Explain the working of McLeod gauge for measurement of very low pressure.
1 answer:
Explanation:
McLeod gauge:
It is used to measure very low pressure of gas.It measure gas pressure by the help of mercury(Hg).it measure absolute pressure of gas.
McLeod gauge on the principle of Boyle's law.Boyle's law state that ,at constant temperature the pressure of gas is directly proportional to its volume.From Boyle's law
A sample of gas is taken from vacuum then its pressure is measured by help Hg.
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Answer:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a =
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist =
velocity of motorist =
velocity of motorist = 42.857 km/h
Answer:
work done = 48.88 × J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
Answer:
flow(m) = 54.45 kg/s
thermal efficiency u = 44.48%
Explanation:
Given:
- P_1 = P_8 = 10 KPa
- P_2 = P_3 = P_6 = P_7 = 800 KPa
- P_4 = P_5 = 10,000 KPa
- T_5 = 550 C
- T_7 = 500 C
- Power Output P = 80 MW
Find:
- The mass flow rate of steam through the boiler
- The thermal efficiency of the cycle.
Solution:
State 1:
P_1 = 10 KPa , saturated liquid
h_1 = 192 KJ/kg
v_1 = 0.00101 m^3 / kg
State 2:
P_2 = 800 KPa , constant volume process work done:
h_2 = h_1 + v_1 * ( P_2 - P_1)
h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg
State 3:
P_3 = 800 KPa , saturated liquid
h_3 = 721 KJ/kg
v_3 = 0.00111 m^3 / kg
State 4:
P_4 = 10,000 KPa , constant volume process work done:
h_4 = h_3 + v_3 * ( P_4 - P_3)
h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg
State 5:
P_5 = 10,000 KPa , T_5 = 550 C
h_5 = 3500 KJ/kg
s_5 = 6.760 KJ/kgK
State 6:
P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK
h_6 = 2810 KJ/kg
State 7:
P_7 = 800 KPa , T_7 = 500 C
h_7 = 3480 KJ/kg
s_7 = 7.870 KJ/kgK
State 8:
P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK
h_8 = 2490 KJ/kg
- Fraction of steam y = flow(m_6 / m_3).
- Use energy balance of steam bleed and cold feed-water:
E_6 + E_2 = E_3
flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3
y*h_6 + (1-y)*h_3 = h_3
y*2810 + (1-y)*192.8 = 721
Compute y: y = 0.2018
- Heat produced by the boiler q_b:
q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)
q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)
Compute q_b: q_b = 3303.58 KJ/ kg
-Heat dissipated by the condenser q_c:
q_c = (1-y)*(h_8 - h_1)
q_c= ( 1 + 0.2018)*(2810 - 192)
Compute q_c: q_c = 1834.26 KJ/ kg
- Net power output w_net:
w_net = q_b - q_c
w_net = 3303.58 - 1834.26
w_net = 1469.32 KJ/kg
- Given out put P = 80,000 KW
flow(m) = P / w_net
compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s
- Thermal efficiency u:
u = 1 - (q_c / q_b)
u = 1 - (1834.26/3303.58)
u = 44.48 %
