# Example problems for the calculation and selection of compressors

**Contents:**

## Problem 1. Calculation of dead volume in a piston compressor

**Conditions:**

The piston of a single-stage, single-cylinder and single-action compressor has diameter d = 200 mm and stroke s = 150 mm. Compressor shaft rotates at n = 120 rpm. The air inside is compressed at pressure from P1 = 0.1 MPa to P2 = 0.32 MPa. Compressor throughput is Q = 0.5 m^{3}/min. Polytropic coefficient m is assumed to be equal to 1.3.

**Problem:**

Calculate the dead volume of the cylinder V_{d}.

**Solution:**

First, let’s determine the piston cross-section area F according to the formula below:

F = (π · d²)/4 = (3.14 · 0.2²)/4 = 0.0314 m^{2}

Let’s also calculate V_{p} - the volume displaced by the piston per stroke:

V_{p} = F · s = 0.0314 · 0.15 = 0.00471 m^{3}

From the compressor throughput formula we can find the delivery rate λ (since this is a single-action compressor, coefficient z = 1):

Q = λ · z · F · s · n

λ = Q/(z · F · s · n) = 0.5/(1 · 0.0314 · 0.15 · 120) = 0.88

Now we will use the approximation formula of delivery rate calculation to find the volumetric efficiency of the pump:

λ = λ_{v} · (1.01 - 0.02·P_{2}/P_{1})

λ_{v} = λ / (1.01 - 0.02·P_{2}/P_{1}) = 0.88 / (1.01 - 0.02·0.32/0.1) = 0.93

Then, from the volumetric efficiency formula, we will express and find the cylinder dead space value:

λ_{v} = 1 – с·[(P_{2}/P_{1})^{1/m}-1]

where c = V_{d}/V_{p}

V_{d} = [(1-0.93) / ([0.32/0.1]^{1/1,3}-1)] · 0.00471 = 0.000228 m^{3}

So, the dead volume of the cylinder is 0.000228 m^{3}

## Problem 2. Determination of flow rate and consumed power of compressor equipment

**Conditions:**

A single-stage double-cylinder double-action compressor is equipped with pistons with diameter d = 0.6 m and stroke s = 0.5 m, while dead space is с = 0.036. Compressor shaft rotates at n = 180 rpm. The air inside is compressed at pressure from P1 = 0.1 MPa to P2 = 0.28 MPa and temperature t = 20^{0}. In the following calculations the polytropic coefficient m is assumed to be equal to 1.2, while mechanical efficiency η_{mech} and adiabatic efficiency η_{ad} - to be equal to 0.95 and 0.85 respectively.

**Problem:**

Calculate throughput Q and consumed power N of the compressor.

**Solution:**

First, let’s determine the piston cross-section area F according to the formula below:

F = (π · d²)/4 = (3.14 · 0.6²)/4 = 0.2826 m^{2}

Then, before we can find the compressor throughput, we need to determine the delivery rate. But first let’s calculate volumetric efficiency:

λ_{v} = 1 – с·[(P_{2}/P_{1})^{1/m}-1] = 1 - 0.036·[(0.28/0.1)^{1/1.2}-1] = 0.95

Now that we know the volumetric efficiency, let’s use this value to determine the delivery rate according to the formula:

λ = λ_{v} · (1.01 – 0.02·P_{2}/P_{1}) = 0.95 · (1.01 – 0.02 · 0.28/0.1) = 0.91

So now we can find the compressor throughput Q:

Q = λ · z · F · s · n

Since it is a double-action compressor, coefficient z is 2. Since the compressor has two cylinders, the final throughput value also has to be multiplied by 2. So, we get:

Q = 2 · λ · z · F · s · n = 2 · 0.91 · 2 · 0.2826 · 0.5 · 180 = 92.6 m^{3}/min

Air mass flow G can be determined, where ρ is air density, which at this temperature is equal to 1.189 kg/m^{3}. So let’s calculate this value:

G = Q · ρ = 92.6 · 1.189 = 44 kg/min

Hourly rate is

60·G = 60·44 = 2,640 kg/h.

