A/an_ Oscilloscope uses a cathode ray tube and displays all voltages.
1 answer:
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Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
The mass flow rate of refrigerant is 0.352 kg/s
Explanation:
Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:
<u>At State A</u>:
From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:
ha = 244.5 KJ/kg
Sa = 0.93788 KJ/kg.k
<u>At State B</u>:
Since, the process from state A to B is isentropic. Therefore,
Sb = Sa = 0.93788 KJ/Kg
From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:
hb = 256.85 KJ/kg (By interpolation)
<u>At State C</u>:
From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:
hc = 107.34 KJ/kg
Now, from the diagram it is very clear that:
Heat Loss = m(hb = hc)
m = (Heat Loss)/(hb - hc)
where,
m = mass flow rate = ?
Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)
Heat Loss = 52.753 KW
Therefore,
m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)
<u>m = 0.352 kg/s</u>
