A fluid particle flowing along a stagnation streamline, as shown in Video V4.9 and Fig. P4.35, slows down as it approaches the s
tagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance s = 0.8 ft upstream of the stagnation point at t = 0 is given approximately by s = 0.8e-0.61t where t is in seconds and s is in feet.(a) Factor the relation for the speed as a function of position out of the relation for speed as a function of time. What is the expression for speed V in terms of position s only?(b) For the position as a function of time given in the problem, what is the relation between the speed of the particle V and time t?(c) Symbolically, what does the derivative of position with respect to time equal?
2 answers:
Answer:
See explanations for step by step procedures into getting answer.
Explanation:
Given that:
A fluid particle flowing along a stagnation streamline, as shown in Video V4.9 and Fig. P4.35, slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance s = 0.8 ft upstream of the stagnation point at t = 0 is given approximately by s = 0.8e-0.61t where t is in seconds and s is in feet.(a) Factor the relation for the speed as a function of position out of the relation for speed as a function of time. What is the expression for speed V in terms of position s only
(b) For the position as a function of time given in the problem, what is the relation between the speed of the particle V and time t
(c) Symbolically, what does the derivative of position with respect to time equal.
See comolete solving at attachment
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Answer:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate =
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat =
net boiler heat =
net boiler heat = 301.94 kW