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Ira Lisetskai [31]

2 years ago

12

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

eneration in the wire per unit volume, in Btu/hrft3 , and the heat flux on the outer surface of the wire, in Btu/hrft2 , as a result of this generation.

1 answer:

SSSSS [86.1K]2 years ago

3 0

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

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3 years ago

A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after

Talja [164]

<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

$v_i$ = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = $\frac{v_f \:-\:v_i}{t}$

where: a - acceleration

$v_i$ - initial velocity

$v_f$ - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = $\frac{1700\:N}{1250\:N}$

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>

Convert 30 km/hr to m/s

= $\frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}$

= $8.33 m/s$

Substitute the value of a, $v_i$ and t

a = $\frac{v_f \:-\:v_i}{t}$

$1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}$

$(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s$

$v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s$

$v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s$

$v_f\: =\: 21.93\: m/s$

Convert to km/hr

= $\frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}$

= $78.95\: km/hr$

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

6 0

2 years ago

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

2 years ago