Explanation:
umm can you explain me what i have to answer
Answer:
<h2>
11.65kW</h2>
Explanation:
Step one:
given data
from the table for R-134a
<em>For the given pressure state the following data were extracted.</em>
s=0.94202kJ/kg.K
h1=241.14kJ/kg
∝1=0.12355m^3/kg
<em>By interpolation, the final enthalpy and pressure with data from the table of R-134a</em>
<em>h2=277.12kJ/kg</em>
The minimum power output is then determined from the energy balance
W=m(h2-h1)
W=V1/∝1(h2-h1)
W=2.4/60/0.12355(277.12-241.14)
W=0.04/0.12355(35.98)
W=0.3237(35.98)
W=11.65kW Approx.
<em><u>The minimum power output of the compressor is 11.65kW</u></em>
Answer:
The overflow rate is 4.24×10^-4 m/s.
The detention time is 7069.5 s
Explanation:
Overflow rate is given as volumetric flow rate ÷ area
volumetric flow rate = 0.3 m^3/s
area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2
Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s
Detention time = volume ÷ volumetric flow rate
volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3
Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s
Answer:
$151094
Explanation:
Solution
Recall that:
Acme Chemical in 2002 purchased a large pump worth of = 112,000
The estimation for the pump to the industrial pump index is =100
The index in 2002= 212
The current index is = 286
k = is the reference year for which cost or price is known.
n = the year for which cost or price is to be estimated (n>k).
Cn = the estimated cost or price of item in year n.
Ck = the cost or price of item in reference year k.
Cn = Ck * (In / Ik )
Now,
We find the estimated cost of the new pump which is stated as follows:
Cn = (112,000 * 286) /212
=32032000/212
=$151094
Therefore, the estimated cost of the new pump is $151094
Answer:
3180.86 Nm
Explanation:
Moment of inertia for shaft AB,
Torque in solid shaft AB will be given by
Where is shear stress, is polar moment of inertia for shaft AB, r is the radius of shaft B
The inner diameter of pipe CD can found considering that the thickness of pipe is 0.006 m hence diameter= 0.09-(2*0.006)= 0.078 m
Moment of inertia for shaft CD will be
Torque for shaft CD will be
and here r = 0.045 m
The minimum of the two torques is the largest torque that can be applied. Therefore, the torque to apply equals 3180.86 Nm