Ask question

Ask question

All categories

Alenkasestr [34]

3 years ago

14

An ideal Diesel Cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum air temperature and the rat

e of heat addition to this cycle when it produces 200 hp of power; the cycle is repeated 1200 times per minute; and the state of the air at the beginning of the compression is 95 kPa and 17oC. Use constant specific heats at room temperature.

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

A) Q'_in = 304.66 hp

B) T_max = 1109.29 °C

Explanation:

We are given;

compression ratio; r = 18

cutoff ratio; r_c = 1.5

Power; W' = 200 hp

Temperature; T1 = 17°C = 290 K

A) To find the rate of heat addition, we will use the derived formula;

Q'_in = (W' × r^(k - 1) × k(r_c - 1))/(r^(k - 1) × k(r_c - 1)) - ((r_c)^(k) - 1)

Where k is a constant = 1.4

Thus;

Q'_in = (200 × 18^(1.4 - 1) × 1.4(1.5 - 1))/((18^(1.4 - 1) × 1.4(1.5 - 1)) - ((1.5^(1.4)) - 1)

Q'_in = 304.66 hp

B) To find the maximum temperature, we will use the Formula;

T_max = T1 × r^(k - 1) × r_c

T_max = 290 × 18^(1.4 - 1) × 1.5

T_max = 1382.29 K

Converting to °C gives;

T_max = 1109.29 °C

You might be interested in

One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight l

Effectus [21]

Answer:

time = 5.22 hr

Explanation:

Given data:

Energy of battery = 9400 J

Power consumed by three led bulb is 0.5 watt

we know Power is give as

$Power = \frac{ energy}{time}$

plugging all value and solve for time

$time = \frac{energy}{power}$

$time = \frac{9400}{0.5}$

time = 18,800 sec

in hour

1 hour = 3600 sec

therefore in 18,800 sec

$time = \frac{18800}{3600} = 5.22 hr$

7 0

4 years ago

An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove

Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

$n=Q out / W in$

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

$12000 BTU/hr = 4.72 hp$

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]

6 0

3 years ago

Define the overall heat transfer coefficient.

9966 [12]

Answer and Explanation:

In thermodynamics, the overall heat transfer coefficient also referred as film effectiveness is a constant of proportionality between force drive for the heat flow and the heat flux.

It gives the measure of the heat transfer as a result of convection or conduction. The coefficient of overall heat transfer depends on surface area, resistance of the material, temperature difference, thickness, etc.

It is given by:

Q = UA $\Delta T$

where

U = overall heat transfer coefficient

Its SI units is $W/m^{2}K$ .

6 0

4 years ago

Consider a power plant with a heat rate of 10,800 kJ/kWh burning bituminous coal with 75 percent carbon and a heating value (ene

grandymaker [24]

Answer:

1. $\eta_{p} = 33.34%$

2. $m_{coal} = 0.396 kg/kWh$

3. carbon emission = 0.297 kg/kWh[/tex]

4. carbon-dioxide emission = 1.089 kg/kWh

5. Min flow of cooling water = = 1457.14 kg/kWh

Given:

Heat Rate = 10,800 kJ/kWh

Heating value = 27,300 kJ/kg

Solution:

In order to calculate the plant efficiency, [rex]\eta_{p}[/tex]

Heat rate in coal fired steam power plant = $\frac{3600}{\eta_{p}}$

Therefore,

$\eta_{p} = \frac{3600}{Heat rate}$

$\eta_{p} = \frac{3600}{10800}\times 100 = 33.34%$

Now,

To calculate mass of coal per kWh, $m_{coal}$ :

$m_{coal} = \frac{Required heat rate}{heating value per kg}$

$m_{coal} = \frac{10800}{27300} = 0.396 kg/kWh$

Now,

Rate of emission carbon and carbon-dioxide from the plant:

In accordance to the question, 75% of the bituminous coal being burned is carbon, thus carbon emission is given by:

carbon emission = $0.75\times m_{coal}$

carbon emission = $0.75\times 0.396 = 0.297 kg/kWh = 29.7%$

Now, for carbon-dioxide emission:

Since, molecular weight of carbon-dioxide = 44 kg

Thus carbon-dioxide emission:

$0.297\times \frac{44}{12} = 1.089 kg/kWh$

The minimum flow of cooling water per kWh if the allowance in temperature is $10^{\circ}$ :

$\frac{0.85\times \frac{2}{3}\times 10800}{4.2} = 1457.14 kg/kWh$

8 0

3 years ago

The decimal number 200 converted to binary is<br> Enter the answer.

Natali5045456 [20]

Answer: 11001000

Here u Go!!!!!!!!!!!!!!!!!!

7 0

3 years ago