In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini
tially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg K and 4.4 kJ/kg K, respectively, determine the final equilibrium temperature after quenching, in K.
1 answer:
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Answer:
a. 25! = (Approximately)
b. 24!
Explanation:
a. In a Playfair cipher, there are 25 keys available because it is structured in a 5 * 4 grid. By using permutations to enumerate all potential configurations, we derive: 25! = 1.551121004×10²⁵ =
Although there are 26 letters available, in the Playfair cipher, the letters 'i' and 'j' are treated as a single letter.
b. Considering any configuration of 5x5, each of the four row shifts yields equivalent configurations, amounting to five total equivalencies. Similarly, for each of these five setups, any of the four column shifts also results in equivalent arrangements. Therefore, each configuration corresponds to 25 equivalent arrangements. Consequently, the total count of distinct keys can be expressed as:
25!/25 = 24! = 6.204484017×10²³
Answer:
The power of the pump is 23.09 kW.
Explanation:
Parameters
gravitational constant,
mass flow rate,
flow density,
efficiency of the pump,
output gauge pressure,
input gauge pressure,
cross-sectional area of output pipe,
cross-sectional area of input pipe,
height of discharge, (evaluated at pump’s maximum height of 1.22 m)
input height,
hydraulic power of the pump,
Initially, the volumetric flow (Q) needs to be determined
Next, compute the velocity (v) for both input and output
Subsequently, the total head (H) can be calculated
Finally, the computation of pump power is done as follows
1) The three possible assumptions are
a) All processes are internally reversible
b) Air, as the working fluid, circulates in a closed-loop
cycle
c) Combustion is represented as a heat-adding process
2) Diagrams are included
5) The net work per cycle is 845.88 kJ/kg
The horsepower produced is approximately 45374 hP
Explanation:
1) The three valid assumptions are
a) All processes are internally reversible
b) Air, the working fluid, moves continuously in a closed-loop
cycle
c) The combustion process is set as a heat addition step
2) Diagrams for illustration are provided
5) The cylinder bore diameter measures 3.7 in., equating to 0.09398 m
The stroke length is 3.4 in., approximately 0.08636 m.
The cylinder volume is calculated as v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume amounts to 16% of the cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ is 9.59 × 10⁻⁵ m³
p₁ is 14.5 lbf/in.² = 99973.981 Pa
T₁ equals 60 F = 288.706 K
For the Otto cycle's T-S diagram,
T₂ calculates to 288.706* = 592.984 K
The peak temperature is T₃ = 5200 R = 2888.89 K
T₄ resolves to 2888.89 / = 1406.5 K
Work performed, W = ×(T₃ - T₂) -
×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power generated in the Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.