This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
The user wants me to solve a physics problem from an image. I need to address Question 2 and Question 3. Question 2.1: Determine what FAD stands for. Step 1: Identify the acronym. FAD stands for Free Air Delivered. Free Air Delivered Question 2.2: Calculate the following for a reciprocating compressor. Given data: Suction pressure, p_1 = 98\,kPa = 98 × 10^3\,Pa Suction temperature, T_1 = 313\,K Delivery pressure, p_2 = 620\,kPa = 620 × 10^3\,Pa Polytropic index, n = 1.29 Volumetric efficiency, _v = 0.73 Mass flow rate, m = 1.6\,kg/s Characteristic gas constant, R = 0.287\,kJ/kg.K = 287\,J/kg.K Atmospheric pressure, p_atm = 101\,kPa = 101 × 10^3\,Pa Atmospheric temperature, T_atm = 290\,K 2.2.1 The indicated power in kilowatts The indicated power for a polytropic compression process in a reciprocating compressor is given by: P_ind = (n)/(n-1) m R T_1 [ ((p_2)/(p_1))^(n-1)/(n) - 1 ] Step 1: Calculate the exponent (n-1)/(n). (n-1)/(n) = (1.29-1)/(1.29) = (0.29)/(1.29) ≈ 0.2248062 Step 2: Calculate the pressure ratio term. ((p_2)/(p_1))^(n-1)/(n) = (620\,kPa98\,kPa)^0.2248062 = (6.3265306)^0.2248062 ≈ 1.5000 Step 3: Substitute the values into the indicated power formula. P_ind = (1.29)/(0.29) × 1.6\,kg/s × 287\,J/kg.K × 313\,K [ 1.5000 - 1 ] P_ind = 4.44827586 × 1.6 × 287 × 313 × 0.5000 P_ind = 320000\,J/s P_ind = 320\,kW The indicated power is 320\,kW. 2.2.2 The induced volume per second The induced volume per second (V_ind) is the volume of air drawn into the cylinder per second at suction conditions. It can be calculated using the ideal gas law: p_1 V_ind = m R T_1. Step 1: Rearrange the ideal gas law to solve for V_ind. V_ind = m R T_1p_1 Step 2: Substitute the given values. V_ind = 1.6\,kg/s × 287\,J/kg.K × 313\,K98 × 10^3\,Pa V_ind = (143689.6)/(98000)\,m^3/s V_ind ≈ 1.4662\,m^3/s The induced volume per second is 1.4662\,m^3/s. 2.2.3 The swept volume and atmospheric volume, both in m^3/s The volumetric efficiency (_v) relates the induced volume to the swept volume: _v = V_indV_s. Step 1: Calculate the swept volume per second (V_s). V_s = V_ind_v V_s = 1.4662\,m^3/s0.73 V_s ≈ 2.0085\,m^3/s The atmospheric volume per second (V_atm) is the volume of the mass flow rate of air if it were at atmospheric conditions (p_atm, T_atm). Use the ideal gas law: p_atm V_atm = m R T_atm. Step 2: Rearrange the ideal gas law to solve for V_atm. V_atm = m R T_atmp_atm Step 3: Substitute the given values. V_atm = 1.6\,kg/s × 287\,J/kg.K × 290\,K101 × 10^3\,Pa V_atm = (133024)/(101000)\,m^3/s V_atm ≈ 1.3171\,m^3/s The swept volume per second is 2.0085\,m^3/s and the atmospheric volume per second is 1.3171\,m^3/s. --- Question 3.1: Explain the following in the context of steam plants. 3.1.1 Saturation temperature Saturation temperature is the temperature at which a pure substance, like water, changes phase (boils or condenses) at a given pressure. For a steam plant, it's the temperature at which water turns into steam or steam turns into water at the boiler's operating pressure. 3.1.2 Function of the economiser An economiser is a heat exchanger in a steam plant that preheats the boiler feed water using the hot exhaust gases from the boiler. This recovers waste heat, increasing the overall thermal efficiency of the plant and reducing fuel consumption. 3.1.3 Control rods Control rods are used in nuclear steam plants to regulate the rate of fission reactions in the nuclear reactor core. They are typically made of neutron-absorbing materials like cadmium or boron and are inserted or withdrawn to control the power output and temperature of the reactor. --- Question 3.2: Calculate the following for a boiler. Given data: Steam quality, x = 0.98 Steam pressure, p_steam = 760\,kPa Steam mass flow rate, m_steam = 90\,kg/min = (90)/(60)\,kg/s = 1.5\,kg/s Calorific value of oil, CV = 31\,MJ/kg = 31 × 10^3\,kJ/kg Oil mass flow rate, m_oil = 11\,kg/min = (11)/(60)\,kg/s ≈ 0.1833\,kg/s Feed