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3 years ago

7

Two gases—neon and air—are expanded from P1 to P2 in a closed-system polytropic process with n = 1.2. _____ produces more work w

hen expanded. The gas constants for neon and air are R = 0.4119 and 0.287 kJ/kg·K, respectively.

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

Note that Air requires lesser work

Explanation:

Calculate for general work done

SInce Gas constant 'R' for: Neon = 0.4119KJ/kg.k , and Air = 0.287 kJ/kg·K

Calculate for work done of NEON

Calculate for work done of Air

See solution attached.

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2 years ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0

3 years ago

Determine the resistance of 3km of copper having a diameter of 0,65mm if the resistivity of copper is 1,7x10^8

LUCKY_DIMON [66]

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

<u>Given the following data;</u>

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

$Radius, r = \frac {diameter}{2}$

$Radius = \frac {0.00065}{2}$

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

$Resistance = P \frac {L}{A}$

Where;

P is the resistivity of the material.
L is the length of the material.
A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

$Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}}$

$Resistance = 1.7 * 10^{8} * 903614.46$

<em>Resistance = 1.54 * 10^18 Ohms </em>

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3 years ago