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Building Physics Exam with
Solutions
Building Physics
Density and Specific Volume
A substance has a mass of 3 kg and occupies a volume of 6 m³. To calculate its density and specific volume, we can use the following formulas:
Density (ρ) = Mass (m) / Volume (V) ρ = 3 kg / 6 m³ = 0.5 kg/m³
Specific Volume (v) = Volume (V) / Mass (m) v = 6 m³ / 3 kg = 2 m³/kg
Kinetic Energy
The kinetic energy (KE) of a body with mass (m) and velocity (v) is given by the formula:
KE = 1/2 × m × v²
In this case, the body has a mass of 3 kg and a velocity of 30 m/s. Substituting the values, we get:
KE = 1/2 × 3 kg × (30 m/s)² = 1350 kJ
Energy Change of a Stationary System
For a stationary system, the energy change (ΔE) is equal to the change in internal energy (ΔU), as the kinetic energy (ΔKE) and potential energy (ΔPE) are both zero.
ΔE = ΔU
Energy Transfer in an Adiabatic System
In an adiabatic system, the only form of energy that can cross the boundaries is work (W). Heat (Q) cannot be transferred across the boundaries of an adiabatic system.
Coefficient of Performance (COP) of a Heat Pump
The coefficient of performance (COP) of a heat pump is the ratio of the heat output (Q_out) to the work input (W_in).
COP = Q_out / W_in
Given: - The heat pump absorbs 600 W of electricity. - It provides 10800 kJ of heat working steadily for 2 hours.
To calculate the COP, we need to convert the heat output to watts: Q_out = 10800 kJ / 2 h = 5400 W
COP = Q_out / W_in = 5400 W / 600 W = 9
Reversed Carnot Cycle
The Reversed Carnot Cycle consists of the following transformations, as shown in the PV diagram:
Isothermal compression (heat rejection) Adiabatic compression Isothermal expansion (heat absorption) Adiabatic expansion
The heat exchanges take place during the isothermal transformations, with heat being rejected during the compression and absorbed during the expansion.
Neutrality Temperature
The typical equation that correlates the suitable indoor temperature (neutrality temperature, T_n) with the monthly mean outdoor temperature (T_o,av) is:
T_n = a + b T_o,av
where 'a' and 'b' are parameters that can be specified based on the reference equations or values provided in the book or slides.
Energy Balance of a Room
The energy balance of the room can be calculated as follows:
Q_c = Heat loss through the glazed façade Q_s = Solar heat gain through the glazed façade Q_v = Heat loss due to ventilation and infiltration Q_i = Internal heat gains
The total energy balance is: Q_bal = (Q_c + Q_v) - (Q_s + Q_i)
Given: - Glazed façade area: 10 m² - U-value of the façade: 2.0 W/m²K - Solar gain factor (g): 0.8 - Air changes due to ventilation and infiltration: 0.5 vol/h
- Inside air volume: 50 m³ - Air heat capacity (C_p,air): 0.35 Wh/m³K - Internal gains: 50 W - Outside average temperature (T_o,av): 5°C - Daily solar radiation on the façade: 500 Wh/m²
Calculating the energy balance components: Q_c = 2 W/m²K × 10 m² × 15 K × 24 h = 7.2 kWh Q_s = 10 m² × 0.8 × 500 Wh/m² = 4 kWh Q_v = 0.5 h⁻¹ × 50 m³ × 0.35 Wh/m³K × 15 K × 24 h = 3.15 kWh Q_i = 50 W × 24 h = 1. kWh
