heat transfer through a plane wall exposed to convection from both sides to environments at know

n temperatures T[infinity]1 and T[infinity]2 with known heat transfer coefficients h1 and h2. Once the rate of heat transfer has been evaluated, explain how you would determine the temperature of each surface.

Engineering
1 answer:
Leona [35]3 years ago
4 0

Answer:

Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and  the temperature on the right side of the wall is T∞₂ + Q/h₂A.

Note: kindly find an attached diagram to the complete question given below.

Sources: The diagram/image was researched and taken from Slader website.

Explanation:

Solution

Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:

Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A

= T∞₁  - T∞₂/1/h₁A + L/kA + 1/h₂A

Here

The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as  heat transfer.

Thus

Let us consider the convection heat transfer on the left side of the wall which is given below:

Q = T∞₁ -T₁/1/h₁A

T₁ = T∞₁ - Q/h₁A

Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A

Now

Let us consider the convection heat transfer  on the left side of the wall which is given below:

Q= T₂ -T∞₂/1/h₂A

T₂ = T∞₂ + Q/h₂A

Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A

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Answer:

0.447 s²

Explanation:

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One Newton is kg m/s²:

(0.354 g) (45000 m) / (35.6 kg m/s²)

(0.000354 kg) (45000 m) / (35.6 kg m/s²)

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0.447 s²

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Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
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Answer:

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velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
A piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a pol
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Answer:-0.4199 J/k

Explanation:

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mass of nitrogen(m)=1.329 Kg

Initial pressure\left ( P_1\right )=120KPa

Initial temperature\left ( T_1\right )=27\degree \approx 300k

Final volume is half of initial

R=particular gas constant

Therefore initial volume of gas is given by

PV=mRT

V=0.986\times 10^{-3}

Using PV^{1.49}=constant

P_{1}V^{1.49}=P_2\left (\frac{V}{2}\right )

P_2=337.066KPa

V_2=0.493\times 10^{-3} m^{3}

and entropy is given by

\Delta s=C_v \ln \left (\frac{P_2}{P_1}\right )+C_p \ln \left (\frac{V_2}{V_1}\right )

Where, C_v=\frac{R}{\gamma-1}=0.6059

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Substituting values we get

\Delta s=0.6059\times\ln \left (\frac{337.066}{120}\right )+0.9027 \ln \left (\frac{1}{2}\right )

\Delta s=-0.4199 J/k

4 0
4 years ago
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