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3 years ago

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Refrigerant-134a enters a diffuser steadily as saturated vapor at 600 kPa with a velocity of 160 m/s, and it leaves at 700 kPa a

nd 40∘C. The refrigerant is gaining heat at a rate of 2 kJ/s as it passes through the diffuser. If the exit area is 80 percent greater than the inlet area, determine (a) the exit velocity and (b) the mass flow rate of the refrigerant.

2 answers:

Zina [86]3 years ago

4 0

Answer:

a) V_2 = 82.1 m/s

b) m = 0.298 Kg/s

Explanation:

from A-11 to A-13 we have the following data

P_1 = 600 kpa

V_1 = 0.033925 m^3/kg

h_1 = 262.52 kJ/kg

P_2 = 700 kpa

V_2 = 0.0313 m^3/kg

T_2 = 40°C = 313K

h_2 = 278.66 kJ/kg

Now, from the conversation of mass,

A_2*V_2/u_2 = A_1*V_1/u_1

V_2 = A_1/A_2*u_2/u_1*V_1

V_2 = A_1/1.8*A_1 * 0.0313 /0.033925*160

V_2 = 82.1 m/s

now from the energy balance equation

E_in = E_out

Q_in + m(h_1 + V_1^2/2) = m(h_2 + V_2^2/2)

m = 0.298 Kg/s

grin007 [14]3 years ago

4 0

Answer:

A) V2 = 82.1 m/s

B) m' = 0.298 kg/s

Explanation:

A) From the tables attached,

At P1 = 600 KPa, we have specific volume and specific enthalpy as;

v1 = 0.033925 m³/kg

And h1 = 262.52 Kj/kg or 262520 J/kg

Also at P2 = 700 KPa, and T2 = 40°C = 313K, we have specific volume and specific enthalpy as ;

v2 = 0.03133 m³/kg and h2 = 278.66 Kj/kg or 278660 J/kg

From conservation of mass, we know that;

A1V1/v1 = A2V2/v2

So, making V2 the subject, we have;

(A1V1v2)/(v1A2) = V2

Now, from the question, the exit area (A2) is 80% greater than the inlet Area (A1). Thus, A2 = A1 + 0.8A1 = 1.8A1

So,

(A1V1v2)/(v1 x 1.8 xA1) = V2

So, A1 will cancel out to give;

(V1v2)/(v1 x 1.8) = V2

Plugging in relevant values;

(160 x 0.0313)/(0.033925 x 1.8) = 82.1 m/s

B) From energy balance, we know that;

E'in = E'out

Thus,

Q'in + m'(h1 + (v1²/2)) = m'(h2 + (v2²/2))

From the question, Q'in = 2kj = 2000J

Thus;

2000 + m'(262520 + (160²/2)) = m'(278660 + (82.1²/2))

So,

2000 + 275320m' = 282030 m'

m'(282030 - 275320) = 2000

m' = 2000/6710 = 0.298 kg/s

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Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine

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A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

$\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}$ (Eq. 1)

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$\dot Q_{total} = \frac{T_{i}-T_{o}}{R}$ (Eq. 2)

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$T_{i}$ , $T_{o}$ - Indoor and outdoor temperatures, measured in Celsius.

$R$ - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

$R = R_{cond} + R_{conv, in}+R_{conv, out}$ (Eq. 3)

Where:

$R_{cond}$ - Conductive thermal resistance, measured in Celsius per watt.

$R_{conv, in}$ , $R_{conv, out}$ - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

$R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}$

$R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)$ (Eq. 4)

Where:

$w$ - Width of the glass window, measured in meters.

$d$ - Length of the glass window, measured in meters.

$l$ - Thickness of the glass window, measured in meters.

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$h_{i}$ , $h_{o}$ - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that $w = 2.4\,m$ , $d = 1.5\,m$ , $l = 0.006\,m$ , $k = 0.78\,\frac{W}{m\cdot ^{\circ}C}$ , $h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}$ and $h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}$ , the overall thermal resistance is:

$R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)$

$R = 0.041\,\frac{^{\circ}C}{W}$

Now, we obtain the steady rate of heat transfer from (Eq. 2): ( $R = 0.041\,\frac{^{\circ}C}{W}$ , $T_{i} = -5\,^{\circ}C$ , $T_{o} = 24\,^{\circ}C$ )

$\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }$

$\dot Q_{total} = 707.317\,W$

The steady rate of heat transfer through the glass window is 707.317 watts.

6 0

2 years ago