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4 years ago

7

A copper-constantan thermocouple is to be used to measure temperatures between 0 and 200°C. The e.m.f. at 0°C is 0 mV, at 100°C

it is 4.277 mV and at 200°C it is 9.286 mV. What will be the non-linearity error at 100°C as a percentage of the full range output if a linear relationship is assumed between e.m.f. and temperature over the full range?

1 answer:

Helga [31]4 years ago

5 0

Answer:

3.941%

Explanation:

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A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.

Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature $T_1=8^{\circ}C$

$T_2=28^{\circ}C$

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

$Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ$

We know that $COP=\frac{Q_1}{W}$

$1.5=\frac{Q_1}{450}$

$Q_1=675$ W=0.675 KW

so time required to cool the watermelon is

$t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec$

4 0

3 years ago

Estimate the energy (head) loss a short length of a pipe conveying 300 litres of water per second and suddenly enlarging from a

Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

D1 = 350 mm = 0.35 m

D2 = 700 mm = 0.7 m

g = 9.81 m/s²

Solution :

A1 = πD1² / 4 = π(0.35²) / 4 = 0.096 m²

A2 = πD2² / 4 = π(0.7²) / 4 = 0.385 m²

hL = (kL / 2g) • (U1² - U2²)

hL = (kL / 2g) • Q² (1/A1² - 1/A2²)

hL = (1 / 2(9.81)) • (0.3²) • (1/(0.096²) - 1/(0.385²))

hL = 0.467 m

5 0

3 years ago

A paper clip is made of wire 0.75 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diamete

Bas_tet [7]

Answer:

True strain = 3.7704

Explanation:

Strain is the measure an object that is stretched or deformed. This occurs when a force is applied to an object. Strain deals mostly with the change in length of the object. Strain = Δ L /L = Change in Length over the original Length:

Volume Constancy :

ΔL/L0=A0/ΔA=(D0/ ΔD)=(25mm/0.75mm)^2

ΔL/L0=44.4

Engineering strain:

Engineering strain =ΔL-L0/L0=ΔL/L0-1

Engineering strain =44.4-1=43.4

True strain, ε=In(ΔL/L0)=In(43.4)=3.7704

Note that strain has no unit, so the True strain = 3.7704

8 0

4 years ago

What type of fire extinguisher is used for electrical fires.

AysviL [449]

The all purpose one will work or you could use type C I believe.

8 0

3 years ago

Consider the brass alloy for which the stress-strain behavior is shown in the Animated Figure 7.12. A cylindrical specimen of th

sleet_krkn [62]

Answer:

(a) 0.1509 mm

(b) 0.00525 mm

Explanation:

Stress, $\sigma$ is given by

$\sigma=\frac {F}{A}$ where F is force and A is area and area is given by $\frac {\pi d^{2}}{4}$ hence

$\sigma=\frac {4F}{\pi d^{2}}$ where d is the diameter. Substituting 9970 N for F and 10mm=0.01 m for d hence

$\sigma=\frac {4*9970 N}{\pi 0.01m^{2}}=126941982.6 N/m^{2}$

$\sigma \approx 127 Mpa$

From the attached stress-strain diagram, the stress of 127 Mpa corresponds to strain of 0.0015 and since strain is given by

$\epsilon=\frac {\triangle l}{l}$ where $\epsilon$ is the strain, $\triangle l$ is elongation and l is original length and making elongation the subject

$\triangle l= \epsilon \times l$ and substituting strain with 0.0015 and length l with 100.6 mm then

$\triangle l=0.0015\times 100.6=0.1509 mm$

(b)

Lateral strain is given by

$\epsilon_{lat}=\frac {\triangle d}{d}$ and substituting $-v\epsilon$ for $\epsilon_{lat}$ where v is poisson ratio then

$-v\epsilon=\frac {\triangle d}{d}$ and making $\triangle d$ the subject then

$\triangle d=-vd\epsilon$ and substituting 0.35 for v, 0.0015 for strain and 10 mm for d

$\triangle d=-(0.35)*10*0.0015=-0.00525 mm$ and the negative sign indicates decrease in diameter

8 0

3 years ago