The device shown below mixes steam and cold water and produces warm water. The c
ID: 2994317 • Letter: T
Question
The device shown below mixes steam and cold water and produces warm water. The cold water flows in at 10 kg/s at 50oC. The steam flows in at 0.2 kg/s at 160oC. The resulting water flows out at 60oC. The pressure for all three is a constant 100 kPa. If the pipe out has a 0.03 m diameter, what is the mass flow rate and velocity of the water coming out.
Explanation / Answer
m_hot water = m_cold water + m_steam
m_hot water = 10 + 0.2 = 10.2 kg/s
From water properties,
At 100 kPa and T = 160 deg C, we get enthalpy h_steam = 2800 kJ/kg
At 100 kPa and T = 50 deg C, we get enthalpy h_cold water = 209 kJ/kg
m_hot water*h_hot water = m_cold water*h_cold water + m_steam*h_steam
10.2*h_hot water = 10*209 + 0.2*2800
h_hot water = 259.8 kJ/kg
From water properties at P = 100 kPa and h = 259.8 kJ/kg, we get Temperature T = 62 deg C and density = 982 kg/m^3
Volume flow rate Q = m_hot water / density
Q = 10.2 / 982
Q = 0.010387 m^3/s
Cross-section area A = 3.14/4 *0.03^2 = 0.0007065 m^2
Velocity V = Q / A
V = 0.010387 / 0.0007065
V = 14.7 m/s