Air from inside a house enters a steady-flow heat exchaneger with a volume flow

Question

Air from inside a house enters a steady-flow heat exchaneger with a volume flow rate of 200 m^3/min at 15 C, 100kPa and leaves the that exchanger at 35C. This heat exchanger is located inside the air handler of a house heat pump system and receives heat from the heat pump. The heat pump receives heat from air outside the house that acts as a heat source at 10 C. Assume air has constant specific heats evaluated at 300k.
Determine the Rate of heat supplied to the air in the heat exchangers Q_h in KW.

Explanation / Answer

Assumptions: 1. the house air pressure doesn't significantly change due to this process 2. the air is heated at constant pressure as it flows through the heat exchanger 3. the heat exchanger outer walls are well-insulated 4. air's specific heat capacity is constant. A: To calculate heat supplied to a flowing stream of a fluid with assumed constant specific heats, you use: Q_dot = m_dot*c*(T2 - T1) In this example, we are interested in the constant pressure cp of air, because it is free to expand as it flows through the heat exchanger. Pressure is assumed to not change, because we neglect drag losses and there is no other reason for it to change. Q_dot = m_dot*cp*(T2 - T1) To get mass flow rate, from volumetric flow rate: m_dot = V_dot_1*rho_1 To find initial density of air, use this form of the ideal gas law: rho_1 = P1*M/(R*T1) M is molar mass of air P1 is pressure at state 1 T1 is temperature at state 1 R is the universal Avogadro gas constant Substitute: Q_dot = V_dot_1*(P1*M/(R*T1)) * cp * (T2 - T1) Value of cp for air: cp = k*(R/M)/(k-1) where k, the adiabatic index, for AIR, is 1.4 Substitute: Q_dot = V_dot_1*(P1*M/(R*T1)) * ( k*(R/M)/(k-1))* (T2 - T1) Cancel R/M: Q_dot = V_dot_1*(P1/T1) * ( k/(k-1) ) * (T2 - T1) SAVE THIS FORMULA for later. Part B: This part wants us to assume a Carnot cycle (the most perfect performing cycle that is possible to exist) acts as the refrigeration of the great outdoors, and the consequential heating of the house via the heat rejection component. We have a formula for Carnot coef of performance of a refrigerator. COPr_carnot = 1/(T_H/T_L - 1) Important: T_H and T_L absolutely MUST be in the Kelvin or Rankine scale. However, this isn't a refrigerator per-say, because we aren't desiring to cool off the great outdoors. We are desiring to heat the house. This changes the definition of "what we want", and thus will change the formula that will yield the ratio of "what we want"/"what we pay for". This is still the same gismo as a refrigerator, we've just switched the inside and outside. Still, use the same formula for energy balance on it: Q_in + W_in = Q_out Refrigeration "cooling" COP: COPr = Q_in/W_in "what we want" = Q_in, heat "sucked" from the interior of the house in to the device. Heat pump "warming" COP: COPh = Q_out/W_in "what we want" = Q_out, heat rejected from the device in to the house COPh = (Q_out)/W_in COPh = (Q_in + W_in)/W_in COPh = COPr + 1 COPh_carnot = 1/(T_H/T_L - 1) + 1 COPh_carnot = 1/(1 - T_L/T_H) From earlier, we know the answer to "what we want", which we've called Q_dot then. Call it Q_dot_out, now, because it is heat expelled by the heat pump device. Compare it with what we pay for, W_dot_in COPh = Q_dot_out/W_dot_in Solve for W_dot_in: W_dot_in = Q_dot_out/COPh Best case scenario, COPh = COPh_carnot Thus: W_dot_in = Q_dot_out * (1 - T_L/T_H) Summary: A: Q_dot = V_dot_1*(P1/T1) * ( k/(k-1) ) * (T2 - T1) B: W_dot_in = Q_dot_out * (1 - T_L/T_H) Remember that Q_dot and Q_dot_out are the same concept. Data: V_dot_1 = 3.333 m^3/sec, notice the unit conversion k = 1.4 P1 = 100 kPa T1 = 288.15 K T2 = 308.15 K T_L = 283.15 K T_H = 298.15 K Results: A: Q_dot = 80.97 kW B: W_dot_in = 4.07 kW

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