Question

a perfectly insulated 50 kg salt bath (solid granulated; MW:58.44 g/mol, p=2.0 g/cm^3) is preheated to 1000K. It then has air initially at 400K being heated to 575K, which cools the salt bath to 950K. How many moles of air were sent through the salt bath heat exchanger?

Answer 1

Given :

For Salt bath

Salt bath mass = m1 ; kg =50 kg ; Preheat or Initial temperature T1; K = 1000 K ; Final temperature T2 ;K = 950 K

For Air

Air mass = m2 kg; Initial Air temperature T3; K = 400 K ; Final temperature T4; K= 575K

Assumption

Specifc Heat of salt: Data is missing on the specific heat of salt. It is assumed to be 0.3 kJ / kg K = Cp1.

Specifc Heat of Air: The variation of Cp over the range of temperature is minimal and can be obtained from any standard reference data. It is assumed to be 1.03 kJ / kg K = Cp₂

By heat balance for the heat exchanger

Heat lost by salt bath = Heat gain by air

m1 * Cp₁ * (T1-T2) = m2 * Cp₂ * (T4-T3)

50 * 0.3 * (1000-950) = .m2 * 1.03 * (575-450)

hence m2 = 5.8 kg

Molecular weight of air = MW= 29 kg/kmol;

Moles of air sent through the exchanger ; kmol = m2 / MW = 5.8 / 29 = 0.2 kmoles