To determine the consumed power of the compressor we first need to calculate the gas compression energy. For this we can use the following formula:

A_{comp} = k/(k-1) · R · t · [(P_{2}/P_{1})^{(k-1)/k}-1]

In this formula k is adiabatic exponent which is equal to the ratio of heat capacity at constant pressure to heat capacity at constant volume (k = С_{P}P/C_{V}), and for air this value is 1.4. R is gas constant equal to 8,310/M J/(kg*К), where М is molar mass of gas. In case of air М is assumed to be 29 g/mole, then R = 8,310/29 = 286.6 J/(kg*К).

Let’s insert the resulting values into the compression energy formula and find its value:

A_{comp} = k/(k-1) · R · t · [(P_{2}/P_{1})^{(k-1)/k}-1] = 1.4/(1.4-1) · 286.6 · (273+20) · [(0.28/0.1)^{(1.4-1)/1.4}-1] = 100,523 J/kg

After we have the air compression energy value, it is possible to determine the power consumed by the compressor according to the formula:

N = (G · A_{comp}) / (3,600 · 1,000 · η_{mech} · η_{ad}) = (2,640 · 100,523) / (3,600 · 1,000 · 0.85 · 0.95) = 91.3 kW

So, we get the following results: compressor throughput is 92.6 m^{3}/min, consumed power is 91.3 kW

## Problem 3. Determination of the number of compression stages in a compressor and pressure values at each stage

**Conditions:**

Ammonia has to be delivered at 160 m^{3}/h at the pressure of 4.5 MPa. Initial nitrogen pressure is 0.1 MPa and initial temperature – 20°C. For the calculations the maximum compression rate x is assumed to be equal to 4.

**Problem:**

Determine the number of compression stages in a compressor and pressure values at each stage.

**Solution:**

First, let’s calculate the required number of stages n using the compression rate formula:

x^{n} = P_{f}/P_{i}

Let’s express and calculate n value:

n = log(Pf_{f}/P_{i}) / log(x) = log(4.5/0.1) / log(4) = 2.75

By rounding up the result to the next larger integer we get that there should be n = 3 stages in the compressor. Then let’s determine the compression rate of a single stage by assuming that the compression rate is the same at each individual stage.

x = ^{n}√(P_{f}/P_{i}) = ∛(4.5/0.1) = 3.56

Let’s find the final pressure of the first stage P_{n1} (n = 1), which is also the initial pressure of the second stage.

P_{f1} = P_{i} · x^{n} = 0.1 · 3.56^{1} = 0.356 MPa

Let’s find the final pressure of the second stage P_{n2} (n = 2), which is also the initial pressure of the third stage.

P_{f2} = P_{i} · x^{n} = 0.1 · 3.56² = 1.267 MPa

So, there should be three stages in the compressor. At the first stage the pressure is increased from 0.1 MPa to 0.356 MPa, at the second one – from 0.356 MPa to 1.267 MPa, and at the third one – from 1.267 MPa to 4.5 MPa.

## Problem 4. Compressor selection based on the specified conditions

**Conditions:**

It is necessary to provide for nitrogen supply with the throughput of Q_{n}= 7.2 m^{3}/h from initial pressure P_{1} = 0.1 MPa to pressure Р_{2} = 0.5 MPa. Only a single-stage double-action piston compressor is available. The piston has diameter d = 80 mm and stroke s equal to 110 mm, while the dead space is 7% of the volume displaced by the piston. Compressor shaft rotation speed n is 120 rpm. Polytropic coefficient m is assumed to be equal to 1.3 in the calculations.

**Problem:**

Find out whether the compressor available is suitable for performing the specified task. If the compressor is not suitable, calculate the required increase in shaft rotation speed to be able to use the compressor.

**Solution:**

Since the dead space occupies 7% of the volume displaced by the piston, then by default, the dead space с is equal to 0.07.

