An LNG fueled vehicle consumes about 12L/h of 422 kg/m³ density cryog

Question

An LNG fueled vehicle consumes about 12L/h of 422 kg/m³ density cryogenic liquid<br />methane. To be used in the engine, the liquid methane must be converted to vapor using<br />a closed (fluids do not mix) heat exchanger, similar to that in problem 5-84. Engine coolant<br />with a density of about 1000 kg/m³ and a specific heat of 3.4 kJ/kgK enters the outer tube <br />of the counter flow exchanger at 100^oC and exits at 30^oC. The methane is inside the inner<br />tube. The latent heat of methane for this situation is said to be 512 kJ/kg (liquid to gas - h_fg). <br />First, determine the flow rate of methane (LNG). Then, what engine coolant flow rate is required, <br />assuming an adiabatic heat exchanger?

Explanation / Answer

Flow rate of methane = (12*10^-3)/ (60*60) *422 = 1.4067*10^-3 kg/s

Heat transfer to coolant = m_coolant*Cp_coolant*(Temperature change of coolant)

= m_coolant*3.4*(100 - 30)

Energy required to change liquid methane into vapor = m_methane*Latent heat

= 1.4067*10^-3 * 512

= 0.72 kW

Now energy conservation yields, 0.72 = m_coolant*3.4*(100 - 30)

m_coolant = 0.003026 kg/s

Coolant flow rate = 0.003026 / 1000 m^3/s = 0.003026*10^-3 m^3/s = 0.003026 L/s = 0.003026*60*60 L/h = 10.894 L/hr

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