Identify the following formulas:

What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)

Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1- $\frac{T2}{T1}$

=1- $\frac{383}{873}$

=1-0.43871

=.56

=56%

5 0

3 years ago

To assist in completing this question, you may reference the Animated Technique Video - MALDI-TOF Mass Spectroscopy. Complete th

a_sh-v [17]

Complete Question

The complete question is shown on the first uploaded image.

Answer:

The answer is shown on the second uploaded image

Explanation:

The explanation is also shown on the second uploaded image

3 0

4 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a

AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2) + x(2) * h(fg2)

so

exit quality = $\frac{h1 - h(f2)}{h(fg2)}$

exit quality = $\frac{332.17- 9.04}{1382.73)}$

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0

3 years ago

A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque

Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

τ = 16*T / pi*d^3

T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

T_c = τ_c*pi*d_c^3 / 16

T_c = 12*1000*pi*1^3 / 16

T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

T_s = τ_s*pi*d_s^3 / 16

T_s = 15*1000*pi*0.25^3 / 16

T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

T = min ( 2356.2 , 46.02 )

T = 46.02 lb-in

6 0

3 years ago

Of the cost reduction strategies for workers' compensation mentioned in the required readings, which one do you think would work

Vesnalui [34]

In industries together with production, we want people to address the manufacturing of merchandise and the usage of heavy machinery.

<h3>What is the painting situation?</h3>

In such painting situations, people are at risk of injuries, and this prices the maximum for the company. So so that you can put into effect value discount is such conditions we want to have right coincidence cowl plans for the people and make sure all of the protection precautions are taken withinside the factory.

The people have to be properly educated on using protection measures and in case any injuries arise we have to have coverage claims in order that we not want to make investments extra cash and we also can offer protection and protection to the people.
This approach is excellent for this enterprise due to the fact regardless of what number of precautions we take people are uncovered to fitness risks and as a result having the right coverage insurance is a superb value discount strategy.