Air enters a horizontal, constant-diameter heating duct operating at steady state at 290 K, 1 bar, with a volumetric flow rate of 0.25 m3/s, and exits at 325 K, 0.95 bar. The flow area is 0.04 m2. Assuming the ideal gas model with k 5 1.4 for the air, determine:

(a) the mass flow rate, in kg/s,
(b) the velocity at the inlet and exit, each in m/s, and
(c) the rate of heat transfer, in kW.

A) Mass flow rate = 0.3004 Kg/s

B) Velocity at Inlet = 6.25 m/s

Velocity at exit = 7.3725 m/s

C) Rate of heat transfer = 12.858 Kw

Explanation:

T1 = 290K ;P1 = 1 bar = 100 KPa

T2= 325K ; P2 = 0.95 bar = 95 KPa

A = 0.04 m² ; k = 1.4

Molar mass of air = 28.97 Kg/Kmol

R = 8.314 J/molK

A) Since we are dealing with a steady state mass flow rate through an open system, we will treat this as an ideal gas. Thus ;

PV = nRT

n=m/M where m =1

Thus; V1 = RT1/MP1 = (8.314 X 290)/(28.97 x 100) = 0.8323 m³/Kg

Now, mass flow rate is given by;

Mass flow rate(m') = Volumetric flow rate(V')/Volume(v)

Thus; Mass flow rate(m') = 0.25/0.8323 = 0.3004 Kg/s

From Gay lussacs law; P1V1/T1 = P2V2/T2

So to find the volume at the exit which is V2, let's make V2 the subject of the formula;

(P1V1T2)/(P2T1) = V2

So; V2 = (100 x 0.8323 x 325)/(95 x 290) = 0.9818 m³/kg

B) we know that velocity = volumetric flow rate/area

Thus;

At inlet; Velocity (Vi) = 0.25/0.04 = 6.25m/s

At exit; Velocity (Ve) = Volumetric flow rate at exit/ Area.

We don't know the volumetric flow at exit so let's look for it. from earlier, we saw that;

volumetric flow rate/Volume = mass flow rate

And rearranging, volumetric flow rate(V') = mass flow rate(m) x volume(v)

So V' = 0.3004 x 0.9818 = 0.2949 Kg/s

So, Ve = 0.2949/0.04 = 7.3725 m/s

C) The steady state equation when potential energy is neglected is given by the formula;

Q'= m'Cp(T2 - T1) + (m'/2){(Ve)² - (Vi)²}

Where Q' is the rate of heat transfer.

Cp is unknown. The formula to find Cp is given as ;

Cp = KR/(K-1)

Since we are dealing with change in enthalpy here, the gas constant R will be expressed per molecular mass of the air and so R = 0.287 KJ/kg.k

Cp = (1.4 x 0.287)/(1.4 - 1) = 1.0045 Kj/KgK

And so, Q' = (0.3004 x 1.0045)(325 - 290) + (0.3004/2){(7.3725)² - (6.25)²} = 10.5613 + 2.2967 = 12.858 Kw

b. check your rearview mirror

explanation:

fanswer:

explanation:

f