Respuesta :
Answer:
( a ) The specific volume by ideal gas equation = 0.02632 [tex]\frac{m^{3} }{kg}[/tex]
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142 [tex]\frac{m^{3} }{kg}[/tex]
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49 [tex]\frac{J}{kg k}[/tex]
(a). From ideal gas equation P V = m R T ---------- (1)
⇒ [tex]\frac{V}{m}[/tex] = [tex]\frac{R T}{P}[/tex]
⇒ Here [tex]\frac{V}{m}[/tex] = Specific volume = v
⇒ v = [tex]\frac{R T}{P}[/tex]
Put all the values in the above formula we get
⇒ v = [tex]\frac{323}{10^{6} }[/tex] ×81.49
⇒ v = 0.02632 [tex]\frac{m^{3} }{kg}[/tex]
This is the specific volume by ideal gas equation.
Actual value = 0.021796 [tex]\frac{m^{3} }{kg}[/tex]
Error = 0.02632 - 0.021796 = 0.004524 [tex]\frac{m^{3} }{kg}[/tex]
% Error = [tex]\frac{0.004524}{0.021796}[/tex] × 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
[tex]\frac{V}{m}[/tex] = v = 0.0142 [tex]\frac{m^{3} }{kg}[/tex]
The actual value = 0.021796 [tex]\frac{m^{3} }{kg}[/tex]
Error = 0.0142 - 0.021796 = [tex]\frac{m^{3} }{kg}[/tex]
% Error = [tex]\frac{- 0.0076}{0.021796}[/tex] × 100
% Error = - 34.85 %
(a) The specific volume of the refrigerant at the given condition is 0.0263 m³/kg.
(b) The error involved the calculation using ideal gas equation is 20.65%.
The given parameters:
- pressure, P = 1 MPa
- temperature, T = 50 ⁰C = 323 K
- actual specific volume, V = 0.021796 m³/kg
The specific volume of the refrigerant is calculated as follows;
[tex]PV = mRT\\\\\frac{V}{m} = \frac{RT}{P}[/tex]
where;
- R is gas constant for refrigerant 134a = 81.49 J/kgK
The specific volume is calculated as follows;
[tex]\frac{V}{m} = \frac{81.49 \times 323}{1 \times 10^6} = 0.0263 \ m^3/kg[/tex]
The error involved the calculation using ideal gas equation is calculated as follows;
[tex]error = calculated \ - \ actual\\\\error = 0.0263 \ m^3/kg \ - \ 0.021796 \ m^3/kg\\\\error = 0.0045 \ m^3/kg\\\\error \ \% = \frac{0.0045}{0.021796} \times 100\%\\\\error \ \% = 20.65 \ \%[/tex]
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