Besides, let’s first calculate the piston cross-section area F:

F = (π · d²)/4 = (3.14 · 0.08²)/4 = 0.005 m^{2}

For further calculations it is necessary to calculate the volumetric efficiency of compressor λ_{v}:

λ_{v} = 1 – с·[(P_{2}/P_{1})^{1/m}-1] = 1 – 0.04·[(0.5/0.1)^{1/1,3}-1] = 0.9

Now that we know λ_{v, }let’s find the delivery rate λ:

λ = λ_{v} · (1.01 – 0.02·(P_{2}/P_{1})) = 0.9 · (1.01 – 0.02·0.5/0.1) = 0.82

Now we can determine the compressor throughput Q. Since it is a double-action compressor, coefficient z is 2:

Q = λ · z · F · s · n = 0.82 · 2 · 0.005 · 0.11 · 120 = 0.11 m^{3}/min

By expressing Q as an hourly rate, we get Q = 0.11 · 60 = 6.6 m^{3}/h.

Since the required delivery rate is 7.2 m^{3}/h, we can deduce that the available compressor is not suitable for performing the task. In this case let’s calculate the required increase in shaft rotations to meet the criteria for applicability. For this we will find the needed number of rotations from the ratio:

n_{n}/n = Q_{n}/Q

n_{n} = n · Q_{n}/Q = 120 · 7.2/6.6 = 131

In this case the compressor could be used if we increase its shaft rotation speed by 131-120 = 11 rpm.

## Problem 5. Calculation of actual throughput of a piston compressor

**Conditions:**

A three-cylinder double-action piston compressor is available. Piston diameter d is equal to 120 mm, their stroke s is 160 mm. Shaft rotation speed n is 360 rpm. Methane is compressed from pressure P_{1} = 0.3 MPa to pressure P_{2} = 1.1 MPa. It is known that the volumetric efficiency λ_{v} is 0.92.

**Problem:**

Calculate the actual throughput of a piston compressor.

**Solution:**

First, let’s determine the cross-section area of pistons F according to the formula below:

F = (π · d²)/4 = (3.14 · 0.12²)/4 = 0.0113 m^{2}

Based on the initial data let’s determine the delivery rate λ according to the formula:

λ = λ_{v} · (1.01 – 0.02 ·(P_{2}/P_{1})) = 0.92 · (1.01 – 0.02·(1.1/0.3)) = 0.86

Now we can use the formula for the calculation of a piston compressor throughput:

Q = λ · z · F · s · n

Here z is a coefficient depending on the number of suction lines of an individual piston. Since we have a double-action compressor, z is equal to 2.

Besides, since this is a three-cylinder compressor, i.e. three cylinders are operating in parallel, the total throughput of the entire compressor will be three times higher than that of a single piston, so we need to add a factor of 3 into the calculation.

From all the above, we deduce:

Q = 3 · λ · z · F · s · n = 3 · 0.86 · 2 · 0.0113 · 0.16 · 360 = 3.6 m^{3}/min.

So, we get that the throughput of the given piston compressor is 3.6 m^{3}/min or 216 m^{3}/h.

## Problem 6. Calculation of throughput of a double-stage piston compressor

**Conditions:**

A double-stage single-action piston compressor is available. The piston of low-pressure stage has diameter d_{l} = 100 mm and stroke s_{l} equal to 125 mm. The diameter of high-pressure piston d_{h} is equal to 80 mm, while its stroke is s_{h} = 125 mm. Compressor shaft rotation speed n is 360 rpm. It is known that the delivery rate of the compressor λ is 0.85.

**Problem:**

Calculate the throughput of the compressor.

**Solution:**

In case of multi-stage piston compressors the data of low-pressure stage should be used for correspondence calculations, since it is there that the primary gas suction occurs, which determines the throughput of the entire compressor. The data of subsequent stages are not used for throughput calculations, since there is no additional gas suction of compressed gas at those stages. Hence, to solve this task it is sufficient to know diameter d_{l} and piston stroke s_{l} of low-pressure stage.

Let’s calculate the cross-sectional area of the low-pressure stage piston:

F_{l} = (π · d_{l}²)/4 = (3.14 · 0.1²)/4 = 0.00785 m^{2}

The given compressor is not a multi-piston type and has a simple principle of operation (z = 1), hence, the final formula for the calculation of throughput in a specific case will look as follows:

Q = λ · F_{l} · s_{l} · n = 0.85 · 0.00785 · 0.125 · 360 = 0.3 m^{3}/min

So, the throughput of this piston compressor is 0.3 m^{3}/min or, converted to an hourly rate, 18 m^{3}/h.

