Respuesta :
Answer:
a) m = 0.3003 kg/s
b) Vel1 = 6.25 m/s, Vel2 = 7.3725 m/s
c) 12.845 KW
Explanation:
a)
using ideal gas law:
PV = nRT
since, n = no. of moles = m/M
therefore,
PV = (m/M)RT
P1 v1 = RT1/M
where,
P1 = inlet pressure = 1 bar = 100000 Pa
v1 = specific volume at inlet = ?
R = universal gas constant = 8.314 KJ/Kmol.k
T1 = inlet temperature = 290 K
M = Molecular mass of air = 28.9628 Kg/kmol
Therefore,
v1 = (8.314 KJ/Kmol.k)(290 k)/(28.9628 kg/kmol)(100 KPa)
v1 = 0.83246 m³/kg
Now, the mass flow rate can be given as:
Mass Flow Rate = (Volume Flow Rate)/(v1)
Mass Flow Rate = (0.25 m³/s)/(0.83246 m³/kg)
Mass Flow Rate = 0.3003 kg/s
b)
using general gas equation to find the specific volume at the exit first, we get:
P1v1/T1 = P2v2/T2
v2 = P1 v1 T2/T1 P2
where,
P1 = inlet pressure = 1 bar
P2 = exit pressure = 0.95 bar
T1 = inlet temperature = 290 k
T2 = exit temperature = 325 k
v1 = specific volume at inlet = 0.83246 m³/kg
v2 = specific volume at exit = ?
Therefore,
v2 = (1 bar)(0.83246 m³/kg)(325 k)/(290 k)(0.95 bar)
v2 = 0.98203 m³/kg
Now, for velocity, we use formula:
Vel = v/A
where,
A = Area = 0.04 m²
For inlet:
Vel1 = Inlet Volume flow Rate/A = (0.25 m³/s)/(0.04 m²)
Vel1 = 6.25 m/s
For exit:
Vel2 = Exit Volume flow Rate/A = (Mass Flow Rate)(v2)/(0.04 m²)
Vel2 = (0.98203 m³/kg)(0.3003 kg/s)/0.04 m²
Vel2 = 7.3725 m/s
c)
using first law of thermodynamics, with no work done, we can derive the formula given below:
Q = m(h2 - h1) + (m/2)(Vel2² - Vel1²)
where,
Q = rate of heat transfer
m = mass flow rate = 0.3003 kg/s
m(h2 - h1) = change in enthalpy = mCpΔT , for ideal gas
Vel1 = 6.25 m/s
Vel2 = 7.3725 m/s
Also,
Cp in this case will be function of temperature and given as:
Cp = KR/(K-1)
where,
K = 1.4
R = Gas constant per molecular mass of air = 0.28699 KJ/kg.k
Therefore,
Cp = 1.0045 KJ/kg.k
Now, using the values in the expression of first law of thermodynamics, we get:
Q = (0.3003 kg/s)(1.0045 KJ/kg.k)(35 K) + [(0.3003 kg/s)/2][(7.3725 m/s)² - (6.25 m/s)²]
Q = 10.55 KW + 2.29 KW
Q = 12.845 KW
Answer:
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
