Answer:
The mass flow rate of water is 1.21 kg/s
Explanation:
according to the conditions given in the exercise:
mhcp=4440 W/K
Th1=150°C
Th2=50°C
Tc1=20°C
Tc2=108°C
di=2 cm
do=2.25 cm
Di=6 cm
mh=2 kg/s
Rfo=0.00015 m^2K/W
Rfi=0.0001 m^2K/W
The properties of the water at 45°C are:
p=990.1 kg/m^3
Cp=4180 J/kgK
From the energy balance, we have:
mc*Cpc(Tc2-Tc1)=mh*Cph(Th1-Th2)
Clearing mc:
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , = 1.8
T1 = (1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= = (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = = 135.7*
J/Kg.
Answer:
point B where has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ;
where ;
For point B ;
Let Q be the moment of the Area B;
SO ;
where ;
For point C ;
Let Q be the moment of the Area C;
SO ;
where ;
For point D ;
Let Q be the moment of the Area D;
SO ;
since there is no area about point D
Area = 0
Thus; from the foregoing ; point B where has the largest Q value at section a–a
