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oksano4ka [1.4K]
3 years ago
11

REVIEW QUESTIONS

Engineering
1 answer:
babunello [35]3 years ago
5 0
Snap rings, and bearings can be used to keep a gear on a shaft, hope this helps!!
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Need help<br> What is elasticity?
photoshop1234 [79]

Answer:

hope this helps,

Explanation:

brainliest please? :)

6 0
2 years ago
Read 2 more answers
A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
Speed limit signs are classified as
luda_lava [24]

Answer:

B. Regulatory signs

Explanation:

traffic rules

5 0
3 years ago
Read 2 more answers
T he area of a circle is pr 2. Define r as 5, then find the area of a circle,using MATLAB®.(b) The surface area of a sphere is 4
aksik [14]

Answer:

Area of Circle = 78.5398

Surface Area of Sphere =  1.2566 x 10^3 = 1256.6 ft

Volume of Sphere =  33.5103 ft

Explanation:

Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.

r=5; % define r as 5

a=pi*r^2;% calculate the area of the circle

AreaOfCircle=a

r=10; % define r and 10 ft

sa=4*pi*r^2; %Calculate the surface area of the sphere

SphereSurfaceArea=sa

r=2;% define r as 2 ft

vs=(4/3)*pi*r^3;% Calculate the volume of the sphere

VolumeShere=vs

3 0
3 years ago
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Technician A says that the enable criteria are the criteria that must be met before the PCM completes a monitor test. Technician
DENIUS [597]

Answer:

"A"

Explanation:

5 0
3 years ago
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