Question

Question

Heat flow q = (mass flow rate) x (enthalpy change)

Prove the following with explanations.

               q(Btu/hr) = 4.5(cfm)(Δh Btu/lb dry air)

q(Btu/hr) = 1.08(cfm)(ΔT) (dry bulb temperature in deg.F)

q(Btu/hr) = 4840(cfm)(Δw lb water/lb dry air)

where (cfm) is the air flow rate in cubic feet per minute.

Also, Show following for liquid water

q(Btu/hr) = 500(gpm)(ΔT deg.F)

where (gpm) is water flow rate in US gallons per minute

Answer 1

a)	To prove
	= 4.5 CFM x dh
	We know from heat added formula
	Q = m x Cp x dT
	where
	Q is heat added or subtracted to a substance
	m - mass of the substance
	Cp - specific heat of the substance
	dT - change in temperature
	But for room which is receiving heat from outdoor , where air is
	moving in the room
	Here we are dealing with air
	is flow rate = mass of air flown in unit time
	Specific heat of dry air at 80 F = 0.24 Btu /lb F …..           (1)
	Specific heat of steam at 80 F = 0.445 Btu /lb F…..       (2)
	We know that at normal condition air contains some moisture .
	Hence
	Q = ma Cp a (dT) + mv Cp v (dT)
	dividing RH and LH with some unit of time , to consider flow rate
	= a Cp a (dT) + v Cp v (dT)
	Putting values (1) &(2)
	= a x 0.24 Btu/ lb F x (dT) + v x 0.445 Btu/ lb F x(dT)
	For a typical air conditioning process,
	the ratio mv/ma =0.01
	Hence v = 0.01  a
	= a x 0.24 Btu/ lb F x (dT) + 0.01 a x 0.445 Btu/ lb F x(dT)
	combining terms
	= a x (dT) x (0.24 + 0.00445) Btu/ lb F
	0.00445 is negligible value , Hence we can remove it from qequation
	= a x (dT) x 0.24 Btu/ lb F …..... (3)
	What we intrested in volumetric flow (CFM) not in mass flow rate
	we know,
	mass = volume x density
	mass flow rate = volume flow rate x density of air
	a lb/hour = ft^3/min x 60 min/1 hr x 1 lb/13.3ft^3
	(density of air at standard condition = 1 /13.3 lb/ft^3)
	Therefore ma dot lb/hour = 4.5 CFM
	Putting this value in eq (3)
	= 4.5 CFM x 0.24 Btu/ lb F x dT ….......(4)
	when we simplify equaltion (4) in terms of enthalphy ,
	change in heat content of system,
	= 4.5 CFM x dh
	as specific heat of air into temperature change gives to change in enthalphy change of air
	Thus
	= 4.5 CFM x dh   ------- (a)
b)	To prove
	Qs =1.08 (CFM ) dT
	From equation (4)
	= 4.5 CFM x 0.24 Btu/ lb F x dT in F
	= 4.5 CFM x 0.24 Btu/ lb F x ( dT in F)
	crossing F
	= 4.5 CFM x 0.24 Btu/ lb x dT
	multipling
	(Btu/hr) = 1.08 CFM x dT     ----- (b)
c)	To prove
	Ql = 4840 CFM dW   lb water /lb dry air
	For a room where air with vapor mass mv1 is entering & air with vapor mass mv2 is leaving
	mass voume of vapour condesed in room mw
	mw =mv2- mv1
	flow rates
	w = v2 - v1 ------(5)
	as per defination, ratio w = mass of vapor / mass of dry air
	w = mv/mda
	w = v/ da         in flow rate
	Rearranfing eq (5)
	w = da / da x v2 - da / da x v1
	but 2 / da   = w2
	1 / da   = w1
	w = a (w2 -w1)
	multiplying both sides of equation by latent heat of vaporization
	hv x w= a (w2 -w1) x hv
	but hv x =latent load = Qlatent
	Thus Q latent = hv x w= a (w2 -w1) x hv ------(6)
	Now for LH side
	hv in Btu/lbm & w = lbm/hr
	crossing lbm , we will get LH side = Btu / hr
	And for a
	mass = volume x density
	mass flow rate = volume flow rate x density of air
	a lb/hour = ft^3/min x 60 min/1 hr x 1 lb/13.3ft3
	(density of air at standard condition = 1 /13.3 lb/ft3)
	Therefore a lb/hour = 4.5 CFM
	putting this expression in equation (6)
	Q latent = 4.5 CFM x (w2-w1) x hv    -----(7)
	from enthalphy table,
	the enthalphy change from satured liquid to satured liquid at degree at
	actual condesation process happening at
	( process happens at 32 0F i.e 00C)
	then enthalphy change at 32 F = 1075.2 btu/lbm
	putting this value in equation (7)
	Qlatent = 4.5 (CFM) x dW x 1075.2
	Q latent = 4838.4 (CFM) x dW
	Rounding off
	Q latent = 4840 x (CFM) x dW     ------(c)