Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
e tube surface temperature is maintained at 100°C. For a 100‐m long tube, solve for a) the log mean temperature difference (Approx. 64.5 deg C), and b) the rate of heat addition into the oil.
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Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.
<h3>How to apply the Chvorinov's rule for casting processes</h3>
The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:
<h3>Cylindrical casting</h3>t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²
t = C · [0.25 · D · L/(0.5 · D + L)]² (1)
<h3>Prism casting</h3>t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²
t' = C · [3 · T · L/(8 · L + 6 · T)]² (2)
<h3>Relationship between the cross sections of both castings</h3>3 · T² = 0.25π · D² (3)
Where:
- <em>t</em> - Cooling time of the cylindrical casting, in time unit.
- <em>t'</em> - Cooling time of the prism casting, in time unit.
- <em>C</em> - Cooling factor, in time unit per square meter.
- <em>D</em> - Diameter of the cylinder, in meters.
- <em>L</em> - Length of the casting, in meters.
- <em>T</em> - Width of the cross section of the prism casting, in meters.
If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:
<em>T ≈ 0.153 m</em>
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And (1) and (2) simplified into these forms:
<h3>Cylindrical casting</h3>t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²
t = 0.0329 · C (1b)
<h3>Prism casting</h3>t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²
t' = 0.00218 · C (2b)
Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:
<em>r = (</em>1 <em>- t'/t) ×</em> 100 %
<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %
<em>r =</em> 93.374 %
The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting.
To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247
<u>Answer</u>:
a. r(t) = 6.40 cos (ωt + 38.66°) units
b. r(t) = 6.40 cos (ωt - 38.66°) units
c. r(t) = 6.40 cos (ωt - 38.66°) units
d. r(t) = 6.40 cos (ωt + 38.66°) units
<u>Explanation</u>:
To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:
(i) Convert the phasor to polar form. The polar form is written as;
r∠Ф
Where;
r = magnitude of the phasor =
Ф = direction = tan⁻¹ ()
(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:
r(t) = r cos (ωt + Φ)
Where;
ω = angular frequency of the sinusoid
Φ = phase angle of the sinusoid
(a) 5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
5 + j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
(b) 5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
5 - j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(c) -5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
-5 + j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(d) -5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
-5 - j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
The question is incomplete. We have to calculate :
a). the cutting force
b). volumetric metal removal rate, MRR
c). the horsepower required at the cut
d). if the power efficiency of the machine tool is 90%, determine the motor horsepower
Solution :
Given :
Cutting velocity (v) = 500 ft/min
= 500 x 12 in/min
= 6000 in/min
Feed , f = 0.025 in/rev
Depth of cut, d = 0.2 in
b). Volumetric material removal rate, MRR = v.f.d
= 6000 x 0.025 x 0.2
= 30
c). Horsepower required = MRR x unit horsepower
= 30 x 1.6
= 48 hp
a). Cutting force,
(1 hp = 550 ft lbf /sec)
= 3168 lbf
d). Machine HP required
= 53.33 HP