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sdas [7]
3 years ago
7

A room is kept at −5°C by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling wa

ter that enters the condenser at 20°C at a rate of 0.13 kg/s and leaves at 28°C. The refrigerant enters the condenser at 1.2 MPa and 50°C and leaves as a saturated liquid. If the compressor consumes 1.9 kW of power, determine (a) the refrigeration load, in Btu/h and the COP, (b) the second-law efficiency of the refrigerator and the total exergy destruction in the cycle, and (c) the exergy destruction in the condenser. Take T0 = 20°C and cp,water = 4.18 kJ/kg·°C

Engineering
2 answers:
Fed [463]3 years ago
4 0

Answer:

note:

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

Download docx
kolbaska11 [484]3 years ago
3 0

Answer:

see below explanation

Explanation:

see the attached file

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How do Consumers sometimes interact with a producers?
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3 years ago
Which traits are common in all four career pathways of the Information Technology field? Check all that apply.
Komok [63]

Answer:

Accuracy and attention to detail, Problem solving and critical thinking skills, Knowledge of programming language .

Explanation:

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7 0
2 years ago
The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

3 0
3 years ago
Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l
ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0
3 years ago
Read 2 more answers
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