We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.
For 1bar=1000kPa:
Replacing,
With the specific volume we know can calculate the mass flow, that is
Then the heat required in input is,
With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,
Finally we can apply the steady flow energy equation, that is
Re-arrange for Q,
We can note that consider the Kinetic Energy will decrease the heat input.
B, tech b only, Blow by is describe or defined as air going by the compression rings. Hope it helps
Answer:
The power developed in HP is 2702.7hp
Explanation:
Given details.
P1 = 150 lbf/in^2,
T1 = 1400°R
P2 = 14.8 lbf/in^2,
T2 = 700°R
Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h
Using air table to obtain the values for h1 and h2 at T1 and T2
h1 at T1 = 1400°R = 342.9 Btu/h
h2 at T2 = 700°R = 167.6 Btu/h
Using;
Q - W + m(h1) - m(h2) = 0
W = Q - m (h2 -h1)
W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h
W = (-65000 Btu/h ) - (-1928.3) Btu/s
W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s
W = -18.06Btu/s + 1928.3 Btu/s
W = 1910.24Btu/s
Note; Btu/s = 1.4148532hp
W = 2702.7hp
Answer:
Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.