## Problem 7. Calculation of actual throughput of a double-screw compressor

**Conditions:**

A double-screw compressor is available. The compressor drive shaft rotates at n=750 rpm and has the following channel characteristics: z=4, length L=20 cm. Besides, it is known that the channel cross-section area of the drive shaft is F_{1}=5.2 cm^{2}, and the equivalent value for the idle shaft F_{2} is 5.8 cm^{2}. In the calculations it is assumed that throughput coefficient λ_{th} is equal to 0.9.

**Problem:**

Calculate the actual throughput of a double-screw compressor V_{a}.

**Solution:**

Before the calculation of actual throughput let’s determine the value of theoretical throughput, which does not account for the inevitable backflows of gas through the gaps between compressor rotors and casing.

V_{t} = L·z·n·(F_{1}+F_{2}) = 0.2·4·750·(0.052+0.058) = 66 m^{3}/min

Since the throughput index accounting for gas backflows is known, it is possible o determine the actual throughput of this double-screw compressor:

V_{a} = λ_{th}·V_{t} = 0.9·66 = 59.4 m^{3}/min

So, the throughput of this double-screw compressor is 59.4 m^{3}/min.

## Problem 8. Calculation of power consumed by a screw compressor

**Conditions:**

A screw compressor is available designed for air pressure boosting from P_{1}=0.6 MPa to P_{2}=1.8 MPa. The theoretical throughput of the compressor V_{th} is 3 m^{3}/min. In the calculation adiabatic efficiency η_{ad} is assumed to be equal to 0.76, and adiabatic k-value for air - 1.4.

**Problem:**

Calculate consumed power of the compressor N_{c}.

**Solution:**

For the calculation of theoretical power of adiabatic compression of a screw compressor let’s use the following formula:

N_{ad} = P_{1} · V_{T} · [k/(k-1)] · [(P_{2}/P_{1})^{(k-1)/k} - 1] = 600,000 · 3/60 · 1.4/(1.4-1) · [(1.8/0.6)^{(1.4-1)/1.4} - 1] · 10^{-3} = 38.7 kW

Now that we know N_{ad}, we can calculate consumed power of a dry-type compressor:

N = N_{ad}/η_{ad} = 38.7/0.76 = 51 kW

So, consumed power of this double-screw compressor is equal to 50 kW.

## Problem 9. Calculation of power consumed by a double-screw compressor

**Conditions:**

A double-screw compressor is available with the throughput of Q=10 m^{3}/min. Its medium is air at the temperature of t=20^{0} C. The air is compressed at pressure from P_{1}=0.1 MPa to P_{2}=0.6 MPa. It is known that the backflow value in the compressor β_{bf} is 0.02. The internal adiabatic efficiency of the compressor η_{ad} is equal to 0.8, while mechanical efficiency η_{mech} is 0.95. In the calculations adiabatic k-value for air is assumed to be 1.4, and gas constant for air R - 286 J/(kg*К).

**Problem:**

Calculate consumed power of the compressor N.

**Solution:**

Let’s determine specific energy of the compressor A_{sp}:

A_{sp} = R · T_{a} · [k/(k-1)] · [(P_{2}/P_{1})^{(k-1)/k}-1] = 286 · [20+273] · [1.4/(1.4-1)] · [(0.6/0.1)^{(1.4-1)/1.4}-1] = 196,068 J/kg

Then we can calculate the mass flow rate of air G by assuming that at 20°C air density ρ_{a} is 1.2 kg/m^{3}:

G = Q·ρ_{a} = 10·1.2 = 12 kg/min

When calculating the compressor power it is necessary to take into account the backflows of working media, since additional power is required to compensate for those. Let’s determine the total compressor flow rate G_{tot} considering the backflow leakages:

G_{tot} = G·(1+β_{bf}) = 12·(1+0.02) = 12.24 kg/min

Now it is necessary to determine compressor power considering adiabatic and mechanical efficiency:

N = (G_{tot}·A_{sp}) / (η_{ad}·η_{mech}) = (12.24·196068) / (60·1000·0.8·0.95) = 52.6 kW

So, the power of this compressor is equal to 52.6 kW.

