Answer:
=4/5 because I'm not going to go back in a year meaning that they are you are
Answer
given,
6 lanes divided highway 3 lanes in each direction
rolling terrain
lane width = 10'
shoulder on right = 5'
PHF = 0.9
shoulder on the left direction = 3'
peak hour volume = 3500 veh/hr
large truck = 7 %
tractor trailer = 3 %
speed = 55 mi/h
LOS is determined based on V p
10' lane weight ; f_{Lw}=6.6 mi/h
5' on right ; f_{Lc} = 0.4 mi/hr
3' on left ; no adjustment
3 lanes in each direction f n = 3 mi/h



= 0.877

= 1,555 veh/hr/lane

= (55 + 5) - 6.6 - 0.4 -3 -0
= 50 mi/h


level of service is D using speed flow curves and LOS for basic free moving of vehicle
Answer:
1. High friction
2. High extrusion temperature
Explanation:
Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.
Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.
Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.
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‰