Answer:
the minimum shaft diameter is 35.026 mm
the maximum shaft diameter is 35.042mm
Explanation:
Given data;
D-maximum = 35.020mm and d-minimum = 35.000mm
we have to go through Tables "Descriptions of preferred Fits using the Basic Hole System" so from the table, locational interference fits H7/p6
so From table, Selection of International Trade Grades metric series
the grade tolerance are;
ΔD = IT7(0.025 mm)
Δd = IT6(0.016 mm)
Also from Table "Fundamental Deviations for Shafts" metric series
Sf = 0.026
so
D-maximum
Dmax = d + Sf + Δd
we substitute
Dmax = 35 + 0.026 + 0.016
Dmax = 35.042 mm
therefore the maximum diameter of shaft is 35.042mm
d-minimum
Dmin = d + Sf
Dmin = 35 + 0.026
Dmin = 35.026 mm
therefore the minimum diameter of shaft is 35.026 mm
Answer:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
Please kindly see explaination and attachment.
Explanation:
a) isentropic efficiency = 84.905%
b) rate of entropy generation = .341 kj/(kg.k)
The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.
Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.
Please refer to attachment for step by step solution of the question.
er:
Explanation:Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A
Answer:
As the asteroid falls closer to the Earth's surface its <u>Gravitational</u> <u>Potential</u> energy <em>decreases</em> and its <u>Kinetic</u> energy <em>increases</em>.
Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits = 
= 2¹³ = 8192
