Answer:
308 acre-ft of water
Explanation:
Given:
Area of the watershed = 22 square miles
Depth of Rainfall = 0.75 in =
=0.0625 ft
Percentage rainfall falling in reservoir as runoff = 35%
Now,
1 square mile = 640 acre
Thus,
22 square miles = 22 × 640 = 14,080 acres
Thus,
The total volume of rainfall = Area of watershed × Depth of the rainfall
or
The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft
also,
only 35% of the total rainfall is contributing as runoff
thus,
Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water
Answer:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
Mark brainliest please!
Isothermal work will be less than the adiabatic work for any given compression ratio and set of suction conditions. The ratio of isothermal work to the actual work is the isothermal efficiency. Isothermal paths are not typically used in most industrial compressor calculations.
Compressors
Compressors are used to move gases and vapors in situations where large pressure differences are necessary.
Types of Compressor
Compressors are classified by the way they work: dynamic (centrifugal and axial) or reciprocating. Dynamic compressors use a set of rotating blades to add velocity and pressure to fluid. They operate at high speeds and are driven by steam or gas turbines or electric motors. They tend to be smaller and lighter for a given service than reciprocating machines, and hence have lower costs.
Reciprocating compressors use pistons to push gas to a higher pressure. They are common in natural gas gathering and transmission systems, but are less common in process applications. Reciprocating compressors may be used when very large pressure differences must be achieved; however, since they produce a pulsating flow, they may need to have a receiver vessel to dampen the pulses.
The compression ratio, pout over pin, is a key parameter in understanding compressors and blowers. When the compression ratio is below 4 or so, a blower is usually adequate. Higher ratios require a compressor, or multiple compressor stages, be used.
When the pressure of a gas is increased in an adiabatic system, the temperature of the fluid must rise. Since the temperature change is accompanied by a change in the specific volume, the work necessary to compress a unit of fluid also changes. Consequently, many compressors must be accompanied by cooling to reduce the consequences of the adiabatic temperature rise. The coolant may flow through a jacket which surrounds the housing with liquid coolant. When multiple stage compressors are used, intercooler heat exchangers are often used between the stages.
Dynamic Compressors
Gas enters a centrifugal or axial compressor through a suction nozzle and is directed into the first-stage impeller by a set of guide vanes. The blades push the gas forward and into a diffuser section where the gas velocity is slowed and the kinetic energy transferred from the blades is converted to pressure. In a multistage compressor, the gas encounters another set of guide vanes and the compression step is repeated. If necessary, the gas may pass through a cooling loop between stages.
Compressor Work
To evaluate the work requirements of a compressor, start with the mechanical energy balance. In most compressors, kinetic and potential energy changes are small, so velocity and static head terms may be neglected. As with pumps, friction can be lumped into the work term by using an efficiency. Unlike pumps, the fluid cannot be treated as incompressible, so a differential equation is required:
Compressor Work
Evaluation of the integral requires that the compression path be known - - is it adiabatic, isothermal, or polytropic?
uncooled units -- adiabatic, isentropic compression
complete cooling during compression -- isothermal compression
large compressors or incomplete cooling -- polytropic compression
Before calculating a compressor cycle, gas properties (heat capacity ratio, compressibility, molecular weight, etc.) must be determined for the fluid to be compressed. For mixtures, use an appropriate weighted mean value for the specific heats and molecular weight.
Adiabatic, Isentropic Compression
If there is no heat transfer to or from the gas being compressed, the porocess is adiabatic and isentropic. From thermodynamics and the study of compressible flow, you are supposed to recall that an ideal gas compression path depends on:
Adiabatic Path
This can be rearranged to solve for density in terms of one known pressure and substituted into the work equation, which then can be integrated.
Adiabatic Work
The ratio of the isentropic work to the actual work is called the adiabatic efficiency (or isentropic efficiency). The outlet temperature may be calculated from
Adiabatic Temperature Change
Power is found by multiplying the work by the mass flow rate and adjusting for the units and efficiency.
Isothermal Compression
If heat is removed from the gas during compression, an isothermal compression cycle may be achieved. In this case, the work may be calculated from:
http://facstaff.cbu.edu/rprice/lectures/compress.html
we can see that the correct person here will be Technician B who says that a diesel engine is designed to operate at near or maximum speed for long periods without damage.
<h3>What is a diesel engine?</h3>
A diesel engine is actually known to be an engine that makes use of diesel as its fuel. In other words, diesel engines run on diesel.
We see here that a diesel engine is actually designed to work for a very time on near or maximum speed without damage. This is true because the diesel fuel has that strength.
Also, diesel engines may be designed as two or four stroke cycles.
Learn more about diesel engine on brainly.com/question/13146091
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Answer:
F = 2840.3 N
Explanation:
Given:
- Diameter of window D = 0.3 m
- Midpoint of window from sea level h = 4 m
- Specific gravity of sea water S.G = 1.024
- Density of water p = 1000 kg/m^3
Find:
The hydro-static force F_r acting on the mid-point of the window.
Solution:
- The average pressure P acting on the midpoint of the window:
P = S.G p*g*h
P = 1.024*1000*9.81*4
P = 40181.76
- The hydro-static force F_r acting on the mid-point of the window:
F = P*A = P*pi*D^2 / 4
F = 40181.76*pi*0.3^2 / 4
F = 2840.3 N
