Answer:
[tex]\eta =46\%[/tex]
Explanation:
Hello!
In this case, we compute the heat output from coal, given its heating value and the mass flow:
[tex]Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW[/tex]
Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:
[tex]\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%[/tex]
Best regards!
Simplify the following expression: T =(AB)'B
Tank has an average bearing pressure of 105 kPa. Ground water is deep. Calculate the total, pore and effective stress before tank loading and with tank
Answer:
Hello your question is incomplete attached below is the complete question
answer :
A) At 8 meters below center
Total stress = 228.827 KN/m^2
pore water pressure = 0 KN/m^2
effective stress = 228.827 - 0 = 228.827 KN/m^2
B) At 8 meters below tank edge
Total stress = 186.289 KN/m^2
pore water stress = 0 Kn/m^2
effective stress = 186.289 - 0 = 186.289 KN/m^2
Explanation:
Given data :
Diameter = 21 m
radius = 21 / 2 = 10.5 m
soil moist = 18.4 KN/cubic meter
Average bearing pressure = 105 kPa
i) Before Loading Tank
Total stress = 8 * 18.4 = 147.2 KN/m^2
Pore pressure = 0 KN/m^2
Effective stress = 147.2 - 0 = 147.2 KN/m^2
ii) After loading Tank
No change occurs
A) At 8 meters below center
Total stress = 228.827 KN/m^2
pore water pressure = 0 KN/m^2
effective stress = 228.827 - 0 = 228.827 KN/m^2
B) At 8 meters below tank edge
Total stress = 186.289 KN/m^2
pore water stress = 0 Kn/m^2
effective stress = 186.289 - 0 = 186.289 KN/m^2
attached below is the detailed solution
What does an environmental science and protection technician have in common with a microbiologist
Answer:
Environmental science and protection technicians both use laboratory equipment, such as microscopes, to analyze samples collected in the field. Environmental science and protection technicians also monitor the environment and investigate sources of pollution and contamination, including those affecting public health.
Hope this was helpful ;)
answer is a and b jjjjxjxjxjxjx xxxjxjxj
What major financial flop led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogether?
1: The founder and CEO of Sega was found to be secretly skimming money off of the top of profits, leading to widespread distrust by the public and a sharp decline in sales until they ultimately had to shut down due to making no profit.
2: A small group of employees found a way to drain all of the Sega Corporation funding accounts and flee the country with all of the money, never to be heard from again.
3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.
4: A game called Katamari Damacy that carried a virus that would infect any console that it was played on forced Sega to spend millions of dollars in refunds and bankrupted the company.
Answer:
The major financial flop that led to the end of the Sega Dreamcast and ultimately caused Sega to stop making game consoles altogether is:
3: A game called Shenmue, that cost more than $70 million to make, meant that everyone who owned a Dreamcast needed to buy two copies of the game just for Sega to make back the money they had spent to develop it-which didn't happen.
Explanation:
Shenmue, released on December 29, 1999, was created for Dreamcast by Suzuki. It was widely described as the most expensive video game ever produced. It had an estimated production and marketing cost of between US$47 and $70 million, according to the latest available data.
What simple changes would you make to a basic PID controller if you expected the sensors to be very noisy and the target to change by large steps?
Answer:
tuning the value of Ki termincreasing/changing the Kp termExplanation:
The simple change that would be made to a basic PID controller to prevent it from been noisy is tuning the value of Ki term while
The simple change that would be made to a PID controller if you want to change the target by large steps is increasing/changing the Kp term associated with the PID controller
How does the span of a truss relate to load-carrying capacity and when a truss fails, how much area is involved?
Answer:
For a deeper material truss, increase in distance between the top chord and bottom chord will cause the truss to bear more load with increase in Span
For Cantilever Truss ; increase in span will result in decrease in load bearing capacity of the truss
When a truss fails the areas that are affected are the supports of the truss and its corresponding spam. i.e. 4 crossbars at quarter of the span
Explanation:
The load-carrying capacity of a truss on the span is basically based on the type of truss ,
For a deeper material truss, increase in distance between the top chord and bottom chord will cause the truss to bear more load with increase in Span
For Cantilever Truss ; increase in span will result in decrease in load bearing capacity of the truss
When a truss fails the areas that are affected are the supports of the truss and its corresponding spam. i.e. 4 crossbars at quarter of the span
A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consumes 0.8 kW of electrical power, determine the maximum water pressure at the discharge of the pump.
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
[tex]E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}[/tex]
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet [tex]P_1[/tex] = 96 kPa
the pressure at the exit [tex]P_2[/tex] = ???
the pressure [tex]\rho[/tex] = 1000 kg/m³
∴
[tex]0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}[/tex]
[tex]0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}[/tex]
[tex]800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}[/tex]
[tex]\dfrac{800000}{2} = P_2-96000[/tex]
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
What is the heat loss coefficient that has a symbol Uair and is found from (volumetric flow * density * specific heat capacity) of a building with a volume of 19456 cubic feet, if there is a natural air change per hour of 0.4
Answer:
Uair = 0.0749 KW/k = 74.9 W/k
Explanation:
The natural air change per hour is given by the formula:
Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume
where,
ACPH = 0.4
Volume Flow = ? in ft³/min
Volume = 19456 ft³
Therefore,
0.4 = (60 min)(Volume Flow)/(19456 ft³)
Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)
Volume Flow = (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s
Now, we find heat loss coefficient:
Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air
Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)
Uair = 0.0749 KW/k = 74.9 W/k