water temperature, T_feed = 41.6\,^ We need steam table values for p = 760\,kPa (0.76 MPa). From steam tables at p = 760\,kPa: Saturation temperature, T_sat ≈ 168.36\,^ Specific enthalpy of saturated liquid, h_f ≈ 709.9\,kJ/kg Specific enthalpy of evaporation, h_fg ≈ 2060.0\,kJ/kg Specific enthalpy of saturated vapor, h_g = h_f + h_fg ≈ 709.9 + 2060.0 = 2769.9\,kJ/kg For feed water at T_feed = 41.6\,^: Specific enthalpy of feed water, h_feed ≈ h_f at 41.6\,^. From steam tables at T = 41.6\,^: h_feed ≈ 4.18 × 41.6\,kJ/kg ≈ 173.968\,kJ/kg (using specific heat capacity of water c_p = 4.18\,kJ/kg.K) More precisely from steam tables, h_f at 41.6\,^ is approximately 174.1\,kJ/kg. Let's use 174.1\,kJ/kg. 3.2.1 The steam produced per kilogram of oil burnt This is the steam-to-fuel ratio, m_steamm_oil. Step 1: Calculate the ratio using the given mass flow rates. m_steamm_oil = 90\,kg/min11\,kg/min m_steamm_oil ≈ 8.1818\,kg steam/kg oil The steam produced per kilogram of oil burnt is 8.1818\,kg steam/kg oil. 3.2.2 The specific enthalpy at the boiler inlet and outlet Boiler inlet specific enthalpy (h_in) is the specific enthalpy of the feed water. Boiler outlet specific enthalpy (h_out) is the specific enthalpy of the wet steam. Step 1: Determine h_in. h_in = h_feed at 41.6\,^. From steam tables, h_f at 41.6\,^ ≈ 174.1\,kJ/kg. h_in = 174.1\,kJ/kg Step 2: Determine h_out. For wet steam, h_out = h_f + x · h_fg at 760\,kPa. Using values from steam tables at 760\,kPa: h_f = 709.9\,kJ/kg and h_fg = 2060.0\,kJ/kg. h_out = 709.9\,kJ/kg + 0.98 × 2060.0\,kJ/kg h_out = 709.9\,kJ/kg + 2018.8\,kJ/kg h_out = 2728.7\,kJ/kg The specific enthalpy at the boiler inlet is 174.1\,kJ/kg and at the outlet is 2728.7\,kJ/kg. 3.2.3 The enthalpy produced per hour The enthalpy produced per hour is the total heat absorbed by the steam per hour. Q_out = m_steam (h_out - h_in) The mass flow rate of steam needs to be in kg/hour. m_steam = 90\,kg/min × 60\,min/hour = 5400\,kg/hour. Step 1: Calculate the enthalpy difference. h_out - h_in = 2728.7\,kJ/kg - 174.1\,kJ/kg = 2554.6\,kJ/kg Step 2: Calculate the total enthalpy produced per hour. Q_out = 5400\,kg/hour × 2554.6\,kJ/kg Q_out = 13794840\,kJ/hour Q_out = 13794.84\,MJ/hour The enthalpy produced per hour is 13794.84\,MJ/hour. 3.2.4 The boiler thermal efficiency Boiler thermal efficiency (_boiler) is the ratio of the heat absorbed by the steam to the heat supplied by the fuel. _boiler = m_steam (h_out - h_in)m_oil × CV Step 1: Calculate the heat supplied by the fuel per hour. m_oil = 11\,kg/min × 60\,min/hour = 660\,kg/hour. Heat supplied = m_oil × CV = 660\,kg/hour × 31\,MJ/kg Heat supplied = 20460\,MJ/hour Step 2: Calculate the boiler thermal efficiency. _boiler = 13794.84\,MJ/hour20460\,MJ/hour _boiler ≈ 0.6749 _boiler ≈ 67.49\% The boiler thermal efficiency is 67.49\%. That's 2 down. 3 left today — send the next one.2.1 Determine what FAD stands for. FAD stands for Free Air Delivered. Free Air Delivered 2.2 Calculate the following for a reciprocating compressor. Given data: Suction pressure, p_1 = 98\,kPa = 98 × 10^3\,Pa Suction temperature, T_1 = 313\,K Delivery pressure, p_2 = 620\,kPa = 620 × 10^3\,Pa Polytropic index, n = 1.29 Volumetric efficiency, _v = 0.73 Mass flow rate, m = 1.6\,kg/s Characteristic gas constant, R = 0.287\,kJ/kg.K = 287\,J/kg.K Atmospheric pressure, p_atm = 101\,kPa = 101 × 10^3\,Pa Atmospheric temperature, T_atm = 290\,K 2.2.1 The indicated power in kilowatts The indicated power for a polytropic compression process is given by: P_ind = (n)/(n-1) m R T_1 [ ((p_2)/(p_1))^(n-1)/(n) - 1 ] Step 1: Calculate the exponent (n-1)/(n). (n-1)/(n) = (1.29-1)/(1.29) = (0.29)/(1.29) ≈ 0.2248062 Step 2: Calculate the pressure ratio term. ((p_2)/(p_1))^(n-1)/(n) = (620\,kPa98\,kPa)^0.2248062 = (6.3265306)^0.2248062 ≈ 1.5000 Step 3: Substitute the values into the indicated power formula. P_ind = (1.29)/(0.29) × 1.6\,kg/s × 287\,J/kg.K ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._