solve the MATLAB part please (2nd Part)
Problem Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour), An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of 150 m- and an average height of 3 m at 1000 m elevation where the standard atmospheric pressure is 89.995 kPa. The house is maintained at a temperature of 22 C and the infiltration losses are estimated to amount to 0.7 ACH. Assuming the pressure and the temperature inside and outside the house remain constants, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is 5 C. Also determine the cost of this energy loss for that day if the unit cost of electricity in that area is $0.082/kWh. 22C 0.7 ACH sor AIR If the pressure at different altitude h (1000, 1500, 2000, 2500) m is given by the following formula: where the constant parameters are as described below Parameter ghM P(h) = p. e ToR Description Sea level standard atmospheric pressure Temperature lapse rate, - / for dry air Constant-pressure specific heat Sea level standard temperature Earth-surface gravitational acceleration Molar mass of dry air Universal gas constant Value 101325 Pa -0.00976 Km 1004.68506 KK) 288.16 K 9.80655 m/s2 0.02896968 kg/mol 8.314462618 J/mol K) Determine the amount of energy loss from the house for each one of the above altitudes. Plot the graph (Energy Loss vs Altitude (mandatory to use a Matlab code). Cp = 1.007 kJ/kg. C. Due: 02/17/2020 Problem Infiltration of cold air into a warm house during winter through the cracks around doors, windows, and other openings is a major source of energy loss since the cold air that enters needs to be heated to the room temperature. The infiltration is often expressed in terms of ACH (air changes per hour), An ACH of 2 indicates that the entire air in the house is replaced twice every hour by the cold air outside. Consider an electrically heated house that has a floor space of 150 m- and an average height of 3 m at 1000 m elevation where the standard atmospheric pressure is 89.995 kPa. The house is maintained at a temperature of 22 C and the infiltration losses are estimated to amount to 0.7 ACH. Assuming the pressure and the temperature inside and outside the house remain constants, determine the amount of energy loss from the house due to infiltration for a day during which the average outdoor temperature is 5 C. Also determine the cost of this energy loss for that day if the unit cost of electricity in that area is $0.082/kWh. 22C 0.7 ACH sor AIR If the pressure at different altitude h (1000, 1500, 2000, 2500) m is given by the following formula: where the constant parameters are as described below Parameter ghM P(h) = p. e ToR Description Sea level standard atmospheric pressure Temperature lapse rate, - / for dry air Constant-pressure specific heat Sea level standard temperature Earth-surface gravitational acceleration Molar mass of dry air Universal gas constant Value 101325 Pa -0.00976 Km 1004.68506 KK) 288.16 K 9.80655 m/s2 0.02896968 kg/mol 8.314462618 J/mol K) Determine the amount of energy loss from the house for each one of the above altitudes. Plot the graph (Energy Loss vs Altitude (mandatory to use a Matlab code). Cp = 1.007 kJ/kg. C. Due: 02/17/2020
