1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anna [14]
3 years ago
8

A food department is kept at -12 °C by a refrigerator in an environment at 30 °C. The total heat gain to the food department is

estimated to be 3,300 kJ/h and the heat rejection in the condenser is 4,800 kJ/h. If the refrigeration cycle was as efficient as the Carnot cycle, how much power (in kW) would be required to remove 3,300 kJ/h heat from the cooled space?
Engineering
1 answer:
Nataly [62]3 years ago
5 0

Answer: P = 0.416 kW

Explanation:

taken a step by step process to solving this problem.

we have that from the question;

the amount of heat rejected Qn = 4800 kJ/h

the cooling effect is Ql = 3300 kJ/h

Applying the first law of thermodynamics for this system gives us

Шnet = Qn -Ql

Шnet = 4800 - 3300 = 1500 kJ/h

Next we would calculate the coefficient of performance of the refrigerator;

COPr = Desired Effect / work output = Ql / Шnet  = 3300/1500 = 2.2

COPr = 2.2

The Power as required gives;

P = Qn - Ql  = 4800 - 3300 = 1500 kJ/h = 0.416

P = 0.416 kW

cheers i hope this helps!!!!1

You might be interested in
In digital communication technologies, what is an internal network also known as?
garri49 [273]
Intranet is a network that is internal and use internet technologies. It makes information of any company accessible to its employees and hence facilitates collaboration. Same methods can be used to get information, use resources, and update the data as that of the internet.
Hopefully this helped.
4 0
3 years ago
If the same type of thermoplastic polymer is being tensile tested and the strain rate is increased, it will: g
Serggg [28]

Answer:

It would break I think need to try it out

Explanation:

3 0
2 years ago
Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl
Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio \frac{v_1}{v_2}  = 10

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

\frac{p_1}{p_2}  = (\frac{v_2}{v_1} )^n Therefore p₂ = p₁ ÷ (\frac{v_2}{v_1} )^n = (1 bar) ÷ (\frac{1}{10} )^{1.3} = 19.953 bar

Similarly for a polytropic process we have

\frac{T_1}{T_2}  = (\frac{v_2}{v_1} )^{n-1} or T₂ = T₁ ÷ (\frac{v_2}{v_1} )^{n-1} = \frac{310}{0.1^{0.3}} = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = \frac{8.3145}{28.9628} = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R =  1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = \frac{R(T_2-T_1)}{n-1} = \frac{0.287(618.531-310)}{1.3 - 1}= 295.16 kJ/kg

Q =(\frac{n-\gamma}{\gamma - 1} )W = (\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = \frac{R(T_4-T_3)}{n-1} = Where T₄ is given by  \frac{T_4}{T_3}  = (\frac{v_3}{v_4} )^{n-1} or T₄ = T₃ ×0.1^{0.3}

= 2200 ×0.1^{0.3}  T₄ = 1102.611 K

W =  \frac{0.287(1102.611-2200)}{1.3 - 1}= -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

\eta = 1-\frac{T_4-T_1}{T_3-T_2} =1-\frac{1102.611-310}{2200-618.531} = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

p_{m}  = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}  where r = compression ratio and r_p = \frac{p_3}{p_2}

However p₃ = \frac{p_2T_3}{T_2} =\frac{(19.953)(2200)}{618.531} =70.97 atm

r_p = \frac{p_3}{p_2} = \frac{70.97}{19.953}  = 3.56

Therefore p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9} = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0
3 years ago
For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0
2 years ago
Viscous effects are negligible outside of the hydrodynamic boundary layer. (3 points) a. True b. False
Valentin [98]

Answer:

I would say false but I am not for sure

8 0
3 years ago
Other questions:
  • What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a
    7·1 answer
  • The enthalpy of the water entering an actual pump is 500 kJ/kg and the enthalpy of the water leaving it is 550 kJ/kg. The pump h
    10·1 answer
  • A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
    12·1 answer
  • Determine the average and rms values for the function, y(t)=25+10sino it over the time periods (a) 0 to 0.1 sec and (b) 0 to 1/3
    9·1 answer
  • What are significant figures​
    13·1 answer
  • Determine the wattmeter reading when it is connected to resistor load.​
    11·1 answer
  • Cho biết tác dụng chung của các hệ giằng khung ngang nhà công nghiệp nhẹ 1 tầng 1 nhịp.
    13·1 answer
  • Suppose to build RSA crypto system you picked primes "p" and "q" as 3 and 7 and "e" as 5 what are the public and private keys? W
    11·1 answer
  • In the following scenario, what could the engineers have done to keep the bridge from collapsing?
    8·1 answer
  • Software that is released to have users test out the "bugs" is known as Ransomeware O Break-in software 2 O Flim flam software O
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!