Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
Answer:
ΔQ = 4930.37 BTu
Explanation:
given data
height h = 8ft
Δt = 8 hours
length L = 24 feet
R value = 16.2 hr⋅°F⋅ft² /Btu
inside temperature t1 = 68°F
outside temperature t2 = 16°F
to find out
number of Btu conducted
solution
we get here number of Btu conducted by this expression that s
......................1
here A is area that is = h × L = 8 × 24 = 1492 ft²
put here value we get
solve it we get
ΔQ = 4930.37 BTu
Answer:
If I am not mistaken I believe it is a higher voltage.
Explanation:
Hope this helps
