Answer:
- the fraction of the turbine work output used to drive the compressor is 50%
- the thermal efficiency is 29.6%
Explanation:
given information:
= 300 k
= 700 kPa
= 580 K
Qin = 950 kj/kg
efficiency, η = 86% = 0.86
Assume, Ideal gas specific heat capacities of air, = 1.005
k = 1.4
(a) the fraction of the turbine work output used to drive the compressor
Qin = ( - )
- = Qin /
= Qin / +
thus,
= = 1525 K
= \
=
so,
= \
= \
= 1525 K
= 875 K
= ( - )
= 1.005 kj/kgK (580 K - 300 K)
= 281.4 kJ/kg
= η ( - )
= 0.86 (1.005 kj/kgK) (1525 k - 875 K)
= 562.2 kJ/kg
therefore,
the fraction of turbine =
=
= 50%
(b) the thermal efficiency
η = ΔW/
= ( - )/
= /950
= 29.6%
Answer:
9.14hrs
Explanation:
1 horse power = 745.7 watts
1.5 horse power = 1118.55 watts
1 ft = 0.305 m
24 ft = 7.3152 m
60 ft = 18.288 m
the water tank is spherical, volume of a spherical tank = 3/4 πr³ where is r is the radius of the tank = 7.3152 m / 2 = 3.66 m
volume = 4/3 ( 3.66 m ³) × 3.142 = 205.394 m³
mass of water = density × volume = 205.394 m³ × 1000kg/m³ = 205,394 kg
weight of water = mass × acceleration due to gravity = 205,394 kg × 9.8 = 2012861.2 N
potential energy stored at the height where water was stored = mgh = weight × height = 2012861.2 N × 18.288 m
potential energy stored = energy from the pump = power in watt × t
1118.55 watts × t =2012861.2 N × 18.288 m
t = (2012861.2 N × 18.288 m) / 1118.55 watts = 32846.62 secs = 9.14 hrs
Answer:
lunar eclipse
Explanation:
if the earth is between to two (not bringing the moons orbit into account) a lunar eclipse happens
If the car is accelerating uniformly then,
velocity of car at the fourth second is 16m/s
Then if the car with this velocity stopped after 2s then acceleration is,
a=0-16/2=-8m/s^2
braking distance is given by
d=16*2-0.5*8*2*2
=32-16
=16m
this answer is valid only if the motion is one dimensional
all the best.