Question

Refrigerant 134a enters a horizontal pipe operating at steady state at 40°C, 300 kPa, and a velocity of 25 m/s. At the exit, the temperature is 90°C and the pressure is 240 kPa. The pipe diameter is 0.1 m. Determine: (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity at the exit, in m/s, and (c) the rate of heat transfer between the pipe and its surroundings, in kW.

Answer 1

Answer:

a) 2.42 $kg/s$

b) 37.20 m/s

c) 120.56 kW

Explanation:

Given that:

The fluid in the Refrigerant = R-134a

Diameter (d) = 0.1 m

In the Inlet:

Temperature $T_1 = 40^0C$

Pressure $P_1= 300kPa$

Velocity $V_1$ = 25 m/s

At the exit:

Temperature $T_2 = 90^0C$

Pressure $P_2 = 240 kPa$

From the  Table A-12 for Refrigerant R-134a at $T_1 = 40^0C$ and $P_1= 300kPa$

Specific Volume $v_1 = 0.0809 m^3/kg$

From the  Table A-12 for Refrigerant R-134a at $T_2 = 90^0C$ and $P_2 = 240 kPa$

Specific Volume $v_2 = 0.12038 kJ/kg$

Their corresponding Enthalpy $h_1$ and $h_2$ are as follows:

Enthalpy $h_1$  =284.05 kJ/kg

Enthalpy $h_2$ = 333 kJ/kg

a) The mass flow rate of the refrigerant can be calculated as :

$m_1 = \frac{AV_1}{v_1}$

$m_1 = \frac{\frac{\pi (0.1)^2}{4}*25}{0.08089}$

$m_1 = 2.42 kg/s$

b) The velocity at the exit point:

we knew that:

$m=m_1 =m_2$

∴

$\frac{AV_1}{v_1} =\frac{AV_2}{v_2}$

$V_2 = \frac{v_2}{v_1} V_1$

$V_2 = \frac{0.12038}{0.08089} *25$

$V_2 = 37.20 m/s$

c) Expression for calculating heat transfer (as long as there is no work that is said to be done and the pipe is horizontal) can be represented as:

$Q_{cv}= m[(h_2-h_1)+\frac{1}{2}(V_2^2-V_1^2)]$

$Q_{cv}= 2.42*[(333.49-284.05)+\frac{1}{2}(37.20^2-25^2)]$

$Q_{cv}= 2.42*[49.44+379.42]$

$Q_{cv}= 119.6448kW+918.19W(\frac{1kW}{1000W} )$

$Q_{cv}= 119.6448kW+0.92 kW$

$Q_{cv} = 120.56 kW$

Answer 2

Following are the solution to the given points:

Obtain the following property at pressure $300\ kPa$ and $40^{\circ} \ C$ from the property of superheated $134a$  

Using the interpolation method,  

Specific volume, $v_1 = 0.0866 -(0.0866 -0.07518) (\frac{0.3-0.28}{0.32-0.28})= 0.08089 \frac{kg}{m^3}$

Enthalpy, $h_1 = 284.42 - (284.42 - 283.67) (\frac{0.3-0.28}{0.32-0.28}) = 284.05\ \frac{kJ}{kg}$

Obtain the following property at pressure $240 \ kPa \ and \ 90^{\circ} \ C$ from the property of superheated $134a$.  

Specific volume,$v_2 = 0.12 \ \frac{kg}{m^3}$

Enthalpy of superheated $134a, \ \ h_2 = 333 \ \frac{kJ}{kg}$

For point a:

Calculating the refrigerant weight rate flow:  

$m_1 =\frac{AV_1}{v_1} =\frac{ \frac{ \pi (0.1)^2}{4} \times 25}{0.08089} = \frac{0.196}{0.08089} \\\\ m_1 = 2.42 \frac{kg}{s}$

Thus, refrigerant weight rate flow is $2.42\ \frac{m^3}{kg}$

For point b:

Calculate the exit velocity:

$m_1 = m_2\\\\ \frac{A V_1}{v_1}=\frac{AV_2}{v_2}\\\\ V_2=\frac{V_2}{v_1} V_1\\\\v_2= \frac{0.12038}{ 0.08089} \times 25\\\\ V_2 = 37.20 \frac{m}{s}$

Thus, the exit velocity is $37.20\ \frac{m}{s}$

For point c:

From the energy rate balance Since, there is no work being done and the pipe is horizontal.  So, the above equation can be written as

$\to Q_{cv} = m [(h_2 - h_1)+ \frac{1}{2}(v_{2}^{2}- v_{1}^{2})]$

          $= 2.42 \times [(333.49 - 284.05) +\frac{1}{2}(37.20^2-25^2)]\\\\ = 2.42 \times [49.44 +379.42] \\\\= 119.64\ kW +918.19 \ W |\frac{1 \ KW}{1000\ W}| \\\\= 119.64 \ kW +0.92\ kW\\\\ =120.56\ kW$

Thus, the heat transfer rate among pipe and its surrounding $120.56\ kW$

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