## Problem 10. Calculation of power consumed by a compressor

**Conditions:**

A centrifugal triple-stage single-section compressor is available with identical wheels. The compressor operates at volume flow rate V of 120 m^{3}/min of air at temperature t=20°C (in this case air density ρ will be equal to 1.2 kg/m^{3}). Besides, it is known that the circumferential speed of the wheel is 260 m/s, and theoretical head coefficient of the stage ϕ is 0.85. Total efficiency of the compressor η is 0.9. For the first stage friction loss coefficient β_{f} is 0.007, leakage loss coefficient β_{l} is 0.009, and let’s assume that for subsequent stages the losses will increase by 1%.

**Problem:**

Calculate consumed power of the compressor N.

**Solution:**

The power consumed for gas compression may be calculated as follows:

N_{in} = V · ρ · ∑[u²_{i} · φ_{i} · (1+β_{f}+β_{l})_{i}]

where i is the number of stages. Since the the conditions specify that all the wheels within a section are the same, they have the same circumferential velocities u and theoretical head coefficients ϕ, so the formula may be transformed as follows:

N_{in} = V · ρ · u² · φ · ∑(1+β_{f}+β_{l})_{i}

For the first stage:

1 + β_{f} + β_{l} = 1 + 0.007 + 0.009 = 1.016

Then, assuming that the losses at the next stage increase by 1%, let’s determine 1+β_{f}+β_{l} for the second stage:

1.016·1.01 = 1.026

For the third stage:

1.026·1.01 = 1.036

So, we get that:

N_{in} = 120/60 · 1.2 · 260² · 0.85 · (1.016+1.026+1.036) · 10^{-3} = 424.5 kW

Now it is possible to calculate consumed power of the compressor:

N = N_{in}/η = 424.5/0.9 = 471.7 W

So, the power of this compressor is equal to 471.7 kW.

## Problem 11. Calculation of a centrifugal compressor efficiency

**Conditions:**

A centrifugal double-stage single-section compressor is available with identical wheels. Compressor pumps air at temperature t=20°C (density ρ at these conditions is equal to 1.2 kg/m^{3}) and flow rate V=100 m^{3}/min from initial pressure P_{1}=0.1 MPa to final pressure P_{2}=0.25 MPa. Circumferential speed of wheels u is 245 m/s, theoretical head ϕ is equal to 0.82. The total friction and leakage losses (1+ β_{f} + β_{l}) for the first stage is equal to 1.012, for the second stage - 1.019. Gas compression occurs in isentropic process. In the calculations adiabatic k-value for air is assumed to be 1.4, and gas constant for air R - 286 J/(kg*К). Gas for this problem is assumed to be incompressible (compressibility factor z=1).

**Problem:**

It is necessary to calculate isentropic efficiency of the compressor η_{is}.

**Solution:**

Isentropic efficiency is a relation of gas compression power in isentropic process N_{is} to internal compression power of the compressor N_{in}. From here we can deduce that to find the target value it is first necessary to determine N_{in} and N_{is}.

Gas compression in isentropic mode may be determined according to the formula:

N_{is} = V · ρ · z · R · (273+t) · k/(k-1) · [(P_{2}/P_{1})^{(k-1)/k}-1] =

= 100/60 · 1.2 · 1 · 286 ·(273+20) · 1.4/(1.4-1) · [(0.25/0.1)^{(1.4-1)/1.4}-1] · 10^{-3} = 175.5 kW

Internal power of the compressor may be calculated as follows:

N_{in} = V · ρ · ∑[u_{i}^{2} · φ_{i} · (1+β_{f}+β_{l})_{i}] = 100/60 · 1.2 · 245² · 0.82 · (1.012+1.019) = 200 kW.

Now we can determine the target value:

η_{is} = N_{is}/N_{in} = 175.5/200 = 0.88

So, the isentropic efficiency of this double-stage single-section compressor is equal to 0.88.

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