Question

3. Derive energy balance equation, water vapor mass balance equation, and determine the heat transfer rate for the heating process shown below. Heating or cooling medium w 1 W2. W 1) Energy balance equation 2) Mass balance equation 3) Heat transfer rate

Answer 1

Let $h_d$ be enthalpy of dry air and $h_w$ be enthalpy of water vapour.

Then total enthalpy at temperature

(t°C) = h = malha + w* h) = malCpt + wlhfg+Cpwt))

Where $C_{p_{a}}$ = specific heat of air ,$C_{p_{w}}$ = Specific heat of water, $h_{fg}$ = latent heat of water, = humidity ratio

0 = та

In above equations, reference is taken that at 0C
0 = Py
and also for saturated liquid water
Hy = 0

Energy balance

maihı = mach2 mal(Cpeti + wi(hfg+Cpwtı)) +q = ma2(Cpet2 + w2(hfg + Cpwtz))
-Ans

Mass balance

As there is no other air or water vapour addition into the control volume

For air: Incoming air mass rate = outgoing air mass rate -Ans

> ma = mas

For water vapour: Incoming water mass rate = outgoing water mass rate

> mw, = mw

=> W1*ma, = W2 * max

> W1= wz
-Ans

Putting the values from answers from mass balance into the energy balance we will get

- Modified Energy balance equation

(Cpti + wlhfg+Cpwtı))+q = (Cp t2 + whfg+Cputz))

Heat transfer rate

From the last equation we can obtain q

-Ans

q = (Cp t2 + wlhfg+Cpwtz)) - (Cpti + wlhfg + Cpwti)) ☆q = Cp.(t2 – tı) + Cpu (t2 – tı)

Approx numerical values of the constants used

Cp. = 1.005kJ/kgK, Cp = 1.88kJ/kgk, hfq = 2501kJ/kg

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