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scZoUnD [109]
3 years ago
11

Carbon dioxide steadily flows into a constant pressure heater at 300 K and 100 kPa witha mass flow rate of 9.2 kg/s. Heat transf

er in the rate of 175 kW is supplied to the carbondioxide as it flows through the heater. Determinea) the carbon dioxide temperature at the exit andb) carbon dioxide volume flow rate at the heater exit.Assume constant specific heat, cp, of 1075 J/(kg K).

Engineering
1 answer:
docker41 [41]3 years ago
8 0

Answer:

Carbon dioxide temperature at exit is 317.69 K

Carbon dioxide flow rate at heater exit is 20.25 m³/s

Explanation:

Detailed steps are attached below.

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Plssssssssssssss Alexi is writing a program which prompts users to enter their age. Which function should she use?
aleksandr82 [10.1K]

Answer:

int()

Explanation:

float() is using decimals, so that can't be it, like float(input( "how much does this cost?"))

print() is used to print something, not a user asking, like print("hello")

string() means like a whole, like string( I am good)

By elimination, int() is correct.

Hope this helps!

7 0
2 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

5 0
3 years ago
Consider the time domain waveforms below on the left. Match waveforms (a) - (e) to their respective frequency spectrum represent
lozanna [386]

Answer:

(a) ------(3). (b)------(1) (c)-----(5) (d)------(2) ------ (e) -----4

Note: Kindly find an attached copy of the diagram associated with the solution to the question below.

Sources: the diagram to this question was researched from Quizlet

Explanation:

Solution

(1) Part (a)a waveform has a high frequency components compared to another waveform. the corresponding frequency components should be high.

So for the wave form a  the corresponding frequency spectrum is (3)

(2) For part (b), waveform has three harmonics, the corresponding frequency spectrum is (1)

(3) The time domain waveform plot (c) is a sine wave but there exists a dc component.

Thus x[0] ≠0

For (c) the corresponding frequency spectrum is (5)

(4) For part (d) the corresponding frequency spectrum is (2)

(5) A sine wave is made of a single frequency only and its spectrum is a single point

For (e) the corresponding frequency spectrum is (4)

3 0
3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
3 years ago
A resistor, an inductor, and a capacitor are connected in series to an ac source. What is the condition for resonance to occur?.
vaieri [72.5K]

Answer:if power factor =1 is possible for that.

Explanation:when pf is unity. means 1.

6 0
1 year ago
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