Answer:
a)-Bin stock items free issue
Explanation:
Bin stock items free issue items are similar to the free issue items, but their access is limited.
Bin stock items free issue items are similar to the free issue items, but their access is limited.
B. Determine the overall heat transfer coefficients, U; and U., for the pipe.
C. Would your answer change if the two materials were swapped so that the inside material were carbon steel and the outside material of the pipe were made of AISI 304 stainless steel? If so, calculate new values for UA and q. If not, explain why your answer would not change. (Here, assume the dimensions of the inner material (now plain carbon steel) match those of the AISI 304 schedule 30 stainless steel from part A, and the stainless steel is 0.075 inches thick on the outside.)
Sorry man i dont know the answer to this one
Answer:
Compression ratio(r)
Explanation:
Otto cycle:
Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.
The efficiency of Otto cycle given as follows
Where r is the compression ratio and γ is heat capacity ratio.
So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).
True
False
Answer: true
Explanation:
Answer:
A) Ductility = 11% EL
B) Radius after deformation = 4.27 mm
Explanation:
A) From equations in steel test,
Tensile Strength (Ts) = 3.45 x HB
Where HB is brinell hardness;
Thus, Ts = 3.45 x 250 = 862MPa
From image 1 attached below, for steel at Tensile strength of 862 MPa, %CW = 27%.
Also, from image 2,at CW of 27%,
Ductility is approximately, 11% EL
B) Now we know that formula for %CW is;
%CW = (Ao - Ad)/(Ao)
Where Ao is area with initial radius and Ad is area deformation.
Thus;
%CW = [[π(ro)² - π(rd)²] /π(ro)²] x 100
%CW = [1 - (rd)²/(ro)²]
1 - (%CW/100) = (rd)²/(ro)²
So;
(rd)²[1 - (%CW/100)] = (ro)²
So putting the values as gotten initially ;
(ro)² = 5²([1 - (27/100)]
(ro)² = 25 - 6.75
(ro) ² = 18.25
ro = √18.25
So ro = 4.27 mm
Answer:
a) The velocity is 42.0833 ft/s
b) The flow rate is 3366.664 ft³/s
c) The Froude number is 0.2345
d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft
e) The critical depth is 3.8030 ft
Explanation:
Given data:
80 ft wide channel, L
1 ft and 10 ft water depths, d₁ and d₂
Questions: a) Velocity of the faster moving flow, v = ?
b) The flow rate (discharge), q = ?
c) The Froude number, F = ?
d) The flow energy dissipated, E = ?
e) The critical depth, dc = ?
a) For the velocity:
Solving for F:
F = 7.4162
Here, g = gravity = 32.2 ft/s²
b) The flow rate:
c) The Froude number:
d) The flow energy dissipated:
e) The critical depth:
Answer:
C)185,500 KJ
Explanation:
Given that
Latent heat fusion = 333.23 KJ/kg
Latent heat vaporisation = 333.23 KJ/kg
Mass of ice = 100 kg
Mass of water = 40 kg
Mass of vapor=60 kg
Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .
Sensible heat for water Q
For water
Q=4.178 x 40 x 100 KJ
Q=16,712 KJ
So total heat
Total heat =100 x 333.23+16,712 + 60 x 2257 KJ
Total heat =185,455 KJ
Approx Total heat = 185,500 KJ
So the answer C is correct.