How and why do scientists use models?
1 answer:
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Answer:
W = 5701 KW
Explanation:
From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;
Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s
Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s
Volumetric flow rate = 15 m^(3)/s
Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.
Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).
So from the table,
v2 = 2.087 m^(3)/kg
Now, mass flow rate (m) = (AV) /v
Where AV is the volumetric flow rate.
Thus, the mass flow rate at exit could be calculated as;
m = 15/(2.087) = 7.17 kg/s
We also know energy equation could be defined as;
Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]
Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;
-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}
From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg
Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;
h2 = 2665.8 KJ/Kg
So we now calculate power developed;
W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW
Since the sign is not negative but positive, it means that the power is developed from the system.
Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
