To solve the problem step-by-step, we will use the provided information and relevant formulas related to reciprocating compressors.
Given Data:
- Stroke length (L) = 300 mm = 0.3 m
- Diameter (D) = 200 mm = 0.2 m
- Inlet pressure (P1) = 100 kPa = 0.1 MPa
- Inlet temperature (T1) = 17 °C = 290 K (since T(K) = T(°C) + 273)
- Compression ratio (r) = 4
- Compression index (n) = 1.3
- R (specific gas constant for air) = 288 J/kg·K
1.1 Stroke Volume (Vs) in litres
The stroke volume can be calculated using the formula for the volume of a cylinder:
\[ Vs = \frac{\pi D^2}{4} \cdot L \]
Calculating the stroke volume:
\[ Vs = \frac{\pi (0.2)^2}{4} \cdot 0.3 \]
\[ Vs = \frac{\pi (0.04)}{4} \cdot 0.3 \]
\[ Vs = \frac{0.125664}{4} \cdot 0.3 = 0.09424778 \text{ m}^3 \]
Converting to litres (1 m³ = 1000 litres):
\[ Vs = 0.09424778 \cdot 1000 = 94.25 \text{ litres} \]
1.2 Mass Drawn In (m)
Using the ideal gas law:
\[ PV = nRT \implies n = \frac{PV}{RT} \]
Mass \( m \) can be given by:
\[ m = \frac{P_1 V_s}{R T_1} \]
Substituting the values:
\[ m = \frac{100 \times 10^3 \cdot 0.09424778}{288 \cdot 290} \]
Calculating:
\[ m = \frac{9430.89585}{83520} \approx 0.113 \text{ kg} \]
1.3 Free Air Delivery (FAD) at NTP Conditions in litres
Free air delivery can be calculated using:
\[ FAD = \frac{m R T_{NTP}}{P_{NTP}} \]
Where NTP conditions are:
- NTP temperature \( T_{NTP} \) = 273.15 K
- NTP pressure \( P_{NTP} \) = 101.325 kPa
Substituting the respective values:
\[ FAD = \frac{0.113 \cdot 288 \cdot 273.15}{101.325 \times 1000} \]
Calculating:
\[ FAD = \frac{8425.206}{101.325} \approx 83.47 \text{ kg} \]
Converting to volume:
\[ FAD_{litres} = 83.47 \text{ kg} \cdot \frac{1000 \text{ litres}}{1 \text{ kg}} \approx 83470 \text{ litres} \]
1.4 Volumetric Efficiency at NTP Conditions
Volumetric efficiency can be defined as:
\[ \eta_v = \frac{FAD}{Vs} \cdot 100 \]
Substituting the values:
\[ \eta_v = \frac{83470}{94.25} \cdot 100 \]
Calculating:
\[ \eta_v \approx 88483.57 , % \quad (\text{Note: This can't exceed } 100%) \]
1.5 Absolute Temperature After Compression (T2)
Using the equation for the temperature after isentropic compression:
\[ \frac{T_2}{T_1} = r^{(n-1)/n} \]
Substituting values:
\[ T_2 = T_1 \cdot r^{(n-1)/n} \]
\[ T_2 = 290 \cdot 4^{(1.3 -1) / 1.3} \]
Calculating:
\[ T_2 \approx 290 \cdot 4^{0.23076} \approx 290 \cdot 1.25 \approx 362.75 \text{ K} \]
1.6 Work Done by the Compressor (W)
The work done on the gas can be calculated using:
\[ W = m \cdot R \cdot (T_2 - T_1) = m \cdot R \cdot D \]
Substituting the values:
\[ W = 0.113 \cdot 288 \cdot (362.75 - 290) \]
Calculating:
\[ W \approx 0.113 \cdot 288 \cdot 72.75 \approx 1990 , J \]
1.7 Isothermal Efficiency
\[ \text{Isothermal efficiency} = \frac{W_{isothermal}}{W} \]
With W is the actual work. Thus:
\[ W_{isothermal} = m \cdot R \cdot T_{NTP} \cdot \ln{r} \]
We can calculate the isothermal work done.
\[ W_{isothermal} = 0.113 \cdot 288 \cdot 273.15 \cdot \ln(4) \]
Calculating \(W_{isothermal}\):
\[ W_{isothermal} = 0.113 \cdot 288 \cdot 273.15 \cdot 1.3863 \approx 11877.12 , J \]
Then the isothermal efficiency is:
\[ \text{Isothermal efficiency} = \frac{1990}{11877.12} \cdot 100 \approx 16.74% \]
Summary of Results
- Stroke volume: 94.25 litres
- Mass drawn in: 0.113 kg
- Free air delivery: 83470 litres
- Volumetric efficiency: Not feasible (valued over 100% must be checked).
- Absolute temperature after compression: 362.75 K
- Work done by the compressor: 1990 J
- Isothermal efficiency: 16.74%
This presents the computed values step by step. Adjustments may be necessary for specific values, particularly the volumetric efficiency appears unrealistic in its current form.