Water is contained in a piston-cylinder assembly undergoes four processes separately, all started from an initial state where th
e pressure is p1, the temperature is T1, and the specific volume is v1:
Process 1→2: Constant temperature (isothermal) to p2 = 2p1; Process 1→3: Constant specific volume to p3 = 2p1;
Process 1→4: Constant pressure (isobaric) to v4 = 2v1; Process 1→5: Constant temperature (isothermal) to v5 = 2v1.
Draw a p-v diagram and a T-v diagram for the water including the saturated liquid line and the saturated vapor line (dome).
The initial state (state 1) is at the saturated vapor state, note down state 2, 3, 4 and state 5 on both diagrams, and draw all processes (Process 1→2, 1→3, 1→4 and 1→5)
Process 1→2: Constant temperature (isothermal) to p2 = 2p1; Process 1→3: Constant specific volume to p3 = 2p1;
Process 1→4: Constant pressure (isobaric) to v4 = 2v1; Process 1→5: Constant temperature (isothermal) to v5 = 2v1.
Draw a p-v diagram and a T-v diagram for the water including the saturated liquid line and the saturated vapor line (dome).
The initial state (state 1) is at the saturated vapor state, note down state 2, 3, 4 and state 5 on both diagrams, and draw all processes (Process 1→2, 1→3, 1→4 and 1→5)
1 answer:
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Solution :
Given :
k = 0.5 per day
Volume, V
Now, input rate = output rate + KCV ------------- (1)
Input rate
The output rate
= ( 40 + 0.5 ) x C x 1000
Decay rate = KCV
∴
= 1.16 C mg/s
Substituting all values in (1)
C = 4.93 mg/L
Answer:
Explanation:
given data:
Diameter =
from continuity equation
by energy flow equation
and q =0, w =0 for nozzle
therefore we have
but we know dh = Cp dt
hence our equation become