Answer:
The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Explanation:
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
assume from specifications that k = 5v/5cm
= 1v/cm
u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2
= 0.01225v
v = 2v * 0.001
= 0.002v
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
= ((0.01225)^2 + (0.002)^2)^(1/2)
= 0.0124 cm
Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer: Pull.
Because it's all about height width and Breadth!
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
