Answer:
f = 628.32 lb
t = 2513.28 lb-inc
Explanation:
given data:
θ = 45°
outside radius = 6 inch
inside radius = 4 inch
coefficient of friction = 0.4
max pressure = 100 psi
a) determine force required for applying one pad
f =
f =
f = 628.32 lb
b) torque capacity (t)
t =
t = 0.4 *628.32*5
torque = 1256.64 lb-inc
for both pad = 2 * 1256.64 =2513.28 lb-inc
Answer:
hello your question is incomplete attached below is the complete question
answer :
a) attached below
b) 63.71°c
Explanation:
A) Given data
The system has ; mass ( M ) , specific heat capacity , and surface area
attached below is a detailed solution
B) Given data
power generated = 60w
weight of heat sink = 0.31 kg
temperature given = 100°c
temp in ambient air = 20°c
lets take
specific heat capacity = 918.18 J /kg-k
density = 2702 kg/m^2
attached below is the detailed solution
Answer:
Pump factor = Fp = 7.854 gal/cycle
Ev = 82.00 %
P_H = 183.29 hp
Explanation:
Given data:
Dimension of duplex pump
6.5 inch liner
2.5 inch rod
18 inch strokes
Pressure 3000 psig
Pit dimension
7 ft wide
20 ft long
Ls = 18 inch
Velocity = (18)/10
volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100
we know that flow rate is given as = Area * velocity
Theoritical flow rate
Ev = 82.00 %
Pump factor
Fp = 1814.22 in^3/cyl
Fp = 7.854 gal/cycle
Flow rate
Power
Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h
b) The coefficient of performance is:
c) The coefficient of performance of a reversible heat pump is:
Th=20+273=293 K
Tl=10+273=283K
Replacing: