Answer:
yes it can
Explanation:
Determine the force in members FD and DB of the frame. Also, find the horizontal and vertical components of reaction the pin at C exerts on member ABC and member EDC.
Answer:
A.
Explanation:
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycle are 295 and 1240 K. Assuming an isentropic efficiency of 83 percent for the compressor and 87 percent for the turbine, determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency.
Answer:
a) 764.45K
b) 210.48 kJ/kg
c) 30.14%
Explanation:
pressure ratio = 10
minimum temperature = 295 k
maximum temperature = 1240 k
isentropic efficiency for compressor = 83%
Isentropic efficiency for turbine = 87%
a) Air temperature at turbine exit
we can achieve this by interpolating for enthalpy
h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air
T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) [tex](\frac{783.05-778.18}{800.13-778.18} )[/tex] = 764.45K
b) The net work output
first we determine the actual work input to compressor
Wc = h2 - h1 ( calculated values )
= 626.57 - 295.17 = 331.4 kJ/kg
next determine the actual work done by Turbine
Wt = h3 - h4 ( calculated values )
= 1324.93 - 783.05 = 541.88 kJ/kg
finally determine the network output of the cycle
Wnet = Wt - Wc
= 541.88 - 331.4 = 210.48 kJ/kg
c) determine thermal efficiency
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
equation 1 becomes
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
35 points and brainiest. A, B, C, D
Which of the following identifies what carbide-tipped bits, lines on machine bolts, and lock washers all have in common?
A. All are very expensive items.
B. All are very recent inventions.
C. All are rather outdated inventions.
D. All provide an added level of strength.
Answer:c
Explanation:
The water in a soil flows from Point K to Point L, a distance of 250 ft. Point K is at elevation 543 ft and Point L is at elevation 461 ft. Piezometers have been installed at both points, and their water levels are 23 ft and 74 ft, respectively, above the points. Compute the average hydraulic gradient between these two points.
Answer:
0.124
Explanation:
We calculate the hydraulic gradient by the formulas below.
I = (change in h)/(change in l)-----eqn 1
I = (hk-hl)/change in L ----- equation 2
At k the headloss = hk,
At L the headloss = hL
The distance of water travel is change in I
Total head at k
hk = 543+23
= 566 ft
Total head at L
hL = 461+74
= 535 ft
Change in L = 250
When we substitute these values in equation 2
566-535/250
= 0.124
The hydraulic gradient is 0.124