What is the heat transfer expected from a ground-loop system that has a 10 gpm flowrate, a heat exchanger inlet temperature of 45°F, and a heat exchanger outlet temperatureof 56°F? Assume that the system uses pure water.A. 52,750 Btu/h C. 55,000 Btu/hB. 53,350 Btu/h D. 56,450 Btu/h
Use the open system calorimetry formula: Q_dot = m_dot*(h2 - h1) h2 and h1 can be looked up as enthalpy based on given temperatures. However, an approximation of calorically perfect water exists, with a constant c_p. Since water is a liquid, you can assume (if you must) that there is no difference between isochoric specific heat (c_v) and isobaric specific heat (c_p). Should it vaporize, you must understand the difference. Including definition of c_p, we can replace h with c_p*T: Q_dot = m_dot*c_p*(T2 - T1) Since we're given V_dot instead of m_dot, include the factor of rho, which relates the two: Q_dot = rho*V_dot*c_p*(T2 - T1) We are given V_dot in gallons per minute. Convert to a m^3/sec because metric units are nicer. Do the same with temperatures. T2:= 13.3333 C T1:= 7.22222 C V_dot := 6.30902e-4 m^3/sec Look up c_p and rho for water: c_p := 4184 Joules/kg-K rho:= 1000 kg/m^3 Calculate to find result: Q_dot = 16131 Watts Convert to BTU/hr: Q_dot = 55,043 BTU/hr Answer is C
The basic engineering formula for the heat load of water is BTU/Hr = 500 GPM (dF) so 500X10x11 =55, 000 BTU